output
stringlengths 52
181k
| instruction
stringlengths 296
182k
|
|---|---|
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
long long n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<long long> matchingId;
for (long long i = 0; i <= n - m; i++) {
string second = s.substr(i, m);
if (second == t) matchingId.push_back(i);
}
while (q--) {
long long l, r;
cin >> l >> r;
l--, r--;
if (r - l + 1 < m) {
cout << 0 << '\n';
continue;
}
l = lower_bound(matchingId.begin(), matchingId.end(), l) -
matchingId.begin();
r = lower_bound(matchingId.begin(), matchingId.end(), r - m + 2) -
matchingId.begin();
cout << r - l << '\n';
}
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
long long n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<long long> matchingId;
for (long long i = 0; i <= n - m; i++) {
string second = s.substr(i, m);
if (second == t) matchingId.push_back(i);
}
while (q--) {
long long l, r;
cin >> l >> r;
l--, r--;
if (r - l + 1 < m) {
cout << 0 << '\n';
continue;
}
l = lower_bound(matchingId.begin(), matchingId.end(), l) -
matchingId.begin();
r = lower_bound(matchingId.begin(), matchingId.end(), r - m + 2) -
matchingId.begin();
cout << r - l << '\n';
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long int n, m, q;
cin >> n >> m >> q;
string s1, s2;
cin >> s1 >> s2;
long long int arr[n], d = 0;
for (long long int i = 0; i <= n - m; i++) {
string s3 = s1.substr(i, m);
if (s3 == s2) {
arr[d] = i;
d++;
}
}
for (long long int i = 0; i < q; i++) {
long long int a, b;
cin >> a >> b;
long long int out = 0;
for (long long int i = 0; i < d; i++) {
if (arr[i] + 1 >= a && arr[i] + m <= b) {
out++;
}
}
cout << out << '\n';
}
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long int n, m, q;
cin >> n >> m >> q;
string s1, s2;
cin >> s1 >> s2;
long long int arr[n], d = 0;
for (long long int i = 0; i <= n - m; i++) {
string s3 = s1.substr(i, m);
if (s3 == s2) {
arr[d] = i;
d++;
}
}
for (long long int i = 0; i < q; i++) {
long long int a, b;
cin >> a >> b;
long long int out = 0;
for (long long int i = 0; i < d; i++) {
if (arr[i] + 1 >= a && arr[i] + m <= b) {
out++;
}
}
cout << out << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
vector<int> left, right;
int n, m, s, l, r;
string kata, q;
cin >> n >> m >> s;
cin >> kata >> q;
size_t pos = kata.find(q, 0);
while (pos != string::npos) {
left.push_back(pos + 1);
right.push_back(pos + m);
pos = kata.find(q, pos + 1);
}
vector<int>::iterator low, up;
for (int i = 0; i < s; i++) {
cin >> l >> r;
int lef = lower_bound(left.begin(), left.end(), l) - left.begin();
int righ = lower_bound(right.begin(), right.end(), r + 1) - right.begin();
if (righ - lef >= 0)
cout << righ - lef << endl;
else
cout << 0 << endl;
}
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
vector<int> left, right;
int n, m, s, l, r;
string kata, q;
cin >> n >> m >> s;
cin >> kata >> q;
size_t pos = kata.find(q, 0);
while (pos != string::npos) {
left.push_back(pos + 1);
right.push_back(pos + m);
pos = kata.find(q, pos + 1);
}
vector<int>::iterator low, up;
for (int i = 0; i < s; i++) {
cin >> l >> r;
int lef = lower_bound(left.begin(), left.end(), l) - left.begin();
int righ = lower_bound(right.begin(), right.end(), r + 1) - right.begin();
if (righ - lef >= 0)
cout << righ - lef << endl;
else
cout << 0 << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int a[10000];
int inds[10000];
int main() {
int n, m, q;
cin >> n >> m >> q;
string str, target;
cin >> str >> target;
for (int i = 0; i <= (n - m); i++) {
if (str[i] == target[0]) {
string temp = str.substr(i, m);
if (temp == target)
a[i] = 1;
else
a[i] = 0;
}
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
l--;
r--;
int count = 0;
for (int b = l; b <= r; b++) {
if (a[b] == 1 && (b + m - 1) <= r) count++;
}
cout << count;
if (i != q - 1) cout << endl;
}
return 0;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[10000];
int inds[10000];
int main() {
int n, m, q;
cin >> n >> m >> q;
string str, target;
cin >> str >> target;
for (int i = 0; i <= (n - m); i++) {
if (str[i] == target[0]) {
string temp = str.substr(i, m);
if (temp == target)
a[i] = 1;
else
a[i] = 0;
}
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
l--;
r--;
int count = 0;
for (int b = l; b <= r; b++) {
if (a[b] == 1 && (b + m - 1) <= r) count++;
}
cout << count;
if (i != q - 1) cout << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void urmi() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
}
int main() {
long long n, m, q, i, c = 0, d = 0, e, f;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
map<long long, long long> m1;
for (i = 0; i < n; i++) {
if (s.compare(i, t.size(), t) == 0) {
d++;
m1[i] = d;
} else
m1[i] = m1[i - 1];
}
while (q--) {
long long l, r;
cin >> l >> r;
r -= m - 1;
l--;
if (r >= l) {
cout << m1[r - 1] - m1[l - 1] << endl;
} else
cout << "0" << endl;
}
}
| ### Prompt
Please create a solution in cpp to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void urmi() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
}
int main() {
long long n, m, q, i, c = 0, d = 0, e, f;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
map<long long, long long> m1;
for (i = 0; i < n; i++) {
if (s.compare(i, t.size(), t) == 0) {
d++;
m1[i] = d;
} else
m1[i] = m1[i - 1];
}
while (q--) {
long long l, r;
cin >> l >> r;
r -= m - 1;
l--;
if (r >= l) {
cout << m1[r - 1] - m1[l - 1] << endl;
} else
cout << "0" << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int pre[1005], p, n, m, t, l, r;
char a[1005], b[1005];
int main() {
cin >> n >> m >> t;
cin >> (a + 1) >> b;
for (int i = 1; i <= n - m + 1; i++) {
p = 0;
for (int j = 0; j < m; j++) {
if (a[i + j] != b[j]) {
p = 1;
break;
}
}
if (p == 1)
pre[i] = pre[i - 1];
else
pre[i] = pre[i - 1] + 1;
}
while (t--) {
cin >> l >> r;
if (r - l < m - 1)
cout << "0" << endl;
else {
cout << pre[r - m + 1] - pre[l - 1] << endl;
}
}
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int pre[1005], p, n, m, t, l, r;
char a[1005], b[1005];
int main() {
cin >> n >> m >> t;
cin >> (a + 1) >> b;
for (int i = 1; i <= n - m + 1; i++) {
p = 0;
for (int j = 0; j < m; j++) {
if (a[i + j] != b[j]) {
p = 1;
break;
}
}
if (p == 1)
pre[i] = pre[i - 1];
else
pre[i] = pre[i - 1] + 1;
}
while (t--) {
cin >> l >> r;
if (r - l < m - 1)
cout << "0" << endl;
else {
cout << pre[r - m + 1] - pre[l - 1] << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
string str, sub;
vector<long long int> v;
int main() {
long long int n, m, q, x, y;
cin >> n >> m >> q;
cin >> str >> sub;
size_t pos = str.find(sub, 0);
while (pos != string::npos) {
v.push_back(pos);
pos = str.find(sub, pos + 1);
}
while (q--) {
cin >> x >> y;
x--;
y--;
long long int ans = 0;
std::vector<long long int>::iterator low, up;
low = std::lower_bound(v.begin(), v.end(), x);
long long int l = low - v.begin();
for (long long int i = low - v.begin(); i < v.size(); i++) {
if (v[i] > y) break;
if (v[i] + m - 1 <= y) ans++;
}
cout << ans << endl;
}
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string str, sub;
vector<long long int> v;
int main() {
long long int n, m, q, x, y;
cin >> n >> m >> q;
cin >> str >> sub;
size_t pos = str.find(sub, 0);
while (pos != string::npos) {
v.push_back(pos);
pos = str.find(sub, pos + 1);
}
while (q--) {
cin >> x >> y;
x--;
y--;
long long int ans = 0;
std::vector<long long int>::iterator low, up;
low = std::lower_bound(v.begin(), v.end(), x);
long long int l = low - v.begin();
for (long long int i = low - v.begin(); i < v.size(); i++) {
if (v[i] > y) break;
if (v[i] + m - 1 <= y) ans++;
}
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int a[1005];
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
for (int i = 0; i + m <= n; ++i) {
if (s.substr(i, m) == t) {
a[i + 1]++;
}
}
for (int i = 1; i <= n; ++i) a[i] += a[i - 1];
int l, r;
while (q--) {
cin >> l >> r;
r -= m - 1;
cout << max(0, a[max(0, r)] - a[l - 1]) << '\n';
}
return 0;
}
| ### Prompt
Develop a solution in cpp to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[1005];
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
for (int i = 0; i + m <= n; ++i) {
if (s.substr(i, m) == t) {
a[i + 1]++;
}
}
for (int i = 1; i <= n; ++i) a[i] += a[i - 1];
int l, r;
while (q--) {
cin >> l >> r;
r -= m - 1;
cout << max(0, a[max(0, r)] - a[l - 1]) << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
;
const double eps = 1e-8;
const int mod = 10007;
const int maxn = 1e6 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f;
const unsigned long long p = 2333;
int n, m, q, l, r, ans;
unsigned long long hs[1007], ht, pw[1007];
char s[1007], t[1007];
void check(int l, int r) {
ans = 0;
for (int i = l; i <= r - m + 1; i++) {
if (hs[i + m - 1] - hs[i - 1] * pw[m] == ht) {
ans++;
}
}
}
int main() {
pw[0] = 1;
for (int i = 1; i < 1007; i++) {
pw[i] = pw[i - 1] * p;
}
scanf("%d%d%d", &n, &m, &q);
scanf("%s%s", s + 1, t + 1);
hs[0] = 0, ht = 0;
for (int i = 1; i <= n; i++) {
hs[i] = hs[i - 1] * p + (s[i] - 'a' + 1);
}
for (int i = 1; i <= m; i++) {
ht = ht * p + (t[i] - 'a' + 1);
}
while (q--) {
scanf("%d%d", &l, &r);
if (r - l + 1 < m) {
printf("0\n");
continue;
}
check(l, r);
printf("%d\n", ans);
}
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
;
const double eps = 1e-8;
const int mod = 10007;
const int maxn = 1e6 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f;
const unsigned long long p = 2333;
int n, m, q, l, r, ans;
unsigned long long hs[1007], ht, pw[1007];
char s[1007], t[1007];
void check(int l, int r) {
ans = 0;
for (int i = l; i <= r - m + 1; i++) {
if (hs[i + m - 1] - hs[i - 1] * pw[m] == ht) {
ans++;
}
}
}
int main() {
pw[0] = 1;
for (int i = 1; i < 1007; i++) {
pw[i] = pw[i - 1] * p;
}
scanf("%d%d%d", &n, &m, &q);
scanf("%s%s", s + 1, t + 1);
hs[0] = 0, ht = 0;
for (int i = 1; i <= n; i++) {
hs[i] = hs[i - 1] * p + (s[i] - 'a' + 1);
}
for (int i = 1; i <= m; i++) {
ht = ht * p + (t[i] - 'a' + 1);
}
while (q--) {
scanf("%d%d", &l, &r);
if (r - l + 1 < m) {
printf("0\n");
continue;
}
check(l, r);
printf("%d\n", ans);
}
return 0;
}
``` |
#include <bits/stdc++.h>
int a, b, c, i, j, ch, x, y, S[1004];
char s[1004], w[1004];
int main() {
scanf("%d%d%d", &a, &b, &c);
scanf("%s%s", s, w);
for (i = 0; i + b - 1 < a; i++) {
ch = 0;
for (j = 0; j < b; j++) {
if (s[i + j] == w[j]) ch++;
}
if (ch == b) {
S[i + 1]++;
}
S[i + 1] += S[i];
}
while (c--) {
scanf("%d%d", &x, &y);
if (y - b + 1 >= x - 1)
printf("%d\n", S[y - b + 1] - S[x - 1]);
else
printf("0\n");
}
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
int a, b, c, i, j, ch, x, y, S[1004];
char s[1004], w[1004];
int main() {
scanf("%d%d%d", &a, &b, &c);
scanf("%s%s", s, w);
for (i = 0; i + b - 1 < a; i++) {
ch = 0;
for (j = 0; j < b; j++) {
if (s[i + j] == w[j]) ch++;
}
if (ch == b) {
S[i + 1]++;
}
S[i + 1] += S[i];
}
while (c--) {
scanf("%d%d", &x, &y);
if (y - b + 1 >= x - 1)
printf("%d\n", S[y - b + 1] - S[x - 1]);
else
printf("0\n");
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
char a[1000010];
char b[1000010];
int f[1000010];
int n, m, t;
int main() {
cin >> n >> m >> t;
cin >> a + 1 >> b;
for (int i = 1; i <= n - m + 1; i++) {
bool mark = 1;
for (int j = 0; j < m; j++) {
if (a[i + j] != b[j]) {
mark = 0;
break;
}
}
if (mark == 0)
f[i] = f[i - 1];
else
f[i] = f[i - 1] + 1;
}
while (t--) {
int l, r;
cin >> l >> r;
if (r - l + 1 < m)
puts("0");
else
cout << f[r - m + 1] - f[l - 1] << "\n";
}
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char a[1000010];
char b[1000010];
int f[1000010];
int n, m, t;
int main() {
cin >> n >> m >> t;
cin >> a + 1 >> b;
for (int i = 1; i <= n - m + 1; i++) {
bool mark = 1;
for (int j = 0; j < m; j++) {
if (a[i + j] != b[j]) {
mark = 0;
break;
}
}
if (mark == 0)
f[i] = f[i - 1];
else
f[i] = f[i - 1] + 1;
}
while (t--) {
int l, r;
cin >> l >> r;
if (r - l + 1 < m)
puts("0");
else
cout << f[r - m + 1] - f[l - 1] << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m, q;
scanf("%lld", &n), scanf("%lld", &m), scanf("%lld", &q);
string s, t;
cin >> s >> t;
long long cs[n + 10];
cs[0] = 0;
memset(cs, 0, sizeof cs);
for (long long i = 0; i < n; i++) {
bool f = true;
if (n < m) {
cs[i + 1] = 0;
continue;
}
for (long long j = i, k = 0; j < n and k < m; j++, k++) {
if (s[j] != t[k]) {
f = false;
}
}
if (f) {
cs[i + 1] = cs[i] + 1;
} else {
cs[i + 1] = cs[i];
}
}
long long l, r;
while (q--) {
scanf("%lld", &l), scanf("%lld", &r);
r = r - m + 1;
if (l > r) {
printf("0\n");
continue;
}
printf("%lld\n", cs[r] - cs[l - 1]);
}
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m, q;
scanf("%lld", &n), scanf("%lld", &m), scanf("%lld", &q);
string s, t;
cin >> s >> t;
long long cs[n + 10];
cs[0] = 0;
memset(cs, 0, sizeof cs);
for (long long i = 0; i < n; i++) {
bool f = true;
if (n < m) {
cs[i + 1] = 0;
continue;
}
for (long long j = i, k = 0; j < n and k < m; j++, k++) {
if (s[j] != t[k]) {
f = false;
}
}
if (f) {
cs[i + 1] = cs[i] + 1;
} else {
cs[i + 1] = cs[i];
}
}
long long l, r;
while (q--) {
scanf("%lld", &l), scanf("%lld", &r);
r = r - m + 1;
if (l > r) {
printf("0\n");
continue;
}
printf("%lld\n", cs[r] - cs[l - 1]);
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int val[1005], rang[1005];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
string s, t;
int n, m, q;
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < s.size(); i++) {
int cnt = 0;
for (int j = 0; j < t.size(); j++) {
if (i + j >= s.size()) break;
if (s[i + j] != t[j]) break;
cnt++;
}
if (cnt == t.size()) val[i + 1]++;
}
rang[0] = 0;
for (int i = 1; i < 1005; i++) {
rang[i] = val[i] + rang[i - 1];
}
while (q--) {
int x, y;
cin >> x >> y;
y = y - (int)t.size() + 1;
if (x > y)
cout << 0 << '\n';
else
cout << (rang[y] - rang[x - 1]) << '\n';
}
return 0;
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int val[1005], rang[1005];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
string s, t;
int n, m, q;
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < s.size(); i++) {
int cnt = 0;
for (int j = 0; j < t.size(); j++) {
if (i + j >= s.size()) break;
if (s[i + j] != t[j]) break;
cnt++;
}
if (cnt == t.size()) val[i + 1]++;
}
rang[0] = 0;
for (int i = 1; i < 1005; i++) {
rang[i] = val[i] + rang[i - 1];
}
while (q--) {
int x, y;
cin >> x >> y;
y = y - (int)t.size() + 1;
if (x > y)
cout << 0 << '\n';
else
cout << (rang[y] - rang[x - 1]) << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
vector<int> z_function(string s) {
int n = (int)s.length();
vector<int> z(n);
for (int i = 1, l = 0, r = 0; i < n; ++i) {
if (i <= r) z[i] = min(r - i + 1, z[i - l]);
while (i + z[i] < n && s[z[i]] == s[i + z[i]]) ++z[i];
if (i + z[i] - 1 > r) {
l = i;
r = i + z[i] - 1;
}
}
return z;
}
int main() {
int n, m, q;
string s, t;
while (cin >> n >> m >> q >> s >> t) {
const auto z = z_function(t + '$' + s);
vector<int> is_occurance(1 + n, 0), cumsum_ocuurances(1 + n);
for (int i = 0; i <= n - m; ++i) {
is_occurance[i + 1] = m == z[m + 1 + i];
}
partial_sum(is_occurance.begin(), is_occurance.end(),
cumsum_ocuurances.begin());
for (int i = 0; i < q; ++i) {
int l, r;
cin >> l >> r;
cout << (r - l + 1 >= m
? cumsum_ocuurances[r - m + 1] - cumsum_ocuurances[l - 1]
: 0)
<< endl;
}
}
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
vector<int> z_function(string s) {
int n = (int)s.length();
vector<int> z(n);
for (int i = 1, l = 0, r = 0; i < n; ++i) {
if (i <= r) z[i] = min(r - i + 1, z[i - l]);
while (i + z[i] < n && s[z[i]] == s[i + z[i]]) ++z[i];
if (i + z[i] - 1 > r) {
l = i;
r = i + z[i] - 1;
}
}
return z;
}
int main() {
int n, m, q;
string s, t;
while (cin >> n >> m >> q >> s >> t) {
const auto z = z_function(t + '$' + s);
vector<int> is_occurance(1 + n, 0), cumsum_ocuurances(1 + n);
for (int i = 0; i <= n - m; ++i) {
is_occurance[i + 1] = m == z[m + 1 + i];
}
partial_sum(is_occurance.begin(), is_occurance.end(),
cumsum_ocuurances.begin());
for (int i = 0; i < q; ++i) {
int l, r;
cin >> l >> r;
cout << (r - l + 1 >= m
? cumsum_ocuurances[r - m + 1] - cumsum_ocuurances[l - 1]
: 0)
<< endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> amijanina;
for (int i = 0; i < s.size(); i++) {
bool iloveuazrin = true;
for (int j = 0; j < t.size(); j++) {
if (s[i + j] != t[j]) {
iloveuazrin = false;
break;
}
}
if (iloveuazrin == true) {
amijanina.push_back(i + 1);
}
}
while (q--) {
int x, y;
cin >> x >> y;
int ans = 0;
for (int i = 0; i < amijanina.size(); i++) {
if (amijanina[i] >= x and amijanina[i] + m - 1 <= y) {
ans++;
}
}
cout << ans << endl;
}
}
| ### Prompt
Please formulate a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> amijanina;
for (int i = 0; i < s.size(); i++) {
bool iloveuazrin = true;
for (int j = 0; j < t.size(); j++) {
if (s[i + j] != t[j]) {
iloveuazrin = false;
break;
}
}
if (iloveuazrin == true) {
amijanina.push_back(i + 1);
}
}
while (q--) {
int x, y;
cin >> x >> y;
int ans = 0;
for (int i = 0; i < amijanina.size(); i++) {
if (amijanina[i] >= x and amijanina[i] + m - 1 <= y) {
ans++;
}
}
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
string s, t;
int a[1024];
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i <= n - m; i++) {
a[i] = 1;
for (int j = 0; j < m; j++) {
if (s[i + j] != t[j]) {
a[i] = 0;
break;
}
}
}
for (int i = 1; i < n; i++) a[i] += a[i - 1];
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
if (r - l + 1 < m) {
cout << "0\n";
continue;
}
if (l == 1)
cout << a[r - m] << '\n';
