#include <bits/stdc++.h>
const int K = 1000, N = 250;
inline long long sqr(int x) { return (long long)x * x; }
long long solve(int a[], int b[]) {
  long long res = 0;
  for (int i = 0; i < 8 * N; ++i) res += sqr(a[i] - b[i]);
  return res;
}
int a[K + 1][8 * N], b[K + 1][8 * N], c[8 * N];
void init() {
  std::random_device rd;
  std::mt19937 gen(rd());
  for (int i = 1; i <= K; ++i) {
    std::poisson_distribution<> d(i);
    for (int j = 0; j < 8 * N; ++j) a[i][j] = d(gen);
    std::uniform_int_distribution<> u(0, 2 * i);
    for (int j = 0; j < 8 * N; ++j) b[i][j] = u(gen);
    std::sort(a[i], a[i] + 8 * N);
    std::sort(b[i], b[i] + 8 * N);
  }
}
int main() {
  init();
  int tcase;
  for (scanf("%d", &tcase); tcase--;) {
    static int temp[N];
    for (int i = 0; i < N; ++i) scanf("%d", &temp[i]);
    for (int i = 0; i < 8 * N; ++i) c[i] = temp[i / 8];
    std::sort(c, c + 8 * N);
    int avg = std::accumulate(c, c + 8 * N, 0) / 8. / N + .5;
    int l = std::max(avg - 50, 10), r = std::min(avg + 50, 1000);
    std::pair<long long, int> p, q;
    p.first = q.first = 1LL << 62;
    for (int i = l; i <= r; ++i) p = std::min(p, {solve(a[i], c), i});
    for (int i = l; i <= r; ++i) q = std::min(q, {solve(b[i], c), i});
    int ans;
    if (p < q) {
      ans = p.second;
    } else {
      ans = (q.second + avg + (c[8 * N - 1] + c[0]) / 2.) / 3. + .5;
      ans = std::max(ans, (c[8 * N - 1] + 1) / 2);
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

Good job! Now that Heidi is able to distinguish between Poisson and uniform distributions, she is in a good position to actually estimate the populations.

Can you help Heidi estimate each village's population?

Input

Same as the easy version.

Output

Output one line per village, in the same order as provided in the input, containing your (integer) population estimate.

Your answer is considered correct if it is an integer that falls into the interval <image>, where P is the real population of the village, used to create the distribution (either Poisson or uniform) from which the marmots drew their answers.

### Response

```cpp
#include <bits/stdc++.h>
const int K = 1000, N = 250;
inline long long sqr(int x) { return (long long)x * x; }
long long solve(int a[], int b[]) {
  long long res = 0;
  for (int i = 0; i < 8 * N; ++i) res += sqr(a[i] - b[i]);
  return res;
}
int a[K + 1][8 * N], b[K + 1][8 * N], c[8 * N];
void init() {
  std::random_device rd;
  std::mt19937 gen(rd());
  for (int i = 1; i <= K; ++i) {
    std::poisson_distribution<> d(i);
    for (int j = 0; j < 8 * N; ++j) a[i][j] = d(gen);
    std::uniform_int_distribution<> u(0, 2 * i);
    for (int j = 0; j < 8 * N; ++j) b[i][j] = u(gen);
    std::sort(a[i], a[i] + 8 * N);
    std::sort(b[i], b[i] + 8 * N);
  }
}
int main() {
  init();
  int tcase;
  for (scanf("%d", &tcase); tcase--;) {
    static int temp[N];
    for (int i = 0; i < N; ++i) scanf("%d", &temp[i]);
    for (int i = 0; i < 8 * N; ++i) c[i] = temp[i / 8];
    std::sort(c, c + 8 * N);
    int avg = std::accumulate(c, c + 8 * N, 0) / 8. / N + .5;
    int l = std::max(avg - 50, 10), r = std::min(avg + 50, 1000);
    std::pair<long long, int> p, q;
    p.first = q.first = 1LL << 62;
    for (int i = l; i <= r; ++i) p = std::min(p, {solve(a[i], c), i});
    for (int i = l; i <= r; ++i) q = std::min(q, {solve(b[i], c), i});
    int ans;
    if (p < q) {
      ans = p.second;
    } else {
      ans = (q.second + avg + (c[8 * N - 1] + c[0]) / 2.) / 3. + .5;
      ans = std::max(ans, (c[8 * N - 1] + 1) / 2);
    }
    printf("%d\n", ans);
  }
  return 0;
}

```