else
cout << a[r - m] - a[l - 2] << '\n';
}
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
string s, t;
int a[1024];
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i <= n - m; i++) {
a[i] = 1;
for (int j = 0; j < m; j++) {
if (s[i + j] != t[j]) {
a[i] = 0;
break;
}
}
}
for (int i = 1; i < n; i++) a[i] += a[i - 1];
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
if (r - l + 1 < m) {
cout << "0\n";
continue;
}
if (l == 1)
cout << a[r - m] << '\n';
else
cout << a[r - m] - a[l - 2] << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 5;
int n, m, q, dp[N][N], p[N + N], used[N + N];
string s, t;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> q >> s >> t;
string x = t + "#" + s;
for (int i = 1; i < (int)x.size(); i++) {
int j = p[i - 1];
while (j > 0 && x[i] != x[j]) {
j = p[j - 1];
}
if (x[i] == x[j]) {
j++;
}
p[i] = j;
if (p[i] == m) {
used[i - m] = true;
}
}
s = '#' + s;
t = '%' + t;
for (int i = 1; i <= n; i++) {
for (int j = i; j <= n; j++) {
dp[i][j] = dp[i][j - 1] + (j - m + 1 >= i && used[j]);
}
}
while (q--) {
int l, r;
cin >> l >> r;
cout << dp[l][r] << "\n";
}
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 5;
int n, m, q, dp[N][N], p[N + N], used[N + N];
string s, t;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> q >> s >> t;
string x = t + "#" + s;
for (int i = 1; i < (int)x.size(); i++) {
int j = p[i - 1];
while (j > 0 && x[i] != x[j]) {
j = p[j - 1];
}
if (x[i] == x[j]) {
j++;
}
p[i] = j;
if (p[i] == m) {
used[i - m] = true;
}
}
s = '#' + s;
t = '%' + t;
for (int i = 1; i <= n; i++) {
for (int j = i; j <= n; j++) {
dp[i][j] = dp[i][j - 1] + (j - m + 1 >= i && used[j]);
}
}
while (q--) {
int l, r;
cin >> l >> r;
cout << dp[l][r] << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long MX = 1e3 + 100, INF = 1e17, INF2 = -1e18, D = 1e9 + 7;
long long par[MX];
int32_t main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
for (long long i = n - m; i >= 0; i--) {
par[i] = par[i + 1] + (s.substr(i, m) == t);
}
while (q--) {
long long l, r;
cin >> l >> r;
l--;
cout << par[l] - par[max(l, r - m + 1)] << endl;
}
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MX = 1e3 + 100, INF = 1e17, INF2 = -1e18, D = 1e9 + 7;
long long par[MX];
int32_t main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
for (long long i = n - m; i >= 0; i--) {
par[i] = par[i + 1] + (s.substr(i, m) == t);
}
while (q--) {
long long l, r;
cin >> l >> r;
l--;
cout << par[l] - par[max(l, r - m + 1)] << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
char s[1005], t[1005];
int n, m, q;
int c[1005];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s", s + 1);
scanf("%s", t + 1);
for (int i = 1; i <= n; i++) {
c[i] = c[i - 1];
int d = 1;
for (int j = 1; j <= m; j++) {
if (s[i + j - 1] != t[j]) d = 0;
}
c[i] += d;
}
for (int i = 0; i < q; i++) {
int l, r;
scanf("%d%d", &l, &r);
if (r - m + 1 <= l - 1) {
puts("0");
continue;
}
printf("%d\n", c[r - m + 1] - c[l - 1]);
}
return 0;
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char s[1005], t[1005];
int n, m, q;
int c[1005];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s", s + 1);
scanf("%s", t + 1);
for (int i = 1; i <= n; i++) {
c[i] = c[i - 1];
int d = 1;
for (int j = 1; j <= m; j++) {
if (s[i + j - 1] != t[j]) d = 0;
}
c[i] += d;
}
for (int i = 0; i < q; i++) {
int l, r;
scanf("%d%d", &l, &r);
if (r - m + 1 <= l - 1) {
puts("0");
continue;
}
printf("%d\n", c[r - m + 1] - c[l - 1]);
}
return 0;
}
``` |
#include <bits/stdc++.h>
const double PI = acos(-1.0);
using namespace std;
int setb(int n, int pos) { return n = n | (1 << pos); }
int resb(int n, int pos) { return n = n & ~(1 << pos); }
bool checkb(long long n, long long pos) { return (bool)(n & (1ll << pos)); }
long long bigmod(long long b, long long p, long long m) {
if (p == 0) return 1;
long long ret = bigmod(b, p / 2, m);
ret = (ret * ret) % m;
if (p & 1) ret = (ret * b) % m;
return ret;
}
string s, t;
bool oka(int idx) {
for (int i = idx, j = 0; j < t.size(); i++, j++)
if (s[i] != t[j]) return false;
return true;
}
int ara[1005];
int n, m, q;
int F(int l, int r) {
int ed = r - m + 1;
int ret = 0;
for (int i = l; i <= ed; i++) ret += ara[i];
return ret;
}
int main() {
int x, y;
scanf("%d %d %d", &n, &m, &q);
cin >> s;
cin >> t;
for (int i = 0; i < n - m + 1; i++) ara[i] = oka(i);
for (int i = 1; i <= q; i++) {
scanf("%d %d", &x, &y);
int ret = F(x - 1, y - 1);
printf("%d", ret);
printf("\n");
}
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
const double PI = acos(-1.0);
using namespace std;
int setb(int n, int pos) { return n = n | (1 << pos); }
int resb(int n, int pos) { return n = n & ~(1 << pos); }
bool checkb(long long n, long long pos) { return (bool)(n & (1ll << pos)); }
long long bigmod(long long b, long long p, long long m) {
if (p == 0) return 1;
long long ret = bigmod(b, p / 2, m);
ret = (ret * ret) % m;
if (p & 1) ret = (ret * b) % m;
return ret;
}
string s, t;
bool oka(int idx) {
for (int i = idx, j = 0; j < t.size(); i++, j++)
if (s[i] != t[j]) return false;
return true;
}
int ara[1005];
int n, m, q;
int F(int l, int r) {
int ed = r - m + 1;
int ret = 0;
for (int i = l; i <= ed; i++) ret += ara[i];
return ret;
}
int main() {
int x, y;
scanf("%d %d %d", &n, &m, &q);
cin >> s;
cin >> t;
for (int i = 0; i < n - m + 1; i++) ara[i] = oka(i);
for (int i = 1; i <= q; i++) {
scanf("%d %d", &x, &y);
int ret = F(x - 1, y - 1);
printf("%d", ret);
printf("\n");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long n, m, f[200000], d[200000], i, j, l, r, q, edd;
string s, t;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> q;
cin >> s >> t;
for (i = 0; i < n; i++) {
if (s[i] == t[0] && i + m <= n) {
edd = 1;
for (j = i + 1; j < i + m; j++)
if (s[j] != t[j - i]) {
edd = 0;
break;
}
f[i + 1] = edd;
}
d[i + 1] = d[i] + f[i + 1];
}
while (q--) {
cin >> l >> r;
if (r - m + 1 < l)
cout << 0 << "\n";
else
cout << d[r - m + 1] - d[l - 1] << "\n";
}
return 0;
}
| ### Prompt
Create a solution in cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, m, f[200000], d[200000], i, j, l, r, q, edd;
string s, t;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> q;
cin >> s >> t;
for (i = 0; i < n; i++) {
if (s[i] == t[0] && i + m <= n) {
edd = 1;
for (j = i + 1; j < i + m; j++)
if (s[j] != t[j - i]) {
edd = 0;
break;
}
f[i + 1] = edd;
}
d[i + 1] = d[i] + f[i + 1];
}
while (q--) {
cin >> l >> r;
if (r - m + 1 < l)
cout << 0 << "\n";
else
cout << d[r - m + 1] - d[l - 1] << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 7;
const int M = 1e5 + 7;
const long double pi = 3.1415926535897;
int n, m, q, l, r, cnt[N];
string s, t;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> q;
cin >> s;
cin >> t;
s = ' ' + s;
t = ' ' + t;
for (int i = m; i <= n; ++i) {
cnt[i] = cnt[i - 1];
int flag = 1;
for (int j = i - m + 1; j <= i; ++j) {
if (s[j] != t[j - (i - m)]) {
flag = false;
break;
}
}
if (flag) ++cnt[i];
}
for (int i = 1; i <= q; ++i) {
cin >> l >> r;
if (r - l + 1 < m)
cout << 0 << '\n';
else
cout << cnt[r] - cnt[l + m - 2] << '\n';
}
}
| ### Prompt
Develop a solution in cpp to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 7;
const int M = 1e5 + 7;
const long double pi = 3.1415926535897;
int n, m, q, l, r, cnt[N];
string s, t;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> q;
cin >> s;
cin >> t;
s = ' ' + s;
t = ' ' + t;
for (int i = m; i <= n; ++i) {
cnt[i] = cnt[i - 1];
int flag = 1;
for (int j = i - m + 1; j <= i; ++j) {
if (s[j] != t[j - (i - m)]) {
flag = false;
break;
}
}
if (flag) ++cnt[i];
}
for (int i = 1; i <= q; ++i) {
cin >> l >> r;
if (r - l + 1 < m)
cout << 0 << '\n';
else
cout << cnt[r] - cnt[l + m - 2] << '\n';
}
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize "Ofast"
#pragma GCC optimize "unroll-loops"
#pragma GCC target "sse,sse2,sse3,sse4,abm,avx,mmx,popcnt,tune=native"
char _;
using namespace std;
string a, b;
int n, m, q, c, d;
int PSA[3010];
int main() {
cin.sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> q >> a >> b;
for (int i = 0; i + m <= n; i++)
if (a.substr(i, m) == b) PSA[i + 1]++;
for (int i = 2; i <= n; i++) PSA[i] += PSA[i - 1];
while (q--) {
cin >> c >> d;
d = d - m + 1;
cout << (max(0, PSA[max(0, d)] - PSA[c - 1])) << "\n";
}
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize "Ofast"
#pragma GCC optimize "unroll-loops"
#pragma GCC target "sse,sse2,sse3,sse4,abm,avx,mmx,popcnt,tune=native"
char _;
using namespace std;
string a, b;
int n, m, q, c, d;
int PSA[3010];
int main() {
cin.sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> q >> a >> b;
for (int i = 0; i + m <= n; i++)
if (a.substr(i, m) == b) PSA[i + 1]++;
for (int i = 2; i <= n; i++) PSA[i] += PSA[i - 1];
while (q--) {
cin >> c >> d;
d = d - m + 1;
cout << (max(0, PSA[max(0, d)] - PSA[c - 1])) << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long n, m, q, i, j;
cin >> n >> m >> q;
string a, b;
cin >> a >> b;
vector<long long> v(n + 1);
for (i = 0; i <= n - m; i++) {
string tmp = a.substr(i, m);
if (tmp == b) v[i + 1] = 1;
}
for (i = 1; i <= n; i++) v[i] += v[i - 1];
while (q--) {
long long l, r, tot = 0;
cin >> l >> r;
r = r - m + 1;
long long ans = 0;
if (r >= l) ans = v[r] - v[l - 1];
cout << ans << endl;
}
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long n, m, q, i, j;
cin >> n >> m >> q;
string a, b;
cin >> a >> b;
vector<long long> v(n + 1);
for (i = 0; i <= n - m; i++) {
string tmp = a.substr(i, m);
if (tmp == b) v[i + 1] = 1;
}
for (i = 1; i <= n; i++) v[i] += v[i - 1];
while (q--) {
long long l, r, tot = 0;
cin >> l >> r;
r = r - m + 1;
long long ans = 0;
if (r >= l) ans = v[r] - v[l - 1];
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
;
long long int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
long long int l, r;
long long int cnt;
long long int ans[10005] = {0};
for (long long int i = 0; i < (n - m + 1); i++) {
if (s.substr(i, m) == t) {
ans[i] = 1;
}
}
while (q--) {
cnt = 0;
cin >> l >> r;
for (long long int i = l - 1; i < (r - m + 1); i++) {
if (ans[i] == 1) cnt++;
}
cout << cnt << endl;
}
return 0;
}
| ### Prompt
Create a solution in Cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
;
long long int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
long long int l, r;
long long int cnt;
long long int ans[10005] = {0};
for (long long int i = 0; i < (n - m + 1); i++) {
if (s.substr(i, m) == t) {
ans[i] = 1;
}
}
while (q--) {
cnt = 0;
cin >> l >> r;
for (long long int i = l - 1; i < (r - m + 1); i++) {
if (ans[i] == 1) cnt++;
}
cout << cnt << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
int n, m, q;
cin >> n >> m >> q;
int cnt[n + 1];
string s, s1;
cin >> s >> s1;
s = "1" + s;
int k = s.find(s1);
if (k == string::npos)
fill(cnt, cnt + n, 0);
else {
fill(cnt, cnt + k, 0);
cnt[k] = 1;
}
while (k != n + 1) {
int l = k;
k = s.find(s1, k + 1);
if (k == string::npos) k = n + 1;
fill(cnt + l + 1, cnt + k, cnt[l]);
if (k != n + 1) cnt[k] = cnt[l] + 1;
}
int l, r;
for (int i = 0; i < q; i++) {
cin >> l >> r;
if (r < l + m - 1)
cout << 0 << endl;
else
cout << cnt[r - m + 1] - cnt[l - 1] << endl;
}
return 0;
}
| ### Prompt
Please create a solution in CPP to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
int n, m, q;
cin >> n >> m >> q;
int cnt[n + 1];
string s, s1;
cin >> s >> s1;
s = "1" + s;
int k = s.find(s1);
if (k == string::npos)
fill(cnt, cnt + n, 0);
else {
fill(cnt, cnt + k, 0);
cnt[k] = 1;
}
while (k != n + 1) {
int l = k;
k = s.find(s1, k + 1);
if (k == string::npos) k = n + 1;
fill(cnt + l + 1, cnt + k, cnt[l]);
if (k != n + 1) cnt[k] = cnt[l] + 1;
}
int l, r;
for (int i = 0; i < q; i++) {
cin >> l >> r;
if (r < l + m - 1)
cout << 0 << endl;
else
cout << cnt[r - m + 1] - cnt[l - 1] << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
cin >> n >> m >> q;
string s1, s2;
cin >> s1 >> s2;
vector<pair<int, int> > p;
int ans = 1;
for (int i = 0; i < n; i++) {
if (s1[i] == s2[0]) {
for (int k = 1; k < m && i + k < n; k++) {
if (s1[i + k] == s2[k]) ans++;
}
if (ans == m) p.push_back(make_pair(i + 1, i + m));
ans = 1;
}
}
if (n >= m) {
while (q--) {
int a, b, la = 0;
cin >> a >> b;
for (int i = 0; i < p.size(); i++) {
if (p[i].first >= a && p[i].second <= b) la++;
}
cout << la << endl;
}
} else {
while (q--) {
cout << "0" << endl;
}
}
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
cin >> n >> m >> q;
string s1, s2;
cin >> s1 >> s2;
vector<pair<int, int> > p;
int ans = 1;
for (int i = 0; i < n; i++) {
if (s1[i] == s2[0]) {
for (int k = 1; k < m && i + k < n; k++) {
if (s1[i + k] == s2[k]) ans++;
}
if (ans == m) p.push_back(make_pair(i + 1, i + m));
ans = 1;
}
}
if (n >= m) {
while (q--) {
int a, b, la = 0;
cin >> a >> b;
for (int i = 0; i < p.size(); i++) {
if (p[i].first >= a && p[i].second <= b) la++;
}
cout << la << endl;
}
} else {
while (q--) {
cout << "0" << endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e3;
char s[MAXN];
char t[MAXN];
int Begin[MAXN];
int main() {
int n, m, q;
int ans;
int i, j;
int l, r;
cin >> n >> m >> q;
cin >> s + 1;
cin >> t + 1;
for (i = 1; i + m - 1 <= n; i++) {
for (j = 1; j <= m; j++) {
if (s[i + j - 1] != t[j]) break;
}
if (j > m) Begin[i] = 1;
}
while (q--) {
ans = 0;
cin >> l >> r;
for (i = l; i + m - 1 <= r; i++) ans += Begin[i];
printf("%d\n", ans);
}
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e3;
char s[MAXN];
char t[MAXN];
int Begin[MAXN];
int main() {
int n, m, q;
int ans;
int i, j;
int l, r;
cin >> n >> m >> q;
cin >> s + 1;
cin >> t + 1;
for (i = 1; i + m - 1 <= n; i++) {
for (j = 1; j <= m; j++) {
if (s[i + j - 1] != t[j]) break;
}
if (j > m) Begin[i] = 1;
}
while (q--) {
ans = 0;
cin >> l >> r;
for (i = l; i + m - 1 <= r; i++) ans += Begin[i];
printf("%d\n", ans);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
int i, j;
char s[1005];
char t[1005];
int nx[1005];
int ck[1005];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s", s + 1);
scanf("%s", t + 1);
for (i = 2; i <= m; i++) {
while (j && t[j + 1] != t[i]) j = nx[j];
if (t[j + 1] == t[i]) j++;
nx[i] = j;
}
j = 0;
for (i = 1; i <= n; i++) {
while (j && t[j + 1] != s[i]) j = nx[j];
if (t[j + 1] == s[i]) j++;
if (j == m) {
ck[i] = 1;
j = nx[j];
}
}
for (i = 1; i <= n; i++) ck[i] += ck[i - 1];
while (q--) {
scanf("%d%d", &i, &j);
if (i + m - 2 < j)
printf("%d\n", ck[j] - ck[i + m - 2]);
else
printf("0\n");
}
return 0;
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
int i, j;
char s[1005];
char t[1005];
int nx[1005];
int ck[1005];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s", s + 1);
scanf("%s", t + 1);
for (i = 2; i <= m; i++) {
while (j && t[j + 1] != t[i]) j = nx[j];
if (t[j + 1] == t[i]) j++;
nx[i] = j;
}
j = 0;
for (i = 1; i <= n; i++) {
while (j && t[j + 1] != s[i]) j = nx[j];
if (t[j + 1] == s[i]) j++;
if (j == m) {
ck[i] = 1;
j = nx[j];
}
}
for (i = 1; i <= n; i++) ck[i] += ck[i - 1];
while (q--) {
scanf("%d%d", &i, &j);
if (i + m - 2 < j)
printf("%d\n", ck[j] - ck[i + m - 2]);
else
printf("0\n");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T> >;
const int maxn = (int)1e5 + 5;
const int inf = (int)1e9 + 7;
const ll linf = (ll)1e18 + 7;
const ld pi = acosl(-1.0);
const ld eps = 1e-9;
void S() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
set<int> p;
for (int i = 0; i < n; i++)
if (s.substr(i, m) == t) p.insert(i);
while (q--) {
int l, r;
cin >> l >> r;
int an = 0;
for_each((p).begin(), (p).end(),
[&](int cur) { an += cur >= (l - 1) && cur + m <= r; });
cout << an << '\n';
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
S();
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T> >;
const int maxn = (int)1e5 + 5;
const int inf = (int)1e9 + 7;
const ll linf = (ll)1e18 + 7;
const ld pi = acosl(-1.0);
const ld eps = 1e-9;
void S() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
set<int> p;
for (int i = 0; i < n; i++)
if (s.substr(i, m) == t) p.insert(i);
while (q--) {
int l, r;
cin >> l >> r;
int an = 0;
for_each((p).begin(), (p).end(),
[&](int cur) { an += cur >= (l - 1) && cur + m <= r; });
cout << an << '\n';
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
S();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
char s1[(20ll + (long long int)1e3)];
char s2[(20ll + (long long int)1e3)];
long long int xds[(20ll + (long long int)1e3) << 2];
long long int eds[(20ll + (long long int)1e3) << 2];
long long int n, m;
long long int pushup(long long int cur) {
return xds[cur] = xds[cur << 1] + xds[cur << 1 | 1];
}
long long int init(long long int l, long long int r, long long int cur) {
if (l + 1 == r) return xds[cur] = 0;
long long int m = (l + r) >> 1;
init(l, m, cur << 1);
init(m, r, cur << 1 | 1);
return pushup(cur);
}
long long int modify(long long int l, long long int r, long long int pos,
long long int cur) {
if (l + 1 == r) return xds[cur] += 1;
long long int m = (l + r) >> 1;
if (pos < m)
modify(l, m, pos, cur << 1);
else
modify(m, r, pos, cur << 1 | 1);
return pushup(cur);
}
long long int query(long long int l, long long int r, long long int L,
long long int R, long long int cur) {
if (l + 1 == r) return xds[cur];
if (l >= L && r <= R) return xds[cur];
long long int m = l + r >> 1;
long long int acc = 0;
if (m > L) acc += query(l, m, L, R, cur << 1);
if (m < R) acc += query(m, r, L, R, cur << 1 | 1);
return acc;
}
int main() {
scanf(
"%lld"
"%lld",
&n, &m);
long long int q;
scanf("%lld", &q);
scanf("%s", s1 + 1);
scanf("%s", s2 + 1);
long long int le = n - m + 1;
for (long long int i = 1; i <= le; i++) {
if (!strncmp(&s1[i], s2 + 1, m)) {
modify(1, n + 1, i, 1);
}
}
for (long long int i = 0; i < q; i++) {
long long int l, r;
scanf(
"%lld"
"%lld",
&l, &r);
if (r >= m) {
r += 2;
r -= m;
printf(
"%lld"
"\n",
query(1, n + 1, l, r, 1));
} else {
printf(
"%lld"
"\n",
0ll);
}
}
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char s1[(20ll + (long long int)1e3)];
char s2[(20ll + (long long int)1e3)];
long long int xds[(20ll + (long long int)1e3) << 2];
long long int eds[(20ll + (long long int)1e3) << 2];
long long int n, m;
long long int pushup(long long int cur) {
return xds[cur] = xds[cur << 1] + xds[cur << 1 | 1];
}
long long int init(long long int l, long long int r, long long int cur) {
if (l + 1 == r) return xds[cur] = 0;
long long int m = (l + r) >> 1;
init(l, m, cur << 1);
init(m, r, cur << 1 | 1);
return pushup(cur);
}
long long int modify(long long int l, long long int r, long long int pos,
long long int cur) {
if (l + 1 == r) return xds[cur] += 1;
long long int m = (l + r) >> 1;
if (pos < m)
modify(l, m, pos, cur << 1);
else
modify(m, r, pos, cur << 1 | 1);
return pushup(cur);
}
long long int query(long long int l, long long int r, long long int L,
long long int R, long long int cur) {
if (l + 1 == r) return xds[cur];
if (l >= L && r <= R) return xds[cur];
long long int m = l + r >> 1;
long long int acc = 0;
if (m > L) acc += query(l, m, L, R, cur << 1);
if (m < R) acc += query(m, r, L, R, cur << 1 | 1);
return acc;
}
int main() {
scanf(
"%lld"
"%lld",
&n, &m);
long long int q;
scanf("%lld", &q);
scanf("%s", s1 + 1);
scanf("%s", s2 + 1);
long long int le = n - m + 1;
for (long long int i = 1; i <= le; i++) {
if (!strncmp(&s1[i], s2 + 1, m)) {
modify(1, n + 1, i, 1);
}
}
for (long long int i = 0; i < q; i++) {
long long int l, r;
scanf(
"%lld"
"%lld",
&l, &r);
if (r >= m) {
r += 2;
r -= m;
printf(
"%lld"
"\n",
query(1, n + 1, l, r, 1));
} else {
printf(
"%lld"
"\n",
0ll);
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
int main() {
int n, m, q;
scanf("%d %d %d", &n, &m, &q);
string a, b;
cin >> a >> b;
int ps[1007] = {0};
for (int i = 1; i <= n; i++) {
if (i < m) continue;
bool flag = false;
for (int j = i - m; j < i; j++)
if (a[j] != b[j - i + m]) flag = true;
if (!flag)
ps[i] = ps[i - 1] + 1;
else
ps[i] = ps[i - 1];
}
while (q--) {
int x, y;
scanf("%d %d", &x, &y);
x += (m - 1);
if (x > y)
puts("0");
else
printf("%d\n", ps[y] - ps[x - 1]);
}
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
int main() {
int n, m, q;
scanf("%d %d %d", &n, &m, &q);
string a, b;
cin >> a >> b;
int ps[1007] = {0};
for (int i = 1; i <= n; i++) {
if (i < m) continue;
bool flag = false;
for (int j = i - m; j < i; j++)
if (a[j] != b[j - i + m]) flag = true;
if (!flag)
ps[i] = ps[i - 1] + 1;
else
ps[i] = ps[i - 1];
}
while (q--) {
int x, y;
scanf("%d %d", &x, &y);
x += (m - 1);
if (x > y)
puts("0");
else
printf("%d\n", ps[y] - ps[x - 1]);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q, arr[1001], l, r;
string s, t;
int main() {
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i <= n - m; i++) {
arr[i + 1] = arr[i];
if (s.substr(i, m) == t) {
arr[i + 1]++;
}
}
for (int j = 1; j <= q; j++) {
cin >> l >> r;
if (r - l + 1 < m)
cout << 0 << endl;
else {
cout << arr[r - m + 1] - arr[l - 1] << endl;
}
}
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q, arr[1001], l, r;
string s, t;
int main() {
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i <= n - m; i++) {
arr[i + 1] = arr[i];
if (s.substr(i, m) == t) {
arr[i + 1]++;
}
}
for (int j = 1; j <= q; j++) {
cin >> l >> r;
if (r - l + 1 < m)
cout << 0 << endl;
else {
cout << arr[r - m + 1] - arr[l - 1] << endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MX = 1e3 + 10, MD = 1e9 + 7;
long long n, m, a[MX], ans, tmp, ps[MX], ps2[MX], q;
string s, t, k;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> n >> m >> q;
cin >> s >> t;
int flag = 0;
for (int i = 0; i < n; i++) {
ps[i + 1] += ps[i];
ps2[i + 1] += ps2[i];
k = s.substr(i, m);
if (k == t) {
ps2[i + 1]++;
ps[i + m]++;
}
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
if (ps[r] - ps2[l - 1] < 0) {
cout << 0 << "\n";
} else
cout << ps[r] - ps2[l - 1] << "\n";
}
}
| ### Prompt
Create a solution in CPP for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MX = 1e3 + 10, MD = 1e9 + 7;
long long n, m, a[MX], ans, tmp, ps[MX], ps2[MX], q;
string s, t, k;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> n >> m >> q;
cin >> s >> t;
int flag = 0;
for (int i = 0; i < n; i++) {
ps[i + 1] += ps[i];
ps2[i + 1] += ps2[i];
k = s.substr(i, m);
if (k == t) {
ps2[i + 1]++;
ps[i + m]++;
}
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
if (ps[r] - ps2[l - 1] < 0) {
cout << 0 << "\n";
} else
cout << ps[r] - ps2[l - 1] << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int kmp(string T, string &P) {
if (P.empty()) return 0;
if (T.size() < P.size()) return 0;
vector<int> prefix(P.size(), 0);
for (int i = 1, k = 0; i < P.size(); ++i) {
while (k && P[k] != P[i]) k = prefix[k - 1];
if (P[k] == P[i]) ++k;
prefix[i] = k;
}
int count = 0;
for (int i = 0, k = 0; i < T.size(); ++i) {
while (k && P[k] != T[i]) k = prefix[k - 1];
if (P[k] == T[i]) ++k;
if (k == P.size()) count++;
}
return count;
}
int main() {
int n, m, q;
string s, t;
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
string sub = s.substr(l - 1, r - l + 1);
int ret = kmp(sub, t);
cout << ret << endl;
}
return 0;
}
| ### Prompt
Please create a solution in CPP to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int kmp(string T, string &P) {
if (P.empty()) return 0;
if (T.size() < P.size()) return 0;
vector<int> prefix(P.size(), 0);
for (int i = 1, k = 0; i < P.size(); ++i) {
while (k && P[k] != P[i]) k = prefix[k - 1];
if (P[k] == P[i]) ++k;
prefix[i] = k;
}
int count = 0;
for (int i = 0, k = 0; i < T.size(); ++i) {
while (k && P[k] != T[i]) k = prefix[k - 1];
if (P[k] == T[i]) ++k;
if (k == P.size()) count++;
}
return count;
}
int main() {
int n, m, q;
string s, t;
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
string sub = s.substr(l - 1, r - l + 1);
int ret = kmp(sub, t);
cout << ret << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int N, M, Q;
char S[1001], T[1001];
int A[1001], P[1001];
void f() {
for (int i = 0; i + M <= N; i++) {
int ok = 1;
for (int j = 0; j < M && ok; j++) ok = (S[i + j] == T[j]);
A[i] = ok;
}
}
int main(int argc, char **argv) {
ios::sync_with_stdio(false);
cin >> N >> M >> Q;
cin >> S >> T;
f();
P[0] = A[0];
for (int i = 1; i < N; i++) P[i] = P[i - 1] + A[i];
while (Q--) {
int l, r;
cin >> l >> r;
l--;
r--;
r -= M - 1;
int ans = 0;
if (l <= r) {
ans = P[r];
if (l) ans -= P[l - 1];
}
cout << ans << endl;
}
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int N, M, Q;
char S[1001], T[1001];
int A[1001], P[1001];
void f() {
for (int i = 0; i + M <= N; i++) {
int ok = 1;
for (int j = 0; j < M && ok; j++) ok = (S[i + j] == T[j]);
A[i] = ok;
}
}
int main(int argc, char **argv) {
ios::sync_with_stdio(false);
cin >> N >> M >> Q;
cin >> S >> T;
f();
P[0] = A[0];
for (int i = 1; i < N; i++) P[i] = P[i - 1] + A[i];
while (Q--) {
int l, r;
cin >> l >> r;
l--;
r--;
r -= M - 1;
int ans = 0;
if (l <= r) {
ans = P[r];
if (l) ans -= P[l - 1];
}
cout << ans << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
char a[1005];
char b[1005];
int t[1005];
int sum[1005];
int main() {
memset(t, 0, sizeof(t));
int p, n, m;
cin >> n >> m >> p;
cin >> a >> b;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (a[i + j] != b[j])
break;
else if (j == m - 1)
t[i] = 1;
}
}
sum[0] = t[0];
for (int i = 1; i < n; i++) sum[i] = sum[i - 1] + t[i];
while (p--) {
int l, r;
scanf("%d%d", &l, &r);
l--, r--;
if (n < m || r - l + 1 < m)
cout << 0 << endl;
else
cout << sum[r - m + 1] - sum[l - 1] << endl;
}
return 0;
}
| ### Prompt
Create a solution in CPP for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char a[1005];
char b[1005];
int t[1005];
int sum[1005];
int main() {
memset(t, 0, sizeof(t));
int p, n, m;
cin >> n >> m >> p;
cin >> a >> b;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (a[i + j] != b[j])
break;
else if (j == m - 1)
t[i] = 1;
}
}
sum[0] = t[0];
for (int i = 1; i < n; i++) sum[i] = sum[i - 1] + t[i];
while (p--) {
int l, r;
scanf("%d%d", &l, &r);
l--, r--;
if (n < m || r - l + 1 < m)
cout << 0 << endl;
else
cout << sum[r - m + 1] - sum[l - 1] << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
const int maxn = 1005;
char s[maxn], t[maxn];
int sum[maxn];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s", s + 1);
scanf("%s", t + 1);
for (int i = 1; i <= n - m + 1; i++) {
bool f = 1;
for (int j = 1; j <= m; j++)
if (s[i + j - 1] != t[j]) f = 0;
if (f == 1) sum[i] = 1;
}
for (int i = 1; i <= n; i++) sum[i] += sum[i - 1];
for (int i = 1; i <= q; i++) {
int l, r;
scanf("%d%d", &l, &r);
printf("%d\n",
max(0, (r >= m ? sum[r - m + 1] : 0) - (l >= 1 ? sum[l - 1] : 0)));
}
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
const int maxn = 1005;
char s[maxn], t[maxn];
int sum[maxn];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s", s + 1);
scanf("%s", t + 1);
for (int i = 1; i <= n - m + 1; i++) {
bool f = 1;
for (int j = 1; j <= m; j++)
if (s[i + j - 1] != t[j]) f = 0;
if (f == 1) sum[i] = 1;
}
for (int i = 1; i <= n; i++) sum[i] += sum[i - 1];
for (int i = 1; i <= q; i++) {
int l, r;
scanf("%d%d", &l, &r);
printf("%d\n",
max(0, (r >= m ? sum[r - m + 1] : 0) - (l >= 1 ? sum[l - 1] : 0)));
}
return 0;
}
``` |
#include <bits/stdc++.h>
int n, m, q, nxt[1005], a[1005], b[1005], cnt;
char s[1005], t[1005];
inline void getnxt() {
nxt[0] = -1;
for (register int i = 1, tmp; i < m; i++) {
tmp = nxt[i - 1];
while (tmp >= 0 && t[tmp + 1] != t[i]) tmp = nxt[tmp];
if (t[tmp + 1] == t[i])
nxt[i] = tmp + 1;
else
nxt[i] = -1;
}
}
int main() {
scanf("%d%d%d", &n, &m, &q);
if (n < m) {
while (q--) puts("0");
return 0;
}
scanf("%s%s", s, t);
getnxt();
register int i = 0, j = 0, l, r, ans = 0;
while (j < n) {
if (t[i] == s[j]) {
i++;
j++;
if (i == m) a[++cnt] = j - m + 1, b[cnt] = j, i = nxt[i - 1] + 1;
} else if (!i)
j++;
else
i = nxt[i - 1] + 1;
}
while (q--) {
scanf("%d%d", &l, &r);
ans = 0;
for (i = 1; i <= cnt; i++)
if (l <= a[i] && r >= b[i]) ans++;
printf("%d\n", ans);
}
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
int n, m, q, nxt[1005], a[1005], b[1005], cnt;
char s[1005], t[1005];
inline void getnxt() {
nxt[0] = -1;
for (register int i = 1, tmp; i < m; i++) {
tmp = nxt[i - 1];
while (tmp >= 0 && t[tmp + 1] != t[i]) tmp = nxt[tmp];
if (t[tmp + 1] == t[i])
nxt[i] = tmp + 1;
else
nxt[i] = -1;
}
}
int main() {
scanf("%d%d%d", &n, &m, &q);
if (n < m) {
while (q--) puts("0");
return 0;
}
scanf("%s%s", s, t);
getnxt();
register int i = 0, j = 0, l, r, ans = 0;
while (j < n) {
if (t[i] == s[j]) {
i++;
j++;
if (i == m) a[++cnt] = j - m + 1, b[cnt] = j, i = nxt[i - 1] + 1;
} else if (!i)
j++;
else
i = nxt[i - 1] + 1;
}
while (q--) {
scanf("%d%d", &l, &r);
ans = 0;
for (i = 1; i <= cnt; i++)
if (l <= a[i] && r >= b[i]) ans++;
printf("%d\n", ans);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 7;
const int mod = 1e9 + 7;
char str[N];
char s[N];
int a[1000 + 7];
int main() {
int n, m, q;
scanf("%d%d%d", &n, &m, &q);
scanf("%s", str + 1);
scanf("%s", s + 1);
int ans = 0;
for (int i = 1; i <= n; i++) {
int cnt = 1;
while (str[i + cnt - 1] == s[cnt] && str[i + cnt - 1] != '\0') {
cnt++;
}
if (cnt - 1 == m) {
a[ans] = i;
ans++;
}
}
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
int cc = 0;
for (int i = 0; i < ans; i++) {
if (a[i] >= l && a[i] + m - 1 <= r) cc++;
}
printf("%d\n", cc);
}
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 7;
const int mod = 1e9 + 7;
char str[N];
char s[N];
int a[1000 + 7];
int main() {
int n, m, q;
scanf("%d%d%d", &n, &m, &q);
scanf("%s", str + 1);
scanf("%s", s + 1);
int ans = 0;
for (int i = 1; i <= n; i++) {
int cnt = 1;
while (str[i + cnt - 1] == s[cnt] && str[i + cnt - 1] != '\0') {
cnt++;
}
if (cnt - 1 == m) {
a[ans] = i;
ans++;
}
}
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
int cc = 0;
for (int i = 0; i < ans; i++) {
if (a[i] >= l && a[i] + m - 1 <= r) cc++;
}
printf("%d\n", cc);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 2;
const long long N1 = 1e2 + 2;
const long long mod = 1e9 + 7;
const int MASK = 1 << 17 + 1;
const int who = 6 * N * log2(N);
int a[N + 1];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, m, q;
cin >> n >> m >> q;
string s, ser;
cin >> s >> ser;
for (int i = 0; i < n - m + 1; i++) {
if (s.substr(i, m) == ser) {
a[i + 1] = 1;
}
}
a[0] = 0;
for (int i = 1; i <= n - m + 1; i++) {
a[i] += a[i - 1];
}
while (q--) {
int l, r;
cin >> l >> r;
l--;
r -= m - 1;
if (r < l) {
r = l;
}
cout << a[r] - a[l] << "\n";
}
}
| ### Prompt
Please create a solution in cpp to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 2;
const long long N1 = 1e2 + 2;
const long long mod = 1e9 + 7;
const int MASK = 1 << 17 + 1;
const int who = 6 * N * log2(N);
int a[N + 1];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, m, q;
cin >> n >> m >> q;
string s, ser;
cin >> s >> ser;
for (int i = 0; i < n - m + 1; i++) {
if (s.substr(i, m) == ser) {
a[i + 1] = 1;
}
}
a[0] = 0;
for (int i = 1; i <= n - m + 1; i++) {
a[i] += a[i - 1];
}
while (q--) {
int l, r;
cin >> l >> r;
l--;
r -= m - 1;
if (r < l) {
r = l;
}
cout << a[r] - a[l] << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
char a[1005];
char b[1005];
int t[1005];
int sum[1005];
int main() {
memset(t, 0, sizeof(t));
int p, n, m;
cin >> n >> m >> p;
cin >> a >> b;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (a[i + j] != b[j])
break;
else if (j == m - 1)
t[i] = 1;
}
}
sum[0] = t[0];
for (int i = 1; i < n; i++) sum[i] = sum[i - 1] + t[i];
while (p--) {
int l, r;
scanf("%d%d", &l, &r);
l--, r--;
if (n < m || r - l + 1 < m)
cout << 0 << endl;
else
cout << sum[r - m + 1] - sum[l - 1] << endl;
}
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
char a[1005];
char b[1005];
int t[1005];
int sum[1005];
int main() {
memset(t, 0, sizeof(t));
int p, n, m;
cin >> n >> m >> p;
cin >> a >> b;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (a[i + j] != b[j])
break;
else if (j == m - 1)
t[i] = 1;
}
}
sum[0] = t[0];
for (int i = 1; i < n; i++) sum[i] = sum[i - 1] + t[i];
while (p--) {
int l, r;
scanf("%d%d", &l, &r);
l--, r--;
if (n < m || r - l + 1 < m)
cout << 0 << endl;
else
cout << sum[r - m + 1] - sum[l - 1] << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int base = 317;
const int mod1 = 1e9 + 2277;
const int mod2 = 1e9 + 5277;
const int N = 1e3 + 5;
int n, m, q, dp[N][N];
string s, t;
pair<long long, long long> Hash_s[N], Hash_t[N], pw[N];
pair<long long, long long> get_hash_s(int l, int r) {
long long fi =
(Hash_s[r].first - Hash_s[l - 1].first * pw[r - l + 1].first) % mod1;
long long se =
(Hash_s[r].second - Hash_s[l - 1].second * pw[r - l + 1].second) % mod2;
if (fi < 0) fi += mod1;
if (se < 0) se += mod2;
return make_pair(fi, se);
}
pair<long long, long long> get_hash_t(int l, int r) {
long long fi =
(Hash_t[r].first - Hash_t[l - 1].first * pw[r - l + 1].first) % mod1;
long long se =
(Hash_t[r].second - Hash_t[l - 1].second * pw[r - l + 1].second) % mod2;
if (fi < 0) fi += mod1;
if (se < 0) se += mod2;
return make_pair(fi, se);
}
bool check(pair<long long, long long> p1, pair<long long, long long> p2) {
return p1.first == p2.first && p1.second == p2.second;
}
void input() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> q;
cin >> s >> t;
s = "#" + s;
t = "#" + t;
}
void solve() {
pw[0] = make_pair(1, 1);
Hash_s[0] = Hash_t[0] = make_pair(0, 0);
for (int i = 1; i < N; i++)
pw[i] = make_pair((pw[i - 1].first * base) % mod1,
(pw[i - 1].second * base) % mod2);
for (int i = 1; i <= n; i++)
Hash_s[i] = make_pair((Hash_s[i - 1].first * base + s[i]) % mod1,
(Hash_s[i - 1].second * base + s[i]) % mod2);
for (int i = 1; i <= m; i++)
Hash_t[i] = make_pair((Hash_t[i - 1].first * base + t[i]) % mod1,
(Hash_t[i - 1].second * base + t[i]) % mod2);
memset(dp, 0, sizeof(dp));
for (int i = 1; i <= n; i++) {
int r = i + m - 1;
if (r <= n) dp[i][r] = check(get_hash_s(i, r), get_hash_t(1, m));
for (int j = i + m; j <= n; j++) {
dp[i][j] = dp[i][j - 1];
if (check(get_hash_s(j - m + 1, j), get_hash_t(1, m))) dp[i][j]++;
}
}
while (q--) {
int l, r;
cin >> l >> r;
cout << dp[l][r] << "\n";
}
}
int main() {
input();
solve();
return 0;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int base = 317;
const int mod1 = 1e9 + 2277;
const int mod2 = 1e9 + 5277;
const int N = 1e3 + 5;
int n, m, q, dp[N][N];
string s, t;
pair<long long, long long> Hash_s[N], Hash_t[N], pw[N];
pair<long long, long long> get_hash_s(int l, int r) {
long long fi =
(Hash_s[r].first - Hash_s[l - 1].first * pw[r - l + 1].first) % mod1;
long long se =
(Hash_s[r].second - Hash_s[l - 1].second * pw[r - l + 1].second) % mod2;
if (fi < 0) fi += mod1;
if (se < 0) se += mod2;
return make_pair(fi, se);
}
pair<long long, long long> get_hash_t(int l, int r) {
long long fi =
(Hash_t[r].first - Hash_t[l - 1].first * pw[r - l + 1].first) % mod1;
long long se =
(Hash_t[r].second - Hash_t[l - 1].second * pw[r - l + 1].second) % mod2;
if (fi < 0) fi += mod1;
if (se < 0) se += mod2;
return make_pair(fi, se);
}
bool check(pair<long long, long long> p1, pair<long long, long long> p2) {
return p1.first == p2.first && p1.second == p2.second;
}
void input() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> q;
cin >> s >> t;
s = "#" + s;
t = "#" + t;
}
void solve() {
pw[0] = make_pair(1, 1);
Hash_s[0] = Hash_t[0] = make_pair(0, 0);
for (int i = 1; i < N; i++)
pw[i] = make_pair((pw[i - 1].first * base) % mod1,
(pw[i - 1].second * base) % mod2);
for (int i = 1; i <= n; i++)
Hash_s[i] = make_pair((Hash_s[i - 1].first * base + s[i]) % mod1,
(Hash_s[i - 1].second * base + s[i]) % mod2);
for (int i = 1; i <= m; i++)
Hash_t[i] = make_pair((Hash_t[i - 1].first * base + t[i]) % mod1,
(Hash_t[i - 1].second * base + t[i]) % mod2);
memset(dp, 0, sizeof(dp));
for (int i = 1; i <= n; i++) {
int r = i + m - 1;
if (r <= n) dp[i][r] = check(get_hash_s(i, r), get_hash_t(1, m));
for (int j = i + m; j <= n; j++) {
dp[i][j] = dp[i][j - 1];
if (check(get_hash_s(j - m + 1, j), get_hash_t(1, m))) dp[i][j]++;
}
}
while (q--) {
int l, r;
cin >> l >> r;
cout << dp[l][r] << "\n";
}
}
int main() {
input();
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int mxN = 1e3 + 3;
bool cnt[mxN];
int sum[mxN];
int main() {
int n, m, a, i, j;
string s, ss;
cin >> n >> m >> a >> s >> ss;
for (i = 0; i < n - m + 1; i++) {
bool f = true;
for (j = 0; j < m; j++)
if (s[i + j] != ss[j]) {
f = false;
break;
}
cnt[i] = f;
}
for (i = 0; i < s.size(); i++) sum[i + 1] = sum[i] + (cnt[i] ? 1 : 0);
while (a--) {
int b, c;
cin >> b >> c;
if (c - b + 1 < m) {
cout << 0 << endl;
continue;
}
cout << sum[c - m + 1] - sum[b - 1] << endl;
}
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int mxN = 1e3 + 3;
bool cnt[mxN];
int sum[mxN];
int main() {
int n, m, a, i, j;
string s, ss;
cin >> n >> m >> a >> s >> ss;
for (i = 0; i < n - m + 1; i++) {
bool f = true;
for (j = 0; j < m; j++)
if (s[i + j] != ss[j]) {
f = false;
break;
}
cnt[i] = f;
}
for (i = 0; i < s.size(); i++) sum[i + 1] = sum[i] + (cnt[i] ? 1 : 0);
while (a--) {
int b, c;
cin >> b >> c;
if (c - b + 1 < m) {
cout << 0 << endl;
continue;
}
cout << sum[c - m + 1] - sum[b - 1] << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, q, m, i, j;
int dp[1010][1010];
int dp2[1010];
for (i = 0; i < 1010; i++) {
for (j = 0; j < 1010; j++) dp[i][j] = 0;
}
cin >> n >> m >> q;
string str1, str2;
cin >> str1;
cin >> str2;
int flag, k;
for (i = 0; i < n; i++) {
flag = 0;
k = i;
for (j = 0; j < m; j++) {
if (k > n - 1 || str2[j] != str1[k]) {
flag = 1;
break;
}
k++;
}
if (flag == 0)
dp2[i] = 1;
else
dp2[i] = 0;
}
int val1, val2;
for (j = 0; j < n; j++) {
for (i = j; i >= 0; i--) {
val1 = 0;
val2 = 0;
if (i + m - 1 <= j) val1 = dp2[i];
if (i + 1 <= j) val2 = dp[i + 1][j];
dp[i][j] = val1 + val2;
}
}
int num1, num2;
while (q--) {
cin >> num1 >> num2;
cout << dp[num1 - 1][num2 - 1] << endl;
}
return 0;
}
| ### Prompt
Create a solution in Cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, q, m, i, j;
int dp[1010][1010];
int dp2[1010];
for (i = 0; i < 1010; i++) {
for (j = 0; j < 1010; j++) dp[i][j] = 0;
}
cin >> n >> m >> q;
string str1, str2;
cin >> str1;
cin >> str2;
int flag, k;
for (i = 0; i < n; i++) {
flag = 0;
k = i;
for (j = 0; j < m; j++) {
if (k > n - 1 || str2[j] != str1[k]) {
flag = 1;
break;
}
k++;
}
if (flag == 0)
dp2[i] = 1;
else
dp2[i] = 0;
}
int val1, val2;
for (j = 0; j < n; j++) {
for (i = j; i >= 0; i--) {
val1 = 0;
val2 = 0;
if (i + m - 1 <= j) val1 = dp2[i];
if (i + 1 <= j) val2 = dp[i + 1][j];
dp[i][j] = val1 + val2;
}
}
int num1, num2;
while (q--) {
cin >> num1 >> num2;
cout << dp[num1 - 1][num2 - 1] << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long n, m, l, r, ss[10000001], q, a[10000001];
string second, t;
int main() {
cin >> n >> m >> q;
cin >> second >> t;
for (int i = 0; i < second.size(); i++) {
string st = "";
for (int j = i; j < second.size(); j++) {
st += second[j];
if (st == t) {
a[j] = 1;
}
}
}
for (int i = 0; i < second.size() + 100000; i++) ss[i] = ss[i - 1] + a[i];
for (int yy = 0; yy < q; yy++) {
cin >> l >> r;
l--;
r--;
cout << max(ss[r] - ss[l + t.size() - 2], 0LL) << endl;
}
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, m, l, r, ss[10000001], q, a[10000001];
string second, t;
int main() {
cin >> n >> m >> q;
cin >> second >> t;
for (int i = 0; i < second.size(); i++) {
string st = "";
for (int j = i; j < second.size(); j++) {
st += second[j];
if (st == t) {
a[j] = 1;
}
}
}
for (int i = 0; i < second.size() + 100000; i++) ss[i] = ss[i - 1] + a[i];
for (int yy = 0; yy < q; yy++) {
cin >> l >> r;
l--;
r--;
cout << max(ss[r] - ss[l + t.size() - 2], 0LL) << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q, l, r;
cin >> n >> m >> q;
string a, b;
cin >> a >> b;
bool p[n + 1];
for (int i = 0; i < n + 1; i++) {
bool f = true;
for (int j = 0; j < m; j++)
if (a[i + j] != b[j]) {
f = false;
break;
}
p[i + 1] = f;
}
while (q--) {
cin >> l >> r;
int c = 0;
for (int i = l; i + m - 1 <= r; i++)
if (p[i] && i + m - 1 <= r) c++;
cout << c << endl;
}
}
| ### Prompt
Please create a solution in CPP to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q, l, r;
cin >> n >> m >> q;
string a, b;
cin >> a >> b;
bool p[n + 1];
for (int i = 0; i < n + 1; i++) {
bool f = true;
for (int j = 0; j < m; j++)
if (a[i + j] != b[j]) {
f = false;
break;
}
p[i + 1] = f;
}
while (q--) {
cin >> l >> r;
int c = 0;
for (int i = l; i + m - 1 <= r; i++)
if (p[i] && i + m - 1 <= r) c++;
cout << c << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
char s[1005], t[1005];
int pre[1005], q, n, m, a, b, prel[1005];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s%s", &s[1], &t[1]);
for (int i = 1; i <= n; ++i) {
int j = 1, temp = i;
while (temp <= n && j <= m && t[j] == s[temp]) {
j++, temp++;
}
if (j == m + 1) {
pre[i + m - 1] = pre[i + m - 2] + 1;
prel[i] = prel[i - 1] + 1;
} else {
pre[i + m - 1] = pre[i + m - 2];
prel[i] = prel[i - 1];
}
}
for (int i = 1; i <= q; ++i) {
scanf("%d%d", &a, &b);
if (b - a + 1 >= m)
printf("%d\n", pre[b] - prel[a - 1]);
else
printf("0\n");
}
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char s[1005], t[1005];
int pre[1005], q, n, m, a, b, prel[1005];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s%s", &s[1], &t[1]);
for (int i = 1; i <= n; ++i) {
int j = 1, temp = i;
while (temp <= n && j <= m && t[j] == s[temp]) {
j++, temp++;
}
if (j == m + 1) {
pre[i + m - 1] = pre[i + m - 2] + 1;
prel[i] = prel[i - 1] + 1;
} else {
pre[i + m - 1] = pre[i + m - 2];
prel[i] = prel[i - 1];
}
}
for (int i = 1; i <= q; ++i) {
scanf("%d%d", &a, &b);
if (b - a + 1 >= m)
printf("%d\n", pre[b] - prel[a - 1]);
else
printf("0\n");
}
}
``` |
#include <bits/stdc++.h>
int occ[10000];
using namespace std;
int main() {
int n, m, q;
int sol;
cin >> n >> m >> q;
string s1, s2;
cin >> s1 >> s2;
for (int i = 0; i < n; i++) {
bool test = true;
for (int j = 0; j < m; j++) {
if (s1[i + j] != s2[j]) {
test = false;
break;
}
}
occ[i] = test;
}
while (q--) {
int sol = 0;
int a, b;
cin >> a >> b;
a--;
b--;
for (int i = a; i <= b; i++) {
sol += occ[i] && (b - i + 1) >= s2.length();
}
cout << sol << endl;
}
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
int occ[10000];
using namespace std;
int main() {
int n, m, q;
int sol;
cin >> n >> m >> q;
string s1, s2;
cin >> s1 >> s2;
for (int i = 0; i < n; i++) {
bool test = true;
for (int j = 0; j < m; j++) {
if (s1[i + j] != s2[j]) {
test = false;
break;
}
}
occ[i] = test;
}
while (q--) {
int sol = 0;
int a, b;
cin >> a >> b;
a--;
b--;
for (int i = a; i <= b; i++) {
sol += occ[i] && (b - i + 1) >= s2.length();
}
cout << sol << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q, l, r, index = 0, c = 0;
string s, t;
cin >> n >> m >> q >> s >> t;
bool a[n];
for (int i = 0; i < n; i++) {
index = 0;
for (int j = i; j < i + m; j++) {
if (s[j] == t[index])
index++;
else
break;
}
if (index == m)
a[i] = 1;
else
a[i] = 0;
}
for (int i = 0; i < q; i++) {
cin >> l >> r;
c = 0;
for (int j = l - 1; j <= r - 1; j++) {
if (a[j] && j + m - 1 <= (r - 1)) c++;
}
cout << c << endl;
}
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q, l, r, index = 0, c = 0;
string s, t;
cin >> n >> m >> q >> s >> t;
bool a[n];
for (int i = 0; i < n; i++) {
index = 0;
for (int j = i; j < i + m; j++) {
if (s[j] == t[index])
index++;
else
break;
}
if (index == m)
a[i] = 1;
else
a[i] = 0;
}
for (int i = 0; i < q; i++) {
cin >> l >> r;
c = 0;
for (int j = l - 1; j <= r - 1; j++) {
if (a[j] && j + m - 1 <= (r - 1)) c++;
}
cout << c << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long const MOD = 1e9 + 7;
long long const N = 1e3 + 10;
long long ara[N + 1];
long long bra[N + 1];
int main() {
(ios_base::sync_with_stdio(false), cin.tie(NULL));
long long n, m, q;
cin >> n >> m >> q;
string str, s;
cin >> str >> s;
for (long long i = 0; i < n - m + 1; i++) {
if (str.substr(i, m) == s) ara[i + 1]++;
}
for (long long i = 1; i <= n + 10; i++) {
ara[i] += ara[i - 1];
}
while (q--) {
long long a, b;
cin >> a >> b;
b = b - m + 1;
cout << max(0ll, ara[max(b, 0ll)] - ara[a - 1]) << endl;
}
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long const MOD = 1e9 + 7;
long long const N = 1e3 + 10;
long long ara[N + 1];
long long bra[N + 1];
int main() {
(ios_base::sync_with_stdio(false), cin.tie(NULL));
long long n, m, q;
cin >> n >> m >> q;
string str, s;
cin >> str >> s;
for (long long i = 0; i < n - m + 1; i++) {
if (str.substr(i, m) == s) ara[i + 1]++;
}
for (long long i = 1; i <= n + 10; i++) {
ara[i] += ara[i - 1];
}
while (q--) {
long long a, b;
cin >> a >> b;
b = b - m + 1;
cout << max(0ll, ara[max(b, 0ll)] - ara[a - 1]) << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
char s1[1005], s2[1005];
int s1len, s2len;
int Next[1005];
void kmp() {
int i = 0;
Next[i] = -1;
int j = -1;
while (i < s1len) {
if (j == -1 || s1[i] == s1[j]) {
i++;
j++;
Next[i] = j;
} else {
j = Next[j];
}
}
}
int main(void) {
int Q, L, R;
scanf("%d%d%d", &s2len, &s1len, &Q);
scanf("%s", s2);
scanf("%s", s1);
kmp();
while (Q--) {
scanf("%d%d", &L, &R);
int i = L - 1, j = 0;
int ans = 0;
while (i < R) {
if (j == -1 || s2[i] == s1[j]) {
i++;
j++;
} else {
j = Next[j];
}
if (j == s1len) {
ans++;
j = Next[j];
}
}
printf("%d\n", ans);
}
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char s1[1005], s2[1005];
int s1len, s2len;
int Next[1005];
void kmp() {
int i = 0;
Next[i] = -1;
int j = -1;
while (i < s1len) {
if (j == -1 || s1[i] == s1[j]) {
i++;
j++;
Next[i] = j;
} else {
j = Next[j];
}
}
}
int main(void) {
int Q, L, R;
scanf("%d%d%d", &s2len, &s1len, &Q);
scanf("%s", s2);
scanf("%s", s1);
kmp();
while (Q--) {
scanf("%d%d", &L, &R);
int i = L - 1, j = 0;
int ans = 0;
while (i < R) {
if (j == -1 || s2[i] == s1[j]) {
i++;
j++;
} else {
j = Next[j];
}
if (j == s1len) {
ans++;
j = Next[j];
}
}
printf("%d\n", ans);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long int countSetBits(long long int n) {
long long int count = 0;
while (n) {
n &= (n - 1);
count++;
}
return count;
}
long long int power(long long int x, long long int y) {
if (y == 0)
return 1;
else if (y % 2 == 0)
return power(x, y / 2) * power(x, y / 2);
else
return x * power(x, y / 2) * power(x, y / 2);
}
long long int gcd(long long int a, long long int b) {
long long int r;
while (b) {
r = a % b;
a = b;
b = r;
}
return a;
}
bool pr[1000007];
void sieve() {
pr[0] = 1;
pr[1] = 1;
for (int i = 2; i < sqrt(1000007); i++) {
for (int j = 2 * i; j <= 1000007; j += i) {
pr[j] = 1;
}
}
}
const int MOD = 1e9 + 7;
const int SIZE = 4e6 + 10;
const int MAX = 1e9 + 1;
int main() {
int n, m, q;
scanf("%d%d%d", &n, &m, &q);
string s, t;
cin >> s >> t;
int a[n + 1];
memset((a), 0, sizeof((a)));
for (int i = (0); i < (n - m + 1); ++i) {
if (s.substr(i, m) == t) a[i] = 1;
}
while (q--) {
int cnt = 0;
int l, r;
scanf("%d%d", &l, &r);
for (int i = (l - 1); i < (r - m + 1); ++i) {
if (a[i]) {
cnt++;
}
}
cout << cnt;
cout << endl;
}
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int countSetBits(long long int n) {
long long int count = 0;
while (n) {
n &= (n - 1);
count++;
}
return count;
}
long long int power(long long int x, long long int y) {
if (y == 0)
return 1;
else if (y % 2 == 0)
return power(x, y / 2) * power(x, y / 2);
else
return x * power(x, y / 2) * power(x, y / 2);
}
long long int gcd(long long int a, long long int b) {
long long int r;
while (b) {
r = a % b;
a = b;
b = r;
}
return a;
}
bool pr[1000007];
void sieve() {
pr[0] = 1;
pr[1] = 1;
for (int i = 2; i < sqrt(1000007); i++) {
for (int j = 2 * i; j <= 1000007; j += i) {
pr[j] = 1;
}
}
}
const int MOD = 1e9 + 7;
const int SIZE = 4e6 + 10;
const int MAX = 1e9 + 1;
int main() {
int n, m, q;
scanf("%d%d%d", &n, &m, &q);
string s, t;
cin >> s >> t;
int a[n + 1];
memset((a), 0, sizeof((a)));
for (int i = (0); i < (n - m + 1); ++i) {
if (s.substr(i, m) == t) a[i] = 1;
}
while (q--) {
int cnt = 0;
int l, r;
scanf("%d%d", &l, &r);
for (int i = (l - 1); i < (r - m + 1); ++i) {
if (a[i]) {
cnt++;
}
}
cout << cnt;
cout << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
struct re {
int x, y, cs;
};
const long long module = 1000000000 + 7;
const int maxN = 1000 + 10;
int n, m, q, kq[maxN];
string s, t;
void nhap() { cin >> n >> m >> q >> s >> t; }
int kt(int x) {
for (int i = 1; i <= m - 1; ++i)
if (s[x + i] != t[i]) return (0);
return (1);
}
void xuly() {
for (int i = 0; i <= n - m; ++i) {
if (i > 0) kq[i] = kq[i - 1];
if (s[i] == t[0]) kq[i] += kt(i);
}
}
void xuat() {
int x, y;
for (int i = 1; i <= q; ++i) {
cin >> x >> y;
x--;
y--;
if (y - x + 1 < m)
cout << "0";
else if (x == 0)
cout << kq[y - m + 1];
else
cout << kq[y - m + 1] - kq[x - 1];
cout << "\n";
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
nhap();
xuly();
xuat();
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
struct re {
int x, y, cs;
};
const long long module = 1000000000 + 7;
const int maxN = 1000 + 10;
int n, m, q, kq[maxN];
string s, t;
void nhap() { cin >> n >> m >> q >> s >> t; }
int kt(int x) {
for (int i = 1; i <= m - 1; ++i)
if (s[x + i] != t[i]) return (0);
return (1);
}
void xuly() {
for (int i = 0; i <= n - m; ++i) {
if (i > 0) kq[i] = kq[i - 1];
if (s[i] == t[0]) kq[i] += kt(i);
}
}
void xuat() {
int x, y;
for (int i = 1; i <= q; ++i) {
cin >> x >> y;
x--;
y--;
if (y - x + 1 < m)
cout << "0";
else if (x == 0)
cout << kq[y - m + 1];
else
cout << kq[y - m + 1] - kq[x - 1];
cout << "\n";
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
nhap();
xuly();
xuat();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, t;
cin >> n >> m >> t;
string a, b;
cin >> a >> b;
int begin[n];
int cumm[n];
memset(begin, 0, sizeof(begin));
for (int i = 0; i <= n - m; i++) {
bool match = true;
for (int j = i; j < i + m; j++) {
if (b[j - i] != a[j]) {
match = false;
break;
}
}
if (match) {
begin[i] = 1;
}
}
cumm[0] = begin[0];
for (int i = 1; i < n; i++) {
cumm[i] = cumm[i - 1] + begin[i];
}
for (int p = 0; p < t; p++) {
int l, r;
cin >> l >> r;
l--;
r--;
if (l > 0) {
if (r - m + 1 >= 0) {
cout << max(0, cumm[r - m + 1] - cumm[l - 1]) << endl;
} else {
cout << 0 << endl;
}
} else {
if (r - m + 1 >= 0) {
cout << cumm[r - m + 1] << endl;
} else {
cout << 0 << endl;
}
}
}
return 0;
}
| ### Prompt
Develop a solution in CPP to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, t;
cin >> n >> m >> t;
string a, b;
cin >> a >> b;
int begin[n];
int cumm[n];
memset(begin, 0, sizeof(begin));
for (int i = 0; i <= n - m; i++) {
bool match = true;
for (int j = i; j < i + m; j++) {
if (b[j - i] != a[j]) {
match = false;
break;
}
}
if (match) {
begin[i] = 1;
}
}
cumm[0] = begin[0];
for (int i = 1; i < n; i++) {
cumm[i] = cumm[i - 1] + begin[i];
}
for (int p = 0; p < t; p++) {
int l, r;
cin >> l >> r;
l--;
r--;
if (l > 0) {
if (r - m + 1 >= 0) {
cout << max(0, cumm[r - m + 1] - cumm[l - 1]) << endl;
} else {
cout << 0 << endl;
}
} else {
if (r - m + 1 >= 0) {
cout << cumm[r - m + 1] << endl;
} else {
cout << 0 << endl;
}
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
vector<int> v;
int main() {
int n, m, q;
string s, t;
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < n; i++) {
if (s.substr(i, m) == t)
v.push_back(1);
else
v.push_back(0);
}
while (q--) {
int left, right, ans = 0;
cin >> left >> right;
for (int i = left - 1; i < right; i++) {
if (right - i >= m) ans += v[i];
}
cout << ans << endl;
}
return 0;
}
| ### Prompt
Create a solution in Cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
vector<int> v;
int main() {
int n, m, q;
string s, t;
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < n; i++) {
if (s.substr(i, m) == t)
v.push_back(1);
else
v.push_back(0);
}
while (q--) {
int left, right, ans = 0;
cin >> left >> right;
for (int i = left - 1; i < right; i++) {
if (right - i >= m) ans += v[i];
}
cout << ans << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 5;
char s[maxn];
char t[maxn];
int ans[maxn];
int main() {
int n, m, q;
scanf("%d%d%d", &n, &m, &q);
int l, r;
int cnt = 0;
scanf("%s%s", s, t);
if (n < m) {
while (q--) {
printf("0\n");
}
return 0;
}
int flag;
for (int i = 0; i < n - m + 1; i++) {
flag = 1;
for (int j = i, pp = 0; pp < m; j++, pp++) {
if (s[j] != t[pp]) {
flag = 0;
break;
}
}
if (flag) {
ans[cnt++] = i + 1;
}
}
int n1, n2;
while (q--) {
scanf("%d%d", &l, &r);
n1 = 0, n2 = 0;
for (int i = 0; i < cnt; i++) {
if (ans[i] < l) {
n1++;
}
if (ans[i] <= r - m + 1) {
n2++;
}
}
printf("%d\n", max(n2 - n1, 0));
}
}
| ### Prompt
Develop a solution in CPP to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 5;
char s[maxn];
char t[maxn];
int ans[maxn];
int main() {
int n, m, q;
scanf("%d%d%d", &n, &m, &q);
int l, r;
int cnt = 0;
scanf("%s%s", s, t);
if (n < m) {
while (q--) {
printf("0\n");
}
return 0;
}
int flag;
for (int i = 0; i < n - m + 1; i++) {
flag = 1;
for (int j = i, pp = 0; pp < m; j++, pp++) {
if (s[j] != t[pp]) {
flag = 0;
break;
}
}
if (flag) {
ans[cnt++] = i + 1;
}
}
int n1, n2;
while (q--) {
scanf("%d%d", &l, &r);
n1 = 0, n2 = 0;
for (int i = 0; i < cnt; i++) {
if (ans[i] < l) {
n1++;
}
if (ans[i] <= r - m + 1) {
n2++;
}
}
printf("%d\n", max(n2 - n1, 0));
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m, q;
cin >> n >> m >> q;
string s;
cin >> s;
string r;
cin >> r;
long long lps[m];
lps[0] = 0;
long long len = 0;
long long i = 1;
while (i < m) {
if (r[i] == r[len]) {
len++;
lps[i] = len;
i++;
} else {
if (len) {
len = lps[len - 1];
} else {
lps[i] = 0;
i++;
}
}
}
for (long long p = 0; p < q; p++) {
long long x, y;
cin >> x >> y;
long long count = 0;
string k = s.substr(x - 1, y - x + 1);
long long j = 0;
i = 0;
while (i < k.size()) {
if (r[j] == k[i]) {
i++;
j++;
}
if (j == m) {
count++;
j = lps[j - 1];
} else if (i < n and r[j] != k[i]) {
if (j != 0) {
j = lps[j - 1];
} else {
i++;
}
}
}
cout << count << endl;
}
return 0;
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m, q;
cin >> n >> m >> q;
string s;
cin >> s;
string r;
cin >> r;
long long lps[m];
lps[0] = 0;
long long len = 0;
long long i = 1;
while (i < m) {
if (r[i] == r[len]) {
len++;
lps[i] = len;
i++;
} else {
if (len) {
len = lps[len - 1];
} else {
lps[i] = 0;
i++;
}
}
}
for (long long p = 0; p < q; p++) {
long long x, y;
cin >> x >> y;
long long count = 0;
string k = s.substr(x - 1, y - x + 1);
long long j = 0;
i = 0;
while (i < k.size()) {
if (r[j] == k[i]) {
i++;
j++;
}
if (j == m) {
count++;
j = lps[j - 1];
} else if (i < n and r[j] != k[i]) {
if (j != 0) {
j = lps[j - 1];
} else {
i++;
}
}
}
cout << count << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> k(n, false);
for (int i = 0; i < n; ++i) {
bool flag = true;
for (int j = 0; j < m; ++j) {
if (i + j >= n || s[i + j] != t[j]) flag = false;
}
if (flag) {
if (i == 0) {
k[i] = 1;
continue;
}
k[i] = k[i - 1] + 1;
} else {
if (i == 0) {
k[i] = 0;
continue;
}
k[i] = k[i - 1];
}
}
for (int i = 0; i < q; ++i) {
int l, r;
cin >> l >> r;
l--;
r--;
if (r - l + 1 < m) {
cout << 0 << "\n";
continue;
}
if (l - 1 < 0) {
cout << k[r - m + 1] << "\n";
continue;
}
cout << k[r - m + 1] - k[l - 1] << "\n";
continue;
}
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> k(n, false);
for (int i = 0; i < n; ++i) {
bool flag = true;
for (int j = 0; j < m; ++j) {
if (i + j >= n || s[i + j] != t[j]) flag = false;
}
if (flag) {
if (i == 0) {
k[i] = 1;
continue;
}
k[i] = k[i - 1] + 1;
} else {
if (i == 0) {
k[i] = 0;
continue;
}
k[i] = k[i - 1];
}
}
for (int i = 0; i < q; ++i) {
int l, r;
cin >> l >> r;
l--;
r--;
if (r - l + 1 < m) {
cout << 0 << "\n";
continue;
}
if (l - 1 < 0) {
cout << k[r - m + 1] << "\n";
continue;
}
cout << k[r - m + 1] - k[l - 1] << "\n";
continue;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s;
string a, b;
int num[200005];
int t = 0;
cin >> n >> m >> s;
cin >> a >> b;
for (int i = 0; i < n; i++) {
int flag = 0;
for (int j = 0; j < m; j++) {
if (a[i + j] != b[j]) {
flag = 1;
break;
}
}
if (flag == 0) num[t++] = i;
}
int q, w;
for (int i = 0; i < s; i++) {
int sum = 0;
cin >> q >> w;
for (int j = 0; j < t; j++) {
if (q <= num[j] + 1 && w >= num[j] + m) sum++;
}
cout << sum << endl;
}
return 0;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s;
string a, b;
int num[200005];
int t = 0;
cin >> n >> m >> s;
cin >> a >> b;
for (int i = 0; i < n; i++) {
int flag = 0;
for (int j = 0; j < m; j++) {
if (a[i + j] != b[j]) {
flag = 1;
break;
}
}
if (flag == 0) num[t++] = i;
}
int q, w;
for (int i = 0; i < s; i++) {
int sum = 0;
cin >> q >> w;
for (int j = 0; j < t; j++) {
if (q <= num[j] + 1 && w >= num[j] + m) sum++;
}
cout << sum << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long M = 1000005;
int occur[1005];
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n, m, q;
string s, t;
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < n - m + 1; i++) {
if (s.substr(i, m) == t) {
occur[i + 1] = occur[i] + 1;
} else
occur[i + 1] = occur[i];
}
while (q--) {
int l, r;
cin >> l >> r;
if ((r - l + 1) >= m) {
cout << occur[r - m + 1] - occur[l - 1] << '\n';
} else
cout << 0 << '\n';
}
return 0;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long M = 1000005;
int occur[1005];
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n, m, q;
string s, t;
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < n - m + 1; i++) {
if (s.substr(i, m) == t) {
occur[i + 1] = occur[i] + 1;
} else
occur[i + 1] = occur[i];
}
while (q--) {
int l, r;
cin >> l >> r;
if ((r - l + 1) >= m) {
cout << occur[r - m + 1] - occur[l - 1] << '\n';
} else
cout << 0 << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
int pi[1002];
void pi_function(char temp[], int len) {
int k = 0;
for (int q = 2; q < len + 1; q++) {
while (k > 0 && temp[k] != temp[q - 1]) k = pi[k];
if (temp[k] == temp[q - 1]) k++;
pi[q] = k;
}
return;
}
int main() {
int n, m, q;
scanf("%d%d%d", &n, &m, &q);
char s[n + 2];
scanf("%s", s);
char t[m + 2];
scanf("%s", t);
pi_function(t, m);
for (int i = 0; i < q; i++) {
int l, r;
scanf("%d%d", &l, &r);
int k = 0, ans = 0;
for (int j = l - 1; j < r; j++) {
if (s[j] != t[k]) {
if (k != 0) j--;
k = pi[k];
} else
k++;
if (k == m) ans++;
}
printf("%d\n", ans);
}
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
int pi[1002];
void pi_function(char temp[], int len) {
int k = 0;
for (int q = 2; q < len + 1; q++) {
while (k > 0 && temp[k] != temp[q - 1]) k = pi[k];
if (temp[k] == temp[q - 1]) k++;
pi[q] = k;
}
return;
}
int main() {
int n, m, q;
scanf("%d%d%d", &n, &m, &q);
char s[n + 2];
scanf("%s", s);
char t[m + 2];
scanf("%s", t);
pi_function(t, m);
for (int i = 0; i < q; i++) {
int l, r;
scanf("%d%d", &l, &r);
int k = 0, ans = 0;
for (int j = l - 1; j < r; j++) {
if (s[j] != t[k]) {
if (k != 0) j--;
k = pi[k];
} else
k++;
if (k == m) ans++;
}
printf("%d\n", ans);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> v(n);
for (int i = 0; i <= n - m; i++) {
bool flag = 0;
for (int j = 0; j < m; j++) {
if (t[j] != s[i + j]) {
flag = 1;
}
}
if (!flag) {
v[i] = 1;
}
}
for (int i = 1; i < n; i++) {
v[i] += v[i - 1];
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
if (r - l + 1 < m) {
cout << 0 << endl;
} else {
if (l == 1) {
cout << v[r - m] << endl;
} else {
cout << v[r - m] - v[l - 2] << endl;
}
}
}
}
| ### Prompt
Please create a solution in Cpp to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> v(n);
for (int i = 0; i <= n - m; i++) {
bool flag = 0;
for (int j = 0; j < m; j++) {
if (t[j] != s[i + j]) {
flag = 1;
}
}
if (!flag) {
v[i] = 1;
}
}
for (int i = 1; i < n; i++) {
v[i] += v[i - 1];
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
if (r - l + 1 < m) {
cout << 0 << endl;
} else {
if (l == 1) {
cout << v[r - m] << endl;
} else {
cout << v[r - m] - v[l - 2] << endl;
}
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
inline int read() {
char c = getchar();
int x = 0, f = 1;
for (; !isdigit(c); c = getchar())
if (c == '-') f = -1;
for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
return x * f;
}
const int MAXN = 1e5 + 5;
char s[MAXN], t[MAXN];
int nxt[MAXN], sum[MAXN];
int N, M, Q;
int main() {
N = read();
M = read();
Q = read();
scanf("%s", s + 1);
scanf("%s", t + 1);
int j = 0;
for (int i = 2; i <= M; i++) {
while (j && t[i] != t[j + 1]) j = nxt[j];
if (t[i] == t[j + 1]) j++;
nxt[i] = j;
}
j = 0;
for (int i = 1; i <= N; i++) {
while (j && s[i] != t[j + 1]) j = nxt[j];
if (s[i] == t[j + 1]) j++;
if (j == M) sum[i]++, j = nxt[j];
}
for (int i = 1; i <= N; i++) sum[i] += sum[i - 1];
while (Q--) {
int l = read(), r = read();
l += M - 1;
if (l > r)
puts("0");
else
printf("%d\n", sum[r] - sum[l - 1]);
}
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
inline int read() {
char c = getchar();
int x = 0, f = 1;
for (; !isdigit(c); c = getchar())
if (c == '-') f = -1;
for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
return x * f;
}
const int MAXN = 1e5 + 5;
char s[MAXN], t[MAXN];
int nxt[MAXN], sum[MAXN];
int N, M, Q;
int main() {
N = read();
M = read();
Q = read();
scanf("%s", s + 1);
scanf("%s", t + 1);
int j = 0;
for (int i = 2; i <= M; i++) {
while (j && t[i] != t[j + 1]) j = nxt[j];
if (t[i] == t[j + 1]) j++;
nxt[i] = j;
}
j = 0;
for (int i = 1; i <= N; i++) {
while (j && s[i] != t[j + 1]) j = nxt[j];
if (s[i] == t[j + 1]) j++;
if (j == M) sum[i]++, j = nxt[j];
}
for (int i = 1; i <= N; i++) sum[i] += sum[i - 1];
while (Q--) {
int l = read(), r = read();
l += M - 1;
if (l > r)
puts("0");
else
printf("%d\n", sum[r] - sum[l - 1]);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 9;
int n, m, q;
string s, t;
int a[N], b[N];
void check(int i) {
for (int j = 1; j <= m; ++j)
if (t[j] != s[j + i - 1]) return;
a[i + m - 1] = 1;
return;
}
int main() {
cin >> n >> m >> q;
cin >> s;
s = "#" + s;
cin >> t;
t = "@" + t;
for (int i = 1; i <= n - m + 1; ++i) check(i);
for (int i = 1; i <= q; ++i) {
int x, y;
cin >> x >> y;
int res = 0;
for (int j = x + m - 1; j <= y; ++j) res += a[j];
cout << res << '\n';
}
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 9;
int n, m, q;
string s, t;
int a[N], b[N];
void check(int i) {
for (int j = 1; j <= m; ++j)
if (t[j] != s[j + i - 1]) return;
a[i + m - 1] = 1;
return;
}
int main() {
cin >> n >> m >> q;
cin >> s;
s = "#" + s;
cin >> t;
t = "@" + t;
for (int i = 1; i <= n - m + 1; ++i) check(i);
for (int i = 1; i <= q; ++i) {
int x, y;
cin >> x >> y;
int res = 0;
for (int j = x + m - 1; j <= y; ++j) res += a[j];
cout << res << '\n';
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, t;
cin >> n >> m >> t;
string a, b;
cin >> a >> b;
std::vector<int> v(n, 0);
int cnt = 0;
int p = 1, index = 0, flag = 0;
for (int i = 0; i <= n - m; ++i) v[i + 1] = v[i] + (a.substr(i, m) == b);
while (t--) {
int l, r;
cin >> l >> r;
if (r - l + 1 < m)
cout << "0\n";
else
cout << v[r - m + 1] - v[l - 1] << endl;
}
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, t;
cin >> n >> m >> t;
string a, b;
cin >> a >> b;
std::vector<int> v(n, 0);
int cnt = 0;
int p = 1, index = 0, flag = 0;
for (int i = 0; i <= n - m; ++i) v[i + 1] = v[i] + (a.substr(i, m) == b);
while (t--) {
int l, r;
cin >> l >> r;
if (r - l + 1 < m)
cout << "0\n";
else
cout << v[r - m + 1] - v[l - 1] << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
vector<int> v[1003];
string text, pat;
vector<int> vv;
int main() {
int n, m, q;
cin >> n >> m >> q;
cin >> text >> pat;
for (int i = 0; i < n; i++) {
int x = 0;
for (int j = 0; j < m; j++) {
if ((j + i) < n && text[j + i] == pat[j]) {
x++;
} else
break;
}
if (x == m) {
vv.push_back(i + 1);
}
}
while (q--) {
int l, r;
scanf("%d %d", &l, &r);
int res = 0;
for (int i = 0; i < vv.size(); i++) {
int st = vv[i];
int en = vv[i] + m - 1;
if (st >= l && en <= r) res++;
}
printf("%d\n", res);
}
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
vector<int> v[1003];
string text, pat;
vector<int> vv;
int main() {
int n, m, q;
cin >> n >> m >> q;
cin >> text >> pat;
for (int i = 0; i < n; i++) {
int x = 0;
for (int j = 0; j < m; j++) {
if ((j + i) < n && text[j + i] == pat[j]) {
x++;
} else
break;
}
if (x == m) {
vv.push_back(i + 1);
}
}
while (q--) {
int l, r;
scanf("%d %d", &l, &r);
int res = 0;
for (int i = 0; i < vv.size(); i++) {
int st = vv[i];
int en = vv[i] + m - 1;
if (st >= l && en <= r) res++;
}
printf("%d\n", res);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
string a, b;
int n, m, q;
int pos[1005];
int main() {
cin >> n >> m >> q;
cin >> a;
cin >> b;
for (int i = 0; i <= n - m; i++) {
if (a.substr(i, m) == b) {
pos[i] = 1;
}
}
while (q--) {
int x, y, ans = 0;
cin >> x >> y;
for (int i = x - 1; i <= y - m; i++) {
if (pos[i] == 1) ans++;
}
cout << ans << endl;
}
}
| ### Prompt
Develop a solution in CPP to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string a, b;
int n, m, q;
int pos[1005];
int main() {
cin >> n >> m >> q;
cin >> a;
cin >> b;
for (int i = 0; i <= n - m; i++) {
if (a.substr(i, m) == b) {
pos[i] = 1;
}
}
while (q--) {
int x, y, ans = 0;
cin >> x >> y;
for (int i = x - 1; i <= y - m; i++) {
if (pos[i] == 1) ans++;
}
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 10;
int n, m, q, l, r, pr[maxn];
bool ok[maxn], flag;
string s, t;
int main() {
cin >> n >> m >> q;
cin >> s >> t;
pr[0] = 0;
for (int i = 0; i < n - m + 1; i++) {
flag = true;
for (int j = 0; j < m; j++)
if (s[i + j] != t[j]) {
flag = false;
break;
}
ok[i] = flag;
pr[i + 1] = pr[i] + ok[i];
}
for (int i = max(0, n - m + 1); i < n; i++) {
pr[i + 1] = pr[i];
}
while (q--) {
cin >> l >> r;
l--, r -= m - 1;
cout << (r >= l ? (pr[r] - pr[l]) : 0) << '\n';
}
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 10;
int n, m, q, l, r, pr[maxn];
bool ok[maxn], flag;
string s, t;
int main() {
cin >> n >> m >> q;
cin >> s >> t;
pr[0] = 0;
for (int i = 0; i < n - m + 1; i++) {
flag = true;
for (int j = 0; j < m; j++)
if (s[i + j] != t[j]) {
flag = false;
break;
}
ok[i] = flag;
pr[i + 1] = pr[i] + ok[i];
}
for (int i = max(0, n - m + 1); i < n; i++) {
pr[i + 1] = pr[i];
}
while (q--) {
cin >> l >> r;
l--, r -= m - 1;
cout << (r >= l ? (pr[r] - pr[l]) : 0) << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
string str1, str2;
int ans[1005];
int main() {
scanf("%d %d %d", &n, &m, &q);
cin >> str1 >> str2;
int f[1005];
str1 = " " + str1;
str2 = " " + str2;
for (int i = 1; i <= n - m + 1; i++) {
f[i] = 1;
for (int j = 0; j < m; j++) {
if (str1[i + j] != str2[j + 1]) {
f[i] = 0;
break;
}
}
ans[i] = ans[i - 1] + f[i];
}
for (int i = 0; i < q; i++) {
int l, r;
scanf("%d %d", &l, &r);
r = r - m + 1;
if (r < l) {
printf("0\n");
continue;
}
printf("%d\n", ans[r] - ans[l - 1]);
}
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
string str1, str2;
int ans[1005];
int main() {
scanf("%d %d %d", &n, &m, &q);
cin >> str1 >> str2;
int f[1005];
str1 = " " + str1;
str2 = " " + str2;
for (int i = 1; i <= n - m + 1; i++) {
f[i] = 1;
for (int j = 0; j < m; j++) {
if (str1[i + j] != str2[j + 1]) {
f[i] = 0;
break;
}
}
ans[i] = ans[i - 1] + f[i];
}
for (int i = 0; i < q; i++) {
int l, r;
scanf("%d %d", &l, &r);
r = r - m + 1;
if (r < l) {
printf("0\n");
continue;
}
printf("%d\n", ans[r] - ans[l - 1]);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int desll[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const long long mod = 1e9 + 7;
const int maxn = 1e6 + 7;
const int maxm = 1e8 + 7;
const double eps = 1e-4;
int m, n;
int ar[maxn];
char ch1[maxn], ch2[maxn];
int main() {
scanf("%d", &n);
scanf("%d", &m);
int q;
scanf("%d", &q);
scanf("%s", ch1 + 1);
scanf("%s", ch2 + 1);
memset(ar, 0, sizeof(ar));
for (int i = 1; i <= (n - m + 1); i++) {
int ins = 1;
for (int j = 1; j <= m; j++) {
if (ch1[i + j - 1] != ch2[j]) {
ins = 0;
break;
}
}
ar[i] = ar[i - 1] + ins;
}
for (int i = n - m + 2; i <= n; i++) {
ar[i] = ar[i - 1];
}
while (q--) {
int a, b;
scanf("%d", &a);
scanf("%d", &b);
int ins = ar[b - m + 1] - ar[a - 1];
if (ins < 0) ins = 0;
printf("%d\n", ins);
}
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int desll[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const long long mod = 1e9 + 7;
const int maxn = 1e6 + 7;
const int maxm = 1e8 + 7;
const double eps = 1e-4;
int m, n;
int ar[maxn];
char ch1[maxn], ch2[maxn];
int main() {
scanf("%d", &n);
scanf("%d", &m);
int q;
scanf("%d", &q);
scanf("%s", ch1 + 1);
scanf("%s", ch2 + 1);
memset(ar, 0, sizeof(ar));
for (int i = 1; i <= (n - m + 1); i++) {
int ins = 1;
for (int j = 1; j <= m; j++) {
if (ch1[i + j - 1] != ch2[j]) {
ins = 0;
break;
}
}
ar[i] = ar[i - 1] + ins;
}
for (int i = n - m + 2; i <= n; i++) {
ar[i] = ar[i - 1];
}
while (q--) {
int a, b;
scanf("%d", &a);
scanf("%d", &b);
int ins = ar[b - m + 1] - ar[a - 1];
if (ins < 0) ins = 0;
printf("%d\n", ins);
}
return 0;
}
``` |
#include <bits/stdc++.h>
const double EPS = 0.00000001;
const long long mod = 1000000000 + 7;
using namespace std;
string s, t;
int a[2000], sum[2000];
int main() {
cout.sync_with_stdio(0);
int n, m, q;
cin >> n >> m >> q >> s >> t;
for (int i = 0; i < n; i++) {
int is = 1;
for (int j = 0; j < m; j++) {
if (i + j == n || s[i + j] != t[j]) {
is = 0;
break;
}
}
if (is) a[i] = 1;
}
sum[0] = a[0];
for (int i = 1; i < n; i++) sum[i] = a[i] + sum[i - 1];
while (q--) {
int i, j;
cin >> i >> j;
if (j - i + 1 < m) {
cout << 0 << endl;
continue;
}
i--, j--;
cout << max(sum[j - m + 1] - sum[i] + a[i], 0) << endl;
}
return 0;
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
const double EPS = 0.00000001;
const long long mod = 1000000000 + 7;
using namespace std;
string s, t;
int a[2000], sum[2000];
int main() {
cout.sync_with_stdio(0);
int n, m, q;
cin >> n >> m >> q >> s >> t;
for (int i = 0; i < n; i++) {
int is = 1;
for (int j = 0; j < m; j++) {
if (i + j == n || s[i + j] != t[j]) {
is = 0;
break;
}
}
if (is) a[i] = 1;
}
sum[0] = a[0];
for (int i = 1; i < n; i++) sum[i] = a[i] + sum[i - 1];
while (q--) {
int i, j;
cin >> i >> j;
if (j - i + 1 < m) {
cout << 0 << endl;
continue;
}
i--, j--;
cout << max(sum[j - m + 1] - sum[i] + a[i], 0) << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 10;
char s1[maxn], s2[maxn];
int nxt[maxn];
int n, m, q;
void getFail(char *P, int *f) {
f[0] = 0;
f[1] = 0;
for (int i = 1; i < m; i++) {
int j = f[i];
while (j && P[i] != P[j]) j = f[j];
f[i + 1] = P[i] == P[j] ? j + 1 : 0;
}
}
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s", s1);
scanf("%s", s2);
getFail(s2, nxt);
int l, r;
while (q--) {
scanf("%d%d", &l, &r);
l--;
r--;
int j = 0, cnt = 0;
for (int i = l; i <= r; i++) {
while (j && s2[j] != s1[i]) j = nxt[j];
if (s2[j] == s1[i]) j++;
if (j == m) {
cnt++;
}
}
printf("%d\n", cnt);
}
return 0;
}
| ### Prompt
Create a solution in cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 10;
char s1[maxn], s2[maxn];
int nxt[maxn];
int n, m, q;
void getFail(char *P, int *f) {
f[0] = 0;
f[1] = 0;
for (int i = 1; i < m; i++) {
int j = f[i];
while (j && P[i] != P[j]) j = f[j];
f[i + 1] = P[i] == P[j] ? j + 1 : 0;
}
}
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s", s1);
scanf("%s", s2);
getFail(s2, nxt);
int l, r;
while (q--) {
scanf("%d%d", &l, &r);
l--;
r--;
int j = 0, cnt = 0;
for (int i = l; i <= r; i++) {
while (j && s2[j] != s1[i]) j = nxt[j];
if (s2[j] == s1[i]) j++;
if (j == m) {
cnt++;
}
}
printf("%d\n", cnt);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
string st, subSt;
int lenSt, lenSubSt, queries;
vector<int> positions;
int main() {
ios::sync_with_stdio(0);
cin >> lenSt >> lenSubSt >> queries;
cin >> st;
cin >> subSt;
bool found;
for (int i = 0; i < lenSt - (lenSubSt - 1); i++) {
found = true;
for (int j = 0; j < lenSubSt; j++) {
if (st[i + j] != subSt[j]) {
found = false;
break;
}
}
if (found) {
positions.push_back(i);
}
}
int l, k;
for (int i = 0; i < queries; i++) {
cin >> l >> k;
auto firstIndex = lower_bound(positions.begin(), positions.end(), l - 1);
auto secondIndex =
upper_bound(positions.begin(), positions.end(), k - lenSubSt);
if (secondIndex - firstIndex < 0) {
cout << 0 << endl;
} else {
cout << secondIndex - firstIndex << endl;
}
}
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string st, subSt;
int lenSt, lenSubSt, queries;
vector<int> positions;
int main() {
ios::sync_with_stdio(0);
cin >> lenSt >> lenSubSt >> queries;
cin >> st;
cin >> subSt;
bool found;
for (int i = 0; i < lenSt - (lenSubSt - 1); i++) {
found = true;
for (int j = 0; j < lenSubSt; j++) {
if (st[i + j] != subSt[j]) {
found = false;
break;
}
}
if (found) {
positions.push_back(i);
}
}
int l, k;
for (int i = 0; i < queries; i++) {
cin >> l >> k;
auto firstIndex = lower_bound(positions.begin(), positions.end(), l - 1);
auto secondIndex =
upper_bound(positions.begin(), positions.end(), k - lenSubSt);
if (secondIndex - firstIndex < 0) {
cout << 0 << endl;
} else {
cout << secondIndex - firstIndex << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int match[1000 + 1];
int psum[1000 + 1];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
memset(match, false, sizeof match);
for (int i = 0; i + m - 1 < n; i++) {
match[i + 1] = true;
for (int j = 0; j < m; j++) {
match[i + 1] &= s[i + j] == t[j];
}
}
partial_sum(match, match + n + 1, psum);
while (q--) {
int L, R;
cin >> L >> R;
R -= m - 1;
cout << (R >= L ? psum[R] - psum[L - 1] : 0) << '\n';
}
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int match[1000 + 1];
int psum[1000 + 1];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
memset(match, false, sizeof match);
for (int i = 0; i + m - 1 < n; i++) {
match[i + 1] = true;
for (int j = 0; j < m; j++) {
match[i + 1] &= s[i + j] == t[j];
}
}
partial_sum(match, match + n + 1, psum);
while (q--) {
int L, R;
cin >> L >> R;
R -= m - 1;
cout << (R >= L ? psum[R] - psum[L - 1] : 0) << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
const int inf = 1000000005;
const long long INF = 3e18;
const double pi = 2 * acos(0.0);
using namespace std;
int gcd(int a, int b) {
a = abs(a);
b = abs(b);
while (b) {
a = a % b;
swap(a, b);
}
return a;
}
int modInverse(int a, int m) {
int m0 = m;
int y = 0, x = 1;
if (m == 1) return 0;
while (a > 1) {
int q = a / m;
int t = m;
m = a % m, a = t;
t = y;
y = x - q * y;
x = t;
}
if (x < 0) x += m0;
return x;
}
int lcm(int a, int b) {
int temp = gcd(a, b);
return temp ? (a / temp * b) : 0;
}
const int maxn = 1e3 + 10;
int n, m, q;
int tree[4 * maxn], arr[maxn];
void build(int node, int b, int e) {
if (b == e) {
tree[node] = arr[b];
return;
}
int mid = (b + e) / 2;
build(2 * node, b, mid);
build((2 * node) + 1, mid + 1, e);
tree[node] = tree[2 * node] + tree[(2 * node) + 1];
}
int query(int node, int b, int e, int i, int j) {
if (i > e or j < b) return 0;
if (b >= i and j >= e) return tree[node] / m;
int mid = (b + e) / 2;
int p1 = query(2 * node, b, mid, i, j);
int p2 = query((2 * node) + 1, mid + 1, e, i, j);
return p1 + p2;
}
int main() {
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> pos;
for (int i = 0; i < s.size(); i++) {
bool check = true;
for (int j = 0; j < t.size(); j++) {
if (s[i + j] != t[j]) {
check = false;
break;
}
}
if (check == true) {
pos.push_back(i + 1);
}
}
build(1, 1, n);
while (q--) {
int x, y;
cin >> x >> y;
int ans = 0;
for (int i = 0; i < pos.size(); i++) {
if (pos[i] >= x and pos[i] + m - 1 <= y) {
ans++;
}
}
cout << ans << endl;
}
}
| ### Prompt
Please create a solution in CPP to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
const int inf = 1000000005;
const long long INF = 3e18;
const double pi = 2 * acos(0.0);
using namespace std;
int gcd(int a, int b) {
a = abs(a);
b = abs(b);
while (b) {
a = a % b;
swap(a, b);
}
return a;
}
int modInverse(int a, int m) {
int m0 = m;
int y = 0, x = 1;
if (m == 1) return 0;
while (a > 1) {
int q = a / m;
int t = m;
m = a % m, a = t;
t = y;
y = x - q * y;
x = t;
}
if (x < 0) x += m0;
return x;
}
int lcm(int a, int b) {
int temp = gcd(a, b);
return temp ? (a / temp * b) : 0;
}
const int maxn = 1e3 + 10;
int n, m, q;
int tree[4 * maxn], arr[maxn];
void build(int node, int b, int e) {
if (b == e) {
tree[node] = arr[b];
return;
}
int mid = (b + e) / 2;
build(2 * node, b, mid);
build((2 * node) + 1, mid + 1, e);
tree[node] = tree[2 * node] + tree[(2 * node) + 1];
}
int query(int node, int b, int e, int i, int j) {
if (i > e or j < b) return 0;
if (b >= i and j >= e) return tree[node] / m;
int mid = (b + e) / 2;
int p1 = query(2 * node, b, mid, i, j);
int p2 = query((2 * node) + 1, mid + 1, e, i, j);
return p1 + p2;
}
int main() {
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> pos;
for (int i = 0; i < s.size(); i++) {
bool check = true;
for (int j = 0; j < t.size(); j++) {
if (s[i + j] != t[j]) {
check = false;
break;
}
}
if (check == true) {
pos.push_back(i + 1);
}
}
build(1, 1, n);
while (q--) {
int x, y;
cin >> x >> y;
int ans = 0;
for (int i = 0; i < pos.size(); i++) {
if (pos[i] >= x and pos[i] + m - 1 <= y) {
ans++;
}
}
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s, t;
int n, m, q, l, r, f[1010] = {0};
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < n; i++) {
if (s.substr(i, m) == t) f[i] = 1;
}
for (int T = 0; T < q; T++) {
cin >> l >> r;
int c = 0;
for (int i = l - 1; i <= (r - m); i++) {
if (f[i] == 1) {
c++;
}
}
cout << c << endl;
}
}
| ### Prompt
Create a solution in CPP for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
string s, t;
int n, m, q, l, r, f[1010] = {0};
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < n; i++) {
if (s.substr(i, m) == t) f[i] = 1;
}
for (int T = 0; T < q; T++) {
cin >> l >> r;
int c = 0;
for (int i = l - 1; i <= (r - m); i++) {
if (f[i] == 1) {
c++;
}
}
cout << c << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int precision = 16;
const int modulo = 1000000007;
using ll = long long;
const double EPS = 1e-9;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
cout.precision(precision);
cout.setf(ios_base::fixed);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> ocurr(n, 0);
for (auto i = 0; i < n; ++i) {
ocurr[i] = (i - 1) >= 0 ? ocurr[i - 1] : 0;
if ((i - m + 1) < 0) continue;
if (s.substr(i - m + 1, m) == t) {
ocurr[i]++;
}
}
while (q--) {
int l, r;
cin >> l >> r;
l--;
r--;
if (l > r or l < 0 or r < 0) {
cout << 0 << "\n";
continue;
}
if ((r - l + 1) < m) {
cout << 0 << "\n";
continue;
}
int ret = ocurr[r];
if ((l + m - 2) < n and (l + m - 2) >= 0) ret -= ocurr[l + m - 2];
cout << ret << "\n";
}
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int precision = 16;
const int modulo = 1000000007;
using ll = long long;
const double EPS = 1e-9;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
cout.precision(precision);
cout.setf(ios_base::fixed);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> ocurr(n, 0);
for (auto i = 0; i < n; ++i) {
ocurr[i] = (i - 1) >= 0 ? ocurr[i - 1] : 0;
if ((i - m + 1) < 0) continue;
if (s.substr(i - m + 1, m) == t) {
ocurr[i]++;
}
}
while (q--) {
int l, r;
cin >> l >> r;
l--;
r--;
if (l > r or l < 0 or r < 0) {
cout << 0 << "\n";
continue;
}
if ((r - l + 1) < m) {
cout << 0 << "\n";
continue;
}
int ret = ocurr[r];
if ((l + m - 2) < n and (l + m - 2) >= 0) ret -= ocurr[l + m - 2];
cout << ret << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
vector<int> v[102];
bool par[102][102];
bool vis[102];
int cyc = 0;
void dfs(int nd, int p) {
vis[nd] = true;
int z = v[nd].size();
int nx;
for (int i = 0; i < z; i++) {
nx = v[nd][i];
if (vis[nx]) {
if (nx != p && p != 0) {
if (par[nx][nd] || par[nd][nx]) continue;
par[nx][nd] = par[nd][nx] = true;
cyc++;
}
} else
dfs(nx, nd);
}
return;
}
int main() {
int n, m;
cin >> n >> m;
int x, y;
for (int i = 0; i < m; i++) {
cin >> x >> y;
v[x].push_back(y);
v[y].push_back(x);
}
dfs(1, 0);
bool complete = true;
for (int i = 1; i <= n; i++) {
if (!vis[i]) {
complete = false;
}
}
if (cyc == 1 && complete) {
cout << "FHTAGN!";
} else
cout << "NO";
return 0;
}
| ### Prompt
Develop a solution in CPP to the problem described below:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
vector<int> v[102];
bool par[102][102];
bool vis[102];
int cyc = 0;
void dfs(int nd, int p) {
vis[nd] = true;
int z = v[nd].size();
int nx;
for (int i = 0; i < z; i++) {
nx = v[nd][i];
if (vis[nx]) {
if (nx != p && p != 0) {
if (par[nx][nd] || par[nd][nx]) continue;
par[nx][nd] = par[nd][nx] = true;
cyc++;
}
} else
dfs(nx, nd);
}
return;
}
int main() {
int n, m;
cin >> n >> m;
int x, y;
for (int i = 0; i < m; i++) {
cin >> x >> y;
v[x].push_back(y);
v[y].push_back(x);
}
dfs(1, 0);
bool complete = true;
for (int i = 1; i <= n; i++) {
if (!vis[i]) {
complete = false;
}
}
if (cyc == 1 && complete) {
cout << "FHTAGN!";
} else
cout << "NO";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long maxx = 1e6 + 5;
bool dd[maxx], check, d1[maxx];
long long tr[maxx], val[maxx];
long long b, n, m, k, q, a[maxx], c, f[maxx], x, y, tong = 0;
long long bs(long long v) {
long long l1 = 1, h1 = n, m1;
while (l1 <= h1) {
m1 = (l1 + h1) / 2;
if (a[m1] <= v)
l1 = m1 + 1;
else
h1 = m1 - 1;
}
return h1;
}
vector<long long> st[maxx];
void dfs(long long v) {
dd[v] = false;
for (long long u : st[v]) {
if (dd[u]) {
++k;
dfs(u);
}
}
}
void cf() {
cin >> n >> m;
for (int i = 1; i <= m; i++) {
cin >> x >> y;
st[x].push_back(y);
st[y].push_back(x);
}
fill_n(dd, maxx, true);
k = 1;
dfs(1);
if (k != n || m != n)
cout << "NO";
else
cout << "FHTAGN!";
}
long long uoc(long long a, long long b) {
long long r;
while (b != 0) {
r = a % b;
a = b;
b = r;
}
return a;
}
bool prime(long long v) {
if (v < 2) return false;
for (int i = 2; i <= v / i; i++) {
if (v % i == 0) return false;
}
return true;
}
int main() {
cin.tie(0);
cout.tie(0);
ios_base::sync_with_stdio(0);
cf();
}
| ### Prompt
Please formulate a CPP solution to the following problem:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long maxx = 1e6 + 5;
bool dd[maxx], check, d1[maxx];
long long tr[maxx], val[maxx];
long long b, n, m, k, q, a[maxx], c, f[maxx], x, y, tong = 0;
long long bs(long long v) {
long long l1 = 1, h1 = n, m1;
while (l1 <= h1) {
m1 = (l1 + h1) / 2;
if (a[m1] <= v)
l1 = m1 + 1;
else
h1 = m1 - 1;
}
return h1;
}
vector<long long> st[maxx];
void dfs(long long v) {
dd[v] = false;
for (long long u : st[v]) {
if (dd[u]) {
++k;
dfs(u);
}
}
}
void cf() {
cin >> n >> m;
for (int i = 1; i <= m; i++) {
cin >> x >> y;
st[x].push_back(y);
st[y].push_back(x);
}
fill_n(dd, maxx, true);
k = 1;
dfs(1);
if (k != n || m != n)
cout << "NO";
else
cout << "FHTAGN!";
}
long long uoc(long long a, long long b) {
long long r;
while (b != 0) {
r = a % b;
a = b;
b = r;
}
return a;
}
bool prime(long long v) {
if (v < 2) return false;
for (int i = 2; i <= v / i; i++) {
if (v % i == 0) return false;
}
return true;
}
int main() {
cin.tie(0);
cout.tie(0);
ios_base::sync_with_stdio(0);
cf();
}
``` |
#include <bits/stdc++.h>
using namespace std;
int P[100001];
void CreateSet(int x) { P[x] = x; }
int FindSet(int x) {
if (x != P[x]) P[x] = FindSet(P[x]);
return P[x];
}
int MergeSets(int x, int y) {
int PX = FindSet(x);
int PY = FindSet(y);
if (PX == PY) return 0;
P[PX] = PY;
return 1;
}
int main() {
int n, m;
int i, j, u, v;
scanf("%d %d", &n, &m);
for (i = 1; i <= n; i++) P[i] = i;
int kq = 0;
for (i = 0; i < m; i++) {
scanf("%d %d", &u, &v);
u = FindSet(u);
v = FindSet(v);
if (u == v)
kq++;
else
P[u] = v;
}
if (kq == 1) {
u = FindSet(1);
for (i = 2; i <= n; i++)
if (FindSet(i) != u) {
printf("NO");
return 0;
}
printf("FHTAGN!");
} else
printf("NO");
return 0;
}
| ### Prompt
Please create a solution in cpp to the following problem:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int P[100001];
void CreateSet(int x) { P[x] = x; }
int FindSet(int x) {
if (x != P[x]) P[x] = FindSet(P[x]);
return P[x];
}
int MergeSets(int x, int y) {
int PX = FindSet(x);
int PY = FindSet(y);
if (PX == PY) return 0;
P[PX] = PY;
return 1;
}
int main() {
int n, m;
int i, j, u, v;
scanf("%d %d", &n, &m);
for (i = 1; i <= n; i++) P[i] = i;
int kq = 0;
for (i = 0; i < m; i++) {
scanf("%d %d", &u, &v);
u = FindSet(u);
v = FindSet(v);
if (u == v)
kq++;
else
P[u] = v;
}
if (kq == 1) {
u = FindSet(1);
for (i = 2; i <= n; i++)
if (FindSet(i) != u) {
printf("NO");
return 0;
}
printf("FHTAGN!");
} else
printf("NO");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long LINF = 1e18;
const int INF = 1e9;
const long double EPS = 1e-9;
const int MOD = 1e9 + 7;
const int N = 105;
vector<int> adj[N];
bitset<N> vis;
int cycle;
void dfs(int v) {
vis[v] = 1;
for (int& u : adj[v]) {
if (vis[u] == 0) dfs(u);
}
}
int main() {
int n, m;
scanf("%d %d", &n, &m);
int a, b;
for (int i = 0; i < (m); ++i) {
scanf("%d %d", &a, &b);
adj[a].push_back(b);
adj[b].push_back(a);
}
int comp = 0;
if (n != m) {
printf("NO\n");
} else {
bool flag = true;
dfs(1);
for (int i = 1; i <= n; i++) {
if (!vis[i]) {
flag = false;
break;
}
}
if (flag)
printf("FHTAGN!\n");
else
printf("NO\n");
}
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long LINF = 1e18;
const int INF = 1e9;
const long double EPS = 1e-9;
const int MOD = 1e9 + 7;
const int N = 105;
vector<int> adj[N];
bitset<N> vis;
int cycle;
void dfs(int v) {
vis[v] = 1;
for (int& u : adj[v]) {
if (vis[u] == 0) dfs(u);
}
}
int main() {
int n, m;
scanf("%d %d", &n, &m);
int a, b;
for (int i = 0; i < (m); ++i) {
scanf("%d %d", &a, &b);
adj[a].push_back(b);
adj[b].push_back(a);
}
int comp = 0;
if (n != m) {
printf("NO\n");
} else {
bool flag = true;
dfs(1);
for (int i = 1; i <= n; i++) {
if (!vis[i]) {
flag = false;
break;
}
}
if (flag)
printf("FHTAGN!\n");
else
printf("NO\n");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const double pi = 2 * acos(0.0);
const int inf = 0x3f3f3f3f;
const double infd = 1.0 / 0.0;
long long power(long long x, long long y, long long MOD) {
long long res = 1;
x = x % MOD;
while (y > 0) {
if (y & 1) res = (res * x) % MOD;
y = y >> 1;
x = (x * x) % MOD;
}
return res;
}
long long mul(long long a, long long b, long long MOD) {
if (b == 1) {
return a;
}
if (b % 2 == 0) {
return 2 * (mul(a, b / 2, MOD) % MOD);
} else {
return (a + (2 * (mul(a, b / 2, MOD)) % MOD)) % MOD;
}
}
bool ispow2(long long n) {
if (((n & (n - 1)) == 0) && n != 0) {
return true;
}
return false;
}
bool prime(int x) {
if (x == 1) {
return false;
}
if (x == 2) {
return true;
}
if (x % 2 == 0) {
return false;
}
for (int i = 3; i * i <= x; i += 2) {
if (x % i == 0) {
return false;
}
}
return true;
}
int __gcd(int a, int b) {
if (b == 0) return a;
return __gcd(b, a % b);
}
int coprime(int a, int b) {
if (a == b) return 0;
if (prime(a) && prime(b)) return 1;
if ((a == 2 && b % 2 != 0) || (b == 2 && a % 2 != 0)) return 1;
if ((a % 2 != 0 && prime(b) && a < b) || (b % 2 != 0 && prime(a) && a > b))
return 1;
if (abs(a - b) == 1) return 1;
if (a == 1 || b == 1) return 1;
return __gcd(a, b);
}
unsigned long long lcm(unsigned a, unsigned b) {
return ((unsigned long long)a) * (b / __gcd(a, b));
}
int n, m, x, y;
vector<int> g[101];
int vis[101];
int cnt = 0;
void dfs(int u) {
vis[u] = 1;
cnt++;
for (int i = 0; i < g[u].size(); i++) {
if (!vis[g[u][i]]) {
dfs(g[u][i]);
}
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
memset(vis, 0, sizeof(vis));
cin >> n >> m;
for (int i = 0; i < m; i++) {
cin >> x >> y;
g[x].push_back(y);
g[y].push_back(x);
}
if (n == m) {
dfs(1);
(cnt == n) ? cout << "FHTAGN!" << endl : cout << "NO" << endl;
} else {
cout << "NO" << endl;
}
return 0;
}
| ### Prompt
Create a solution in cpp for the following problem:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const double pi = 2 * acos(0.0);
const int inf = 0x3f3f3f3f;
const double infd = 1.0 / 0.0;
long long power(long long x, long long y, long long MOD) {
long long res = 1;
x = x % MOD;
while (y > 0) {
if (y & 1) res = (res * x) % MOD;
y = y >> 1;
x = (x * x) % MOD;
}
return res;
}
long long mul(long long a, long long b, long long MOD) {
if (b == 1) {
return a;
}
if (b % 2 == 0) {
return 2 * (mul(a, b / 2, MOD) % MOD);
} else {
return (a + (2 * (mul(a, b / 2, MOD)) % MOD)) % MOD;
}
}
bool ispow2(long long n) {
if (((n & (n - 1)) == 0) && n != 0) {
return true;
}
return false;
}
bool prime(int x) {
if (x == 1) {
return false;
}
if (x == 2) {
return true;
}
if (x % 2 == 0) {
return false;
}
for (int i = 3; i * i <= x; i += 2) {
if (x % i == 0) {
return false;
}
}
return true;
}
int __gcd(int a, int b) {
if (b == 0) return a;
return __gcd(b, a % b);
}
int coprime(int a, int b) {
if (a == b) return 0;
if (prime(a) && prime(b)) return 1;
if ((a == 2 && b % 2 != 0) || (b == 2 && a % 2 != 0)) return 1;
if ((a % 2 != 0 && prime(b) && a < b) || (b % 2 != 0 && prime(a) && a > b))
return 1;
if (abs(a - b) == 1) return 1;
if (a == 1 || b == 1) return 1;
return __gcd(a, b);
}
unsigned long long lcm(unsigned a, unsigned b) {
return ((unsigned long long)a) * (b / __gcd(a, b));
}
int n, m, x, y;
vector<int> g[101];
int vis[101];
int cnt = 0;
void dfs(int u) {
vis[u] = 1;
cnt++;
for (int i = 0; i < g[u].size(); i++) {
if (!vis[g[u][i]]) {
dfs(g[u][i]);
}
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
memset(vis, 0, sizeof(vis));
cin >> n >> m;
for (int i = 0; i < m; i++) {
cin >> x >> y;
g[x].push_back(y);
g[y].push_back(x);
}
if (n == m) {
dfs(1);
(cnt == n) ? cout << "FHTAGN!" << endl : cout << "NO" << endl;
} else {
cout << "NO" << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int fp(int abc, int parent[]) {
if (parent[abc] == abc) return abc;
int def = fp(parent[abc], parent);
parent[abc] = def;
return def;
}
void cycledmerge(vector<int> hub[], int n, int parent[]) {
stack<int> urutan;
for (int i = 0; i < n; i++) parent[i] = i;
stack<pair<int, int> > ngurut;
vector<int> sudah(n, 0);
vector<int> adadiurutan(n, 0);
vector<int> dari(n, -1);
for (int i = 0; i < n; i++) {
if (sudah[i]) continue;
urutan.push(i);
ngurut.push(make_pair(i, 0));
adadiurutan[i] = 1;
sudah[i] = 1;
while (!ngurut.empty()) {
int id = ngurut.top().first;
int val = ngurut.top().second;
ngurut.pop();
if (val == (int)hub[id].size()) {
while (!urutan.empty() && urutan.top() == id) urutan.pop();
adadiurutan[id] = 0;
continue;
}
ngurut.push(make_pair(id, val + 1));
int next = hub[id][val];
if (dari[id] == next) continue;
if (sudah[next] && !adadiurutan[fp(next, parent)]) continue;
if (!sudah[next]) {
sudah[next] = 1;
urutan.push(next);
ngurut.push(make_pair(next, 0));
adadiurutan[next] = 1;
dari[next] = id;
continue;
}
if (parent[fp(id, parent)] == fp(next, parent)) continue;
parent[fp(id, parent)] = fp(next, parent);
while (!urutan.empty() && urutan.top() != fp(next, parent)) {
parent[fp(urutan.top(), parent)] = fp(next, parent);
urutan.pop();
}
}
}
return;
}
int main() {
int p[200];
vector<int> hub[200];
int n, m;
cin >> n >> m;
for (int(i) = 0; (i) < (m); ++(i)) {
int a, b;
cin >> a >> b;
--a;
--b;
hub[a].push_back(b);
hub[b].push_back(a);
}
cycledmerge(hub, n, p);
vector<int> ukuran(n, 0);
for (int(i) = 0; (i) < (n); ++(i)) ukuran[p[i]]++;
int numcyc = 0;
for (int(i) = 0; (i) < (n); ++(i))
if (ukuran[i] >= 3) ++numcyc;
if (numcyc != 1) {
cout << "NO" << endl;
return 0;
}
int besar = 0;
for (int(i) = 0; (i) < (n); ++(i))
if (ukuran[i] >= 3) besar = ukuran[i];
int yesadj = 0;
int noadj = 0;
for (int(i) = 0; (i) < (n); ++(i))
for (int(j) = 0; (j) < (((int)hub[i].size())); ++(j)) {
int k = hub[i][j];
if (k < i) continue;
if (p[i] != p[k])
++noadj;
else
++yesadj;
}
if (yesadj != besar || noadj != (n - besar + 1) - 1) {
cout << "NO" << endl;
} else {
cout << "FHTAGN!" << endl;
}
return 0;
}
| ### Prompt
Create a solution in cpp for the following problem:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int fp(int abc, int parent[]) {
if (parent[abc] == abc) return abc;
int def = fp(parent[abc], parent);
parent[abc] = def;
return def;
}
void cycledmerge(vector<int> hub[], int n, int parent[]) {
stack<int> urutan;
for (int i = 0; i < n; i++) parent[i] = i;
stack<pair<int, int> > ngurut;
vector<int> sudah(n, 0);
vector<int> adadiurutan(n, 0);
vector<int> dari(n, -1);
for (int i = 0; i < n; i++) {
if (sudah[i]) continue;
urutan.push(i);
ngurut.push(make_pair(i, 0));
adadiurutan[i] = 1;
sudah[i] = 1;
while (!ngurut.empty()) {
int id = ngurut.top().first;
int val = ngurut.top().second;
ngurut.pop();
if (val == (int)hub[id].size()) {
while (!urutan.empty() && urutan.top() == id) urutan.pop();
adadiurutan[id] = 0;
continue;
}
ngurut.push(make_pair(id, val + 1));
int next = hub[id][val];
if (dari[id] == next) continue;
if (sudah[next] && !adadiurutan[fp(next, parent)]) continue;
if (!sudah[next]) {
sudah[next] = 1;
urutan.push(next);
ngurut.push(make_pair(next, 0));
adadiurutan[next] = 1;
dari[next] = id;
continue;
}
if (parent[fp(id, parent)] == fp(next, parent)) continue;
parent[fp(id, parent)] = fp(next, parent);
while (!urutan.empty() && urutan.top() != fp(next, parent)) {
parent[fp(urutan.top(), parent)] = fp(next, parent);
urutan.pop();
}
}
}
return;
}
int main() {
int p[200];
vector<int> hub[200];
int n, m;
cin >> n >> m;
for (int(i) = 0; (i) < (m); ++(i)) {
int a, b;
cin >> a >> b;
--a;
--b;
hub[a].push_back(b);
hub[b].push_back(a);
}
cycledmerge(hub, n, p);
vector<int> ukuran(n, 0);
for (int(i) = 0; (i) < (n); ++(i)) ukuran[p[i]]++;
int numcyc = 0;
for (int(i) = 0; (i) < (n); ++(i))
if (ukuran[i] >= 3) ++numcyc;
if (numcyc != 1) {
cout << "NO" << endl;
return 0;
}
int besar = 0;
for (int(i) = 0; (i) < (n); ++(i))
if (ukuran[i] >= 3) besar = ukuran[i];
int yesadj = 0;
int noadj = 0;
for (int(i) = 0; (i) < (n); ++(i))
for (int(j) = 0; (j) < (((int)hub[i].size())); ++(j)) {
int k = hub[i][j];
if (k < i) continue;
if (p[i] != p[k])
++noadj;
else
++yesadj;
}
if (yesadj != besar || noadj != (n - besar + 1) - 1) {
cout << "NO" << endl;
} else {
cout << "FHTAGN!" << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 110;
vector<int> G[N];
int vis[N];
void DFS(int cur) {
vis[cur]++;
int sz = G[cur].size();
int v;
for (int i = 0; i < sz; i++) {
v = G[cur][i];
if (!vis[v]) DFS(v);
}
}
int main() {
int n, m, u, v;
cin >> n >> m;
for (int i = 0; i < m; i++) {
cin >> u >> v;
G[u].push_back(v);
G[v].push_back(u);
}
DFS(1);
for (int i = 1; i <= n; i++)
if (!vis[i]) {
cout << "NO";
return 0;
}
if (n == m)
cout << "FHTAGN!";
else
cout << "NO";
return 0;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 110;
vector<int> G[N];
int vis[N];
void DFS(int cur) {
vis[cur]++;
int sz = G[cur].size();
int v;
for (int i = 0; i < sz; i++) {
v = G[cur][i];
if (!vis[v]) DFS(v);
}
}
int main() {
int n, m, u, v;
cin >> n >> m;
for (int i = 0; i < m; i++) {
cin >> u >> v;
G[u].push_back(v);
G[v].push_back(u);
}
DFS(1);
for (int i = 1; i <= n; i++)
if (!vis[i]) {
cout << "NO";
return 0;
}
if (n == m)
cout << "FHTAGN!";
else
cout << "NO";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, a, b, ciclos, vertice_visited;
int pai[110], vis[110];
vector<int> adj[110];
void dfs(int v) {
vis[v] = 1;
vertice_visited++;
for (int i = 0; i < adj[v].size(); i++) {
int x = adj[v][i];
if (vis[x]) {
if (pai[v] != x) ciclos++;
continue;
}
pai[x] = v;
vis[x] = 1;
dfs(x);
}
}
int main() {
ios_base::sync_with_stdio(0);
cin >> n >> m;
while (m--) {
cin >> a >> b;
adj[a].push_back(b);
adj[b].push_back(a);
}
dfs(1);
cout << (ciclos == 2 && vertice_visited == n ? "FHTAGN!" : "NO") << endl;
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, a, b, ciclos, vertice_visited;
int pai[110], vis[110];
vector<int> adj[110];
void dfs(int v) {
vis[v] = 1;
vertice_visited++;
for (int i = 0; i < adj[v].size(); i++) {
int x = adj[v][i];
if (vis[x]) {
if (pai[v] != x) ciclos++;
continue;
}
pai[x] = v;
vis[x] = 1;
dfs(x);
}
}
int main() {
ios_base::sync_with_stdio(0);
cin >> n >> m;
while (m--) {
cin >> a >> b;
adj[a].push_back(b);
adj[b].push_back(a);
}
dfs(1);
cout << (ciclos == 2 && vertice_visited == n ? "FHTAGN!" : "NO") << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long power(long long x, unsigned long long y) {
long long temp;
if (y == 0) return 1;
temp = power(x, y / 2);
if (y % 2 == 0)
return temp * temp;
else
return x * temp * temp;
}
long long modpow(long long x, unsigned int y, long long p) {
long long res = 1;
x = x % p;
if (y == 0) return 1;
if (x == 0) return 0;
while (y > 0) {
if (y & 1) res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
long long exponentMod(long long A, long long B, long long C) {
if (B == 0) return 1;
if (A == 0) return 0;
long long y;
if (B % 2 == 0) {
y = exponentMod(A, B / 2, C);
y = (y * y) % C;
} else {
y = A % C;
y = (y * exponentMod(A, B - 1, C) % C) % C;
}
return (long long)((y + C) % C);
}
long long gcd(long long a, long long b) {
if (a == 0) return b;
return gcd(b % a, a);
}
int gcdExtended(int a, int b, int *x, int *y) {
if (a == 0) {
*x = 0;
*y = 1;
return b;
}
int x1, y1;
int gcd = gcdExtended(b % a, a, &x1, &y1);
*x = y1 - (b / a) * x1;
*y = x1;
return gcd;
}
void modInverse(int a, int m) {
int x, y;
int g = gcdExtended(a, m, &x, &y);
if (g != 1)
cout << "Inverse doesn't exist";
else {
int res = (x % m + m) % m;
cout << "Modular multiplicative inverse is " << res;
}
}
void SieveOfEratosthenes(int n) {
bool sieve[n + 1];
long long cnt = 0;
memset(sieve, 0, sizeof(sieve));
for (int p = 2; p * p <= n; p++) {
if (!sieve[p]) {
for (int i = 2 * p; i <= n; i += p) sieve[i] = p;
}
}
for (int p = 2; p <= n; p++) {
if (sieve[p]) cnt++;
}
cout << cnt;
}
int phi(unsigned int n) {
float result = n;
for (int p = 2; p * p <= n; ++p) {
if (n % p == 0) {
while (n % p == 0) n /= p;
result *= (1.0 - (1.0 / (float)p));
}
}
if (n > 1) result *= (1.0 - (1.0 / (float)n));
return (int)result;
}
long long floorSqrt(long long x) {
if (x == 0 || x == 1) return x;
unsigned long long start = 1, end = x, ans;
while (start <= end) {
unsigned long long mid = start + (end - start) / 2;
if (mid * mid == x) return mid;
if (mid * mid < x) {
start = mid + 1;
ans = mid;
} else
end = mid - 1;
}
return ans;
}
void start() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
}
vector<long long> adj[105];
bool vis[105];
void dfs(long long k) {
if (!vis[k]) {
vis[k] = true;
for (auto x : adj[k]) {
if (!vis[x]) {
dfs(x);
vis[x] = true;
}
}
}
}
int main() {
start();
long long n, m;
cin >> n >> m;
for (long long i = 0; i < m; i++) {
long long a, b;
cin >> a >> b;
adj[a].push_back(b);
adj[b].push_back(a);
}
long long cnt = 0;
for (long long i = 1; i < n + 1; i++) {
if (!vis[i]) {
dfs(i);
cnt++;
}
}
if (cnt == 1 && n == m) {
cout << "FHTAGN!";
} else
cout << "NO";
}
| ### Prompt
In Cpp, your task is to solve the following problem:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long power(long long x, unsigned long long y) {
long long temp;
if (y == 0) return 1;
temp = power(x, y / 2);
if (y % 2 == 0)
return temp * temp;
else
return x * temp * temp;
}
long long modpow(long long x, unsigned int y, long long p) {
long long res = 1;
x = x % p;
if (y == 0) return 1;
if (x == 0) return 0;
while (y > 0) {
if (y & 1) res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
long long exponentMod(long long A, long long B, long long C) {
if (B == 0) return 1;
if (A == 0) return 0;
long long y;
if (B % 2 == 0) {
y = exponentMod(A, B / 2, C);
y = (y * y) % C;
} else {
y = A % C;
y = (y * exponentMod(A, B - 1, C) % C) % C;
}
return (long long)((y + C) % C);
}
long long gcd(long long a, long long b) {
if (a == 0) return b;
return gcd(b % a, a);
}
int gcdExtended(int a, int b, int *x, int *y) {
if (a == 0) {
*x = 0;
*y = 1;
return b;
}
int x1, y1;
int gcd = gcdExtended(b % a, a, &x1, &y1);
*x = y1 - (b / a) * x1;
*y = x1;
return gcd;
}
void modInverse(int a, int m) {
int x, y;
int g = gcdExtended(a, m, &x, &y);
if (g != 1)
cout << "Inverse doesn't exist";
else {
int res = (x % m + m) % m;
cout << "Modular multiplicative inverse is " << res;
}
}
void SieveOfEratosthenes(int n) {
bool sieve[n + 1];
long long cnt = 0;
memset(sieve, 0, sizeof(sieve));
for (int p = 2; p * p <= n; p++) {
if (!sieve[p]) {
for (int i = 2 * p; i <= n; i += p) sieve[i] = p;
}
}
for (int p = 2; p <= n; p++) {
if (sieve[p]) cnt++;
}
cout << cnt;
}
int phi(unsigned int n) {
float result = n;
for (int p = 2; p * p <= n; ++p) {
if (n % p == 0) {
while (n % p == 0) n /= p;
result *= (1.0 - (1.0 / (float)p));
}
}
if (n > 1) result *= (1.0 - (1.0 / (float)n));
return (int)result;
}
long long floorSqrt(long long x) {
if (x == 0 || x == 1) return x;
unsigned long long start = 1, end = x, ans;
while (start <= end) {
unsigned long long mid = start + (end - start) / 2;
if (mid * mid == x) return mid;
if (mid * mid < x) {
start = mid + 1;
ans = mid;
} else
end = mid - 1;
}
return ans;
}
void start() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
}
vector<long long> adj[105];
bool vis[105];
void dfs(long long k) {
if (!vis[k]) {
vis[k] = true;
for (auto x : adj[k]) {
if (!vis[x]) {
dfs(x);
vis[x] = true;
}
}
}
}
int main() {
start();
long long n, m;
cin >> n >> m;
for (long long i = 0; i < m; i++) {
long long a, b;
cin >> a >> b;
adj[a].push_back(b);
adj[b].push_back(a);
}
long long cnt = 0;
for (long long i = 1; i < n + 1; i++) {
if (!vis[i]) {
dfs(i);
cnt++;
}
}
if (cnt == 1 && n == m) {
cout << "FHTAGN!";
} else
cout << "NO";
}
``` |
#include <bits/stdc++.h>
using namespace std;
vector<int> v[105];
int visit[105];
int c;
void dfs(int i) {
visit[i] = 1;
vector<int>::iterator it;
for (it = v[i].begin(); it != v[i].end(); it++) {
if (visit[(*it)] == 0) {
c++;
dfs((*it));
}
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n, m, a, b, i;
cin >> n >> m;
for (i = 1; i <= m; i++) {
cin >> a >> b;
v[a].push_back(b);
v[b].push_back(a);
}
c++;
dfs(1);
if (c == n && m == n)
cout << "FHTAGN!";
else
cout << "NO";
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
vector<int> v[105];
int visit[105];
int c;
void dfs(int i) {
visit[i] = 1;
vector<int>::iterator it;
for (it = v[i].begin(); it != v[i].end(); it++) {
if (visit[(*it)] == 0) {
c++;
dfs((*it));
}
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n, m, a, b, i;
cin >> n >> m;
for (i = 1; i <= m; i++) {
cin >> a >> b;
v[a].push_back(b);
v[b].push_back(a);
}
c++;
dfs(1);
if (c == n && m == n)
cout << "FHTAGN!";
else
cout << "NO";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 111;
int N, M;
int cnt = 0;
vector<int> node[MAXN];
int visited[MAXN];
void dfs(int cur) {
if (visited[cur]) return;
visited[cur] = true;
cnt++;
for (int i = 0; i < node[cur].size(); i++) dfs(node[cur][i]);
}
int main() {
cin >> N >> M;
for (int i = 0; i < M; i++) {
int a, b;
cin >> a >> b;
node[a].push_back(b);
node[b].push_back(a);
}
for (int i = 1; i <= N; i++) visited[i] = false;
dfs(1);
if (cnt == N && M == N)
printf("FHTAGN!\n");
else
printf("NO\n");
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 111;
int N, M;
int cnt = 0;
vector<int> node[MAXN];
int visited[MAXN];
void dfs(int cur) {
if (visited[cur]) return;
visited[cur] = true;
cnt++;
for (int i = 0; i < node[cur].size(); i++) dfs(node[cur][i]);
}
int main() {
cin >> N >> M;
for (int i = 0; i < M; i++) {
int a, b;
cin >> a >> b;
node[a].push_back(b);
node[b].push_back(a);
}
for (int i = 1; i <= N; i++) visited[i] = false;
dfs(1);
if (cnt == N && M == N)
printf("FHTAGN!\n");
else
printf("NO\n");
return 0;
}
``` |
#include <bits/stdc++.h>
int N, M;
using namespace std;
int cont = 0;
long long parents[101];
long long find(long long n) {
if (n == parents[n]) {
return n;
}
long long aux = find(parents[n]);
parents[n] = aux;
return find(parents[n]);
}
long long Union(long long n1, long long n2) {
long long p1 = find(n1);
long long p2 = find(n2);
if (p1 != p2) {
parents[p1] = p2;
return p2;
}
cont++;
return -1;
}
int main() {
cin >> N >> M;
for (int i = 0; i < 101; i++) {
parents[i] = i;
}
for (int i = 0; i < M; i++) {
long long x, y;
cin >> x >> y;
Union(x, y);
}
if (cont == 1 && M >= N) {
cout << "FHTAGN!";
} else {
cout << "NO";
}
cout << endl;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
int N, M;
using namespace std;
int cont = 0;
long long parents[101];
long long find(long long n) {
if (n == parents[n]) {
return n;
}
long long aux = find(parents[n]);
parents[n] = aux;
return find(parents[n]);
}
long long Union(long long n1, long long n2) {
long long p1 = find(n1);
long long p2 = find(n2);
if (p1 != p2) {
parents[p1] = p2;
return p2;
}
cont++;
return -1;
}
int main() {
cin >> N >> M;
for (int i = 0; i < 101; i++) {
parents[i] = i;
}
for (int i = 0; i < M; i++) {
long long x, y;
cin >> x >> y;
Union(x, y);
}
if (cont == 1 && M >= N) {
cout << "FHTAGN!";
} else {
cout << "NO";
}
cout << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int maxn = 1e2 + 10;
const double eps = 1e-9;
int mp[maxn][maxn];
int deg[maxn];
int vis[maxn];
int n, m;
void dfs(int u) {
vis[u] = 1;
for (int i = 1; i <= n; i++) {
if (!vis[i] && mp[u][i]) {
dfs(i);
}
}
}
int main() {
ios_base::sync_with_stdio(0);
while (scanf("%d%d", &n, &m) != EOF) {
memset(vis, 0, sizeof vis);
for (int i = 0; i < m; i++) {
int u, v;
scanf("%d%d", &u, &v);
mp[u][v] = mp[v][u] = 1;
}
if (n == m) {
dfs(1);
int flag = 1;
for (int i = 1; i <= n; i++) {
if (!vis[i]) flag = 0;
}
if (flag)
puts("FHTAGN!");
else
puts("NO");
} else
puts("NO");
}
return 0;
}
| ### Prompt
Develop a solution in cpp to the problem described below:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int maxn = 1e2 + 10;
const double eps = 1e-9;
int mp[maxn][maxn];
int deg[maxn];
int vis[maxn];
int n, m;
void dfs(int u) {
vis[u] = 1;
for (int i = 1; i <= n; i++) {
if (!vis[i] && mp[u][i]) {
dfs(i);
}
}
}
int main() {
ios_base::sync_with_stdio(0);
while (scanf("%d%d", &n, &m) != EOF) {
memset(vis, 0, sizeof vis);
for (int i = 0; i < m; i++) {
int u, v;
scanf("%d%d", &u, &v);
mp[u][v] = mp[v][u] = 1;
}
if (n == m) {
dfs(1);
int flag = 1;
for (int i = 1; i <= n; i++) {
if (!vis[i]) flag = 0;
}
if (flag)
puts("FHTAGN!");
else
puts("NO");
} else
puts("NO");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 200;
int n, m;
int used[N];
vector<int> g[N];
void travel(int u) {
used[u] = 1;
for (auto to : g[u]) {
if (!used[to]) travel(to);
}
}
int cycle_st = -1, cycle_ed = -1;
int p[N];
int dfs(int u, int pr = -1) {
used[u] = 1;
for (auto to : g[u]) {
if (used[to] == 0) {
p[to] = u;
if (dfs(to, u)) return 1;
} else if (used[to] == 1 && to != pr) {
cycle_ed = u;
cycle_st = to;
return 1;
}
}
used[u] = 2;
return 0;
}
bool part[N];
void bfs(int u, int pr = -1) {
used[u] = 1;
for (auto to : g[u]) {
if (to == pr) continue;
if (used[to] or part[to]) {
cout << "NO";
exit(0);
}
bfs(to, u);
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m;
for (int i = 1; i <= m; i++) {
int u, v;
cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
travel(1);
bool bad = 0;
for (int i = 1; i <= n; i++) bad |= !used[i];
if (bad) return cout << "NO", 0;
memset(used, 0, sizeof(used));
dfs(1);
if (cycle_st == -1) return cout << "NO", 0;
vector<int> core = {cycle_st};
part[cycle_st] = 1;
while (cycle_ed != cycle_st) {
core.push_back(cycle_ed);
part[cycle_ed] = 1;
cycle_ed = p[cycle_ed];
}
memset(used, 0, sizeof(used));
for (int i = 0; i < core.size(); i++) {
int u = core[i];
if (used[u]) return cout << "NO", 0;
used[u] = 1;
for (auto to : g[u])
if (!part[to]) bfs(to, u);
}
for (int i = 1; i <= n; i++) bad |= !used[i];
if (bad) return cout << "NO", 0;
int num = n - core.size();
if (m - core.size() != num) return cout << "NO", 0;
cout << "FHTAGN!";
return 0;
}
| ### Prompt
Please create a solution in CPP to the following problem:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 200;
int n, m;
int used[N];
vector<int> g[N];
void travel(int u) {
used[u] = 1;
for (auto to : g[u]) {
if (!used[to]) travel(to);
}
}
int cycle_st = -1, cycle_ed = -1;
int p[N];
int dfs(int u, int pr = -1) {
used[u] = 1;
for (auto to : g[u]) {
if (used[to] == 0) {
p[to] = u;
if (dfs(to, u)) return 1;
} else if (used[to] == 1 && to != pr) {
cycle_ed = u;
cycle_st = to;
return 1;
}
}
used[u] = 2;
return 0;
}
bool part[N];
void bfs(int u, int pr = -1) {
used[u] = 1;
for (auto to : g[u]) {
if (to == pr) continue;
if (used[to] or part[to]) {
cout << "NO";
exit(0);
}
bfs(to, u);
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m;
for (int i = 1; i <= m; i++) {
int u, v;
cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
travel(1);
bool bad = 0;
for (int i = 1; i <= n; i++) bad |= !used[i];
if (bad) return cout << "NO", 0;
memset(used, 0, sizeof(used));
dfs(1);
if (cycle_st == -1) return cout << "NO", 0;
vector<int> core = {cycle_st};
part[cycle_st] = 1;
while (cycle_ed != cycle_st) {
core.push_back(cycle_ed);
part[cycle_ed] = 1;
cycle_ed = p[cycle_ed];
}
memset(used, 0, sizeof(used));
for (int i = 0; i < core.size(); i++) {
int u = core[i];
if (used[u]) return cout << "NO", 0;
used[u] = 1;
for (auto to : g[u])
if (!part[to]) bfs(to, u);
}
for (int i = 1; i <= n; i++) bad |= !used[i];
if (bad) return cout << "NO", 0;
int num = n - core.size();
if (m - core.size() != num) return cout << "NO", 0;
cout << "FHTAGN!";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int par[105];
int getpar(int x) {
if (x == par[x]) return par[x];
par[x] = getpar(par[x]);
return par[x];
}
int main() {
int n, m;
cin >> n >> m;
for (int i = 1; i <= n; ++i) par[i] = i;
for (int i = 1; i <= m; ++i) {
int a, b;
cin >> a >> b;
int pa = getpar(a);
int pb = getpar(b);
par[max(pa, pb)] = min(pa, pb);
}
for (int i = 1; i <= n; ++i)
if (getpar(i) != 1) {
printf("NO\n");
exit(0);
}
if (n >= 3 && m == n)
printf("FHTAGN!\n");
else
printf("NO\n");
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int par[105];
int getpar(int x) {
if (x == par[x]) return par[x];
par[x] = getpar(par[x]);
return par[x];
}
int main() {
int n, m;
cin >> n >> m;
for (int i = 1; i <= n; ++i) par[i] = i;
for (int i = 1; i <= m; ++i) {
int a, b;
cin >> a >> b;
int pa = getpar(a);
int pb = getpar(b);
par[max(pa, pb)] = min(pa, pb);
}
for (int i = 1; i <= n; ++i)
if (getpar(i) != 1) {
printf("NO\n");
exit(0);
}
if (n >= 3 && m == n)
printf("FHTAGN!\n");
else
printf("NO\n");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m;
int parent[110];
int ranks[110];
int visit[110];
int cycle = 0;
int Findset(int i) {
int root = i;
while (root != parent[root]) root = parent[root];
while (i != root) {
int newp = parent[i];
parent[i] = root;
i = newp;
}
return root;
}
void Unionset(int x, int y) {
int xp = Findset(x);
int yp = Findset(y);
if (xp == yp)
cycle++;
else {
if (ranks[xp] < ranks[yp])
parent[xp] = yp;
else if (ranks[xp] > ranks[yp])
parent[yp] = xp;
else {
parent[yp] = xp;
ranks[xp]++;
}
}
}
int main() {
memset(visit, false, sizeof(visit));
for (int i = 0; i < 110; i++) {
parent[i] = i;
ranks[i] = -1;
}
scanf("%d%d", &n, &m);
for (int i = 0; i < m; i++) {
int x, y;
scanf("%d%d", &x, &y);
visit[x] = visit[y] = true;
Unionset(x, y);
}
for (int i = 1; i <= n; i++) {
if (!visit[i]) {
printf("NO");
return 0;
}
}
if (cycle == 1)
printf("FHTAGN!");
else
printf("NO");
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m;
int parent[110];
int ranks[110];
int visit[110];
int cycle = 0;
int Findset(int i) {
int root = i;
while (root != parent[root]) root = parent[root];
while (i != root) {
int newp = parent[i];
parent[i] = root;
i = newp;
}
return root;
}
void Unionset(int x, int y) {
int xp = Findset(x);
int yp = Findset(y);
if (xp == yp)
cycle++;
else {
if (ranks[xp] < ranks[yp])
parent[xp] = yp;
else if (ranks[xp] > ranks[yp])
parent[yp] = xp;
else {
parent[yp] = xp;
ranks[xp]++;
}
}
}
int main() {
memset(visit, false, sizeof(visit));
for (int i = 0; i < 110; i++) {
parent[i] = i;
ranks[i] = -1;
}
scanf("%d%d", &n, &m);
for (int i = 0; i < m; i++) {
int x, y;
scanf("%d%d", &x, &y);
visit[x] = visit[y] = true;
Unionset(x, y);
}
for (int i = 1; i <= n; i++) {
if (!visit[i]) {
printf("NO");
return 0;
}
}
if (cycle == 1)
printf("FHTAGN!");
else
printf("NO");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
vector<int> e[101];
bool used[101];
int cou, d[101];
void dfs(int x, int y, int last) {
d[y] = x;
cou++;
used[x] = 1;
for (int i = 0; i < e[x].size(); i++) {
if (e[x][i] == last) continue;
if (used[e[x][i]]) continue;
dfs(e[x][i], y + 1, x);
}
}
int main() {
int n, m, an = 1, x, y;
scanf("%d%d", &n, &m);
if (m != n) an = 0;
while (m--) {
scanf("%d%d", &x, &y);
e[x].push_back(y);
e[y].push_back(x);
}
if (an == 0)
puts("NO");
else {
dfs(1, 0, 0);
if (cou == n)
puts("FHTAGN!");
else
puts("NO");
}
}
| ### Prompt
In Cpp, your task is to solve the following problem:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
vector<int> e[101];
bool used[101];
int cou, d[101];
void dfs(int x, int y, int last) {
d[y] = x;
cou++;
used[x] = 1;
for (int i = 0; i < e[x].size(); i++) {
if (e[x][i] == last) continue;
if (used[e[x][i]]) continue;
dfs(e[x][i], y + 1, x);
}
}
int main() {
int n, m, an = 1, x, y;
scanf("%d%d", &n, &m);
if (m != n) an = 0;
while (m--) {
scanf("%d%d", &x, &y);
e[x].push_back(y);
e[y].push_back(x);
}
if (an == 0)
puts("NO");
else {
dfs(1, 0, 0);
if (cou == n)
puts("FHTAGN!");
else
puts("NO");
}
}
``` |
#include <bits/stdc++.h>
int N, M, lv[105], flg = 0, cnt = 0;
struct EDGE {
int s, f, nxt;
EDGE() {}
EDGE(int a, int b, int c) : s(a), f(b), nxt(c) {}
} edge[105 * 105];
int head[105], pE;
void addedge(int s, int f) {
edge[pE] = EDGE(s, f, head[s]);
head[s] = pE++;
edge[pE] = EDGE(f, s, head[f]);
head[f] = pE++;
}
void dfs(int dep, int x, int fa) {
int i;
if (lv[x]) {
if (dep >= lv[x] + 3) flg = 1;
return;
}
lv[x] = dep;
cnt++;
for (i = head[x]; i != -1; i = edge[i].nxt) {
int to = edge[i].f;
if (to == fa) continue;
dfs(dep + 1, to, x);
}
}
int main() {
int i, j;
scanf("%d%d", &N, &M);
memset(head, -1, sizeof(head));
pE = 0;
for (i = 0; i < M; i++) {
int s, f;
scanf("%d%d", &s, &f);
addedge(s, f);
}
if (N != M)
printf("NO\n");
else {
dfs(1, 1, -1);
if (flg && cnt == N)
printf("FHTAGN!\n");
else
printf("NO\n");
}
return 0;
}
| ### Prompt
Please create a solution in cpp to the following problem:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
int N, M, lv[105], flg = 0, cnt = 0;
struct EDGE {
int s, f, nxt;
EDGE() {}
EDGE(int a, int b, int c) : s(a), f(b), nxt(c) {}
} edge[105 * 105];
int head[105], pE;
void addedge(int s, int f) {
edge[pE] = EDGE(s, f, head[s]);
head[s] = pE++;
edge[pE] = EDGE(f, s, head[f]);
head[f] = pE++;
}
void dfs(int dep, int x, int fa) {
int i;
if (lv[x]) {
if (dep >= lv[x] + 3) flg = 1;
return;
}
lv[x] = dep;
cnt++;
for (i = head[x]; i != -1; i = edge[i].nxt) {
int to = edge[i].f;
if (to == fa) continue;
dfs(dep + 1, to, x);
}
}
int main() {
int i, j;
scanf("%d%d", &N, &M);
memset(head, -1, sizeof(head));
pE = 0;
for (i = 0; i < M; i++) {
int s, f;
scanf("%d%d", &s, &f);
addedge(s, f);
}
if (N != M)
printf("NO\n");
else {
dfs(1, 1, -1);
if (flg && cnt == N)
printf("FHTAGN!\n");
else
printf("NO\n");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int cnt, n, mx, p = 0, parent[101];
void Make_set(int v) { parent[v] = v; }
int Find_set(int v) {
if (parent[v] == v) return v;
return Find_set(parent[v]);
}
void Union_sets(int a, int b) {
a = Find_set(a);
b = Find_set(b);
if (a != b)
parent[b] = a;
else
cnt++;
}
int main() {
int m, q, w, l, f, tui = 0;
cin >> n >> m;
cnt = 0;
for (int k = 1; k <= n; k++) Make_set(k);
for (int k = 1; k <= m; k++) {
cin >> q >> w;
Union_sets(q, w);
}
if (cnt != 1) {
cout << "NO";
return 0;
}
cnt = 0;
for (int k = 1; k <= n; k++) {
if (parent[k] == k) cnt++;
}
if (cnt == 1)
cout << "FHTAGN!";
else
cout << "NO";
return 0;
}
| ### Prompt
Develop a solution in CPP to the problem described below:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int cnt, n, mx, p = 0, parent[101];
void Make_set(int v) { parent[v] = v; }
int Find_set(int v) {
if (parent[v] == v) return v;
return Find_set(parent[v]);
}
void Union_sets(int a, int b) {
a = Find_set(a);
b = Find_set(b);
if (a != b)
parent[b] = a;
else
cnt++;
}
int main() {
int m, q, w, l, f, tui = 0;
cin >> n >> m;
cnt = 0;
for (int k = 1; k <= n; k++) Make_set(k);
for (int k = 1; k <= m; k++) {
cin >> q >> w;
Union_sets(q, w);
}
if (cnt != 1) {
cout << "NO";
return 0;
}
cnt = 0;
for (int k = 1; k <= n; k++) {
if (parent[k] == k) cnt++;
}
if (cnt == 1)
cout << "FHTAGN!";
else
cout << "NO";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, a[100][100], vis[100], p[100];
void dfs(int x) {
vis[x] = 1;
for (int i = 0; i < n; i++)
if (a[x][i]) {
if (vis[i]) {
if (p[x] != i) m++;
} else {
p[i] = x;
dfs(i);
}
}
}
int main() {
int x, y;
cin >> n >> m;
memset(p, -1, sizeof(p));
for (int i = 0; i < m; i++) {
cin >> x >> y;
x--;
y--;
a[x][y] = a[y][x] = 1;
}
m = 0;
dfs(0);
y = 0;
for (int i = 0; i < n; i++)
if (!vis[i]) {
y = 1;
break;
}
if (m != 2 || y)
printf("NO");
else
printf("FHTAGN!");
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, a[100][100], vis[100], p[100];
void dfs(int x) {
vis[x] = 1;
for (int i = 0; i < n; i++)
if (a[x][i]) {
if (vis[i]) {
if (p[x] != i) m++;
} else {
p[i] = x;
dfs(i);
}
}
}
int main() {
int x, y;
cin >> n >> m;
memset(p, -1, sizeof(p));
for (int i = 0; i < m; i++) {
cin >> x >> y;
x--;
y--;
a[x][y] = a[y][x] = 1;
}
m = 0;
dfs(0);
y = 0;
for (int i = 0; i < n; i++)
if (!vis[i]) {
y = 1;
break;
}
if (m != 2 || y)
printf("NO");
else
printf("FHTAGN!");
}
``` |
#include <bits/stdc++.h>
using namespace std;
int table[105][105];
int used[105];
int N, M;
int cycle = 0;
int visit[105];
void DFS(int from, int s) {
int i, j, k;
for (i = 1; i <= N; i++) {
if (table[s][i] && i != from && !visit[i]) {
if (used[i]) {
cycle++;
} else {
used[i] = 1;
DFS(s, i);
used[i] = 0;
}
}
}
visit[s] = 1;
}
int main() {
int i, j, k;
int temp1, temp2;
cin >> N >> M;
for (i = 0; i < M; i++) {
cin >> temp1 >> temp2;
table[temp1][temp2] = table[temp2][temp1] = 1;
}
int COM = 0;
used[1] = 1;
DFS(0, 1);
for (i = 1; i <= N; i++) {
if (visit[i]) COM++;
}
if (COM != N || cycle != 1) {
cout << "NO";
} else {
cout << "FHTAGN!";
}
return 0;
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int table[105][105];
int used[105];
int N, M;
int cycle = 0;
int visit[105];
void DFS(int from, int s) {
int i, j, k;
for (i = 1; i <= N; i++) {
if (table[s][i] && i != from && !visit[i]) {
if (used[i]) {
cycle++;
} else {
used[i] = 1;
DFS(s, i);
used[i] = 0;
}
}
}
visit[s] = 1;
}
int main() {
int i, j, k;
int temp1, temp2;
cin >> N >> M;
for (i = 0; i < M; i++) {
cin >> temp1 >> temp2;
table[temp1][temp2] = table[temp2][temp1] = 1;
}
int COM = 0;
used[1] = 1;
DFS(0, 1);
for (i = 1; i <= N; i++) {
if (visit[i]) COM++;
}
if (COM != N || cycle != 1) {
cout << "NO";
} else {
cout << "FHTAGN!";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <class T1>
void deb(T1 e1) {
cout << e1 << endl;
}
template <class T1, class T2>
void deb(T1 e1, T2 e2) {
cout << e1 << " " << e2 << endl;
}
template <class T1, class T2, class T3>
void deb(T1 e1, T2 e2, T3 e3) {
cout << e1 << " " << e2 << " " << e3 << endl;
}
const int maxn = 5000;
vector<int> graph[maxn];
bool used[maxn];
int parent[maxn];
int nodes, edges, from, to, cc, cycle, flag = 0;
bool Hascycle;
void dfs(int u, int parent) {
used[u] = 1;
for (auto child : graph[u]) {
if (used[child] == 0) {
dfs(child, u);
} else if (child != parent) {
Hascycle = true;
return;
}
}
}
int Rank[maxn], p[maxn];
vector<int> setSize;
int numSets;
struct DisjointSet {
void make_set(int n) {
numSets = n;
setSize.assign(n + 1, 1);
for (int i = 1; i <= n; i++) {
Rank[i] = 0;
p[i] = i;
}
}
int findSet(int i) { return (p[i] == i) ? i : (p[i] = findSet(p[i])); }
bool isSameSet(int i, int j) { return findSet(i) == findSet(j); }
void unionSet(int i, int j) {
if (!isSameSet(i, j)) {
numSets--;
int x = findSet(i), y = findSet(j);
if (Rank[x] > Rank[y]) {
p[y] = x;
setSize[x] += setSize[y];
} else {
p[x] = y;
setSize[y] += setSize[x];
if (Rank[x] == Rank[y]) {
Rank[y]++;
}
}
}
}
int numofdisjointset() { return numSets; }
int sizeOfSet(int i) { return setSize[findSet(i)]; }
} UF;
int main() {
scanf("%d %d", &nodes, &edges);
UF.make_set(nodes);
for (int i = 0; i < edges; i++) {
scanf("%d %d", &from, &to);
UF.unionSet(from, to);
graph[from].push_back(to);
graph[to].push_back(from);
}
memset(used, 0, sizeof(used));
for (int i = 1; i <= nodes; i++) {
if (used[i] == 0) {
cc++;
dfs(i, -1);
}
}
if (nodes == edges and UF.numofdisjointset() == 1) {
cout << "FHTAGN!\n";
} else {
cout << "NO\n";
}
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
<image>
Input
The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ <image>).
Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Examples
Input
6 6
6 3
6 4
5 1
2 5
1 4
5 4
Output
FHTAGN!
Input
6 5
5 6
4 6
3 1
5 1
1 2
Output
NO
Note
Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., v - 1 and v, v and 1.
A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).
A rooted tree is a tree where one vertex is selected to be the root.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T1>
void deb(T1 e1) {
cout << e1 << endl;
}
template <class T1, class T2>
void deb(T1 e1, T2 e2) {
cout << e1 << " " << e2 << endl;
}
template <class T1, class T2, class T3>
void deb(T1 e1, T2 e2, T3 e3) {
cout << e1 << " " << e2 << " " << e3 << endl;
}
const int maxn = 5000;
vector<int> graph[maxn];
bool used[maxn];
int parent[maxn];
int nodes, edges, from, to, cc, cycle, flag = 0;
bool Hascycle;
void dfs(int u, int parent) {
used[u] = 1;
for (auto child : graph[u]) {
if (used[child] == 0) {
dfs(child, u);
} else if (child != parent) {
Hascycle = true;
return;
}
}
}
int Rank[maxn], p[maxn];
vector<int> setSize;
int numSets;
struct DisjointSet {
void make_set(int n) {
numSets = n;
setSize.assign(n + 1, 1);
for (int i = 1; i <= n; i++) {
Rank[i] = 0;
p[i] = i;
}
}
int findSet(int i) { return (p[i] == i) ? i : (p[i] = findSet(p[i])); }
bool isSameSet(int i, int j) { return findSet(i) == findSet(j); }
void unionSet(int i, int j) {
if (!isSameSet(i, j)) {
numSets--;
int x = findSet(i), y = findSet(j);
if (Rank[x] > Rank[y]) {
p[y] = x;
setSize[x] += setSize[y];
} else {
p[x] = y;
setSize[y] += setSize[x];
if (Rank[x] == Rank[y]) {
Rank[y]++;
}
}
}
}
int numofdisjointset() { return numSets; }
int sizeOfSet(int i) { return setSize[findSet(i)]; }
} UF;
int main() {
scanf("%d %d", &nodes, &edges);
UF.make_set(nodes);
for (int i = 0; i < edges; i++) {
scanf("%d %d", &from, &to);
UF.unionSet(from, to);
graph[from].push_back(to);
graph[to].push_back(from);
}
memset(used, 0, sizeof(used));
for (int i = 1; i <= nodes; i++) {
if (used[i] == 0) {
cc++;
dfs(i, -1);
}
}
if (nodes == edges and UF.numofdisjointset() == 1) {
cout << "FHTAGN!\n";
} else {
cout << "NO\n";
}
return 0;
}
``` |