<p>You are <strong>building</strong> a string <code>s</code> of length <code>n</code> <strong>one</strong> character at a time, <strong>prepending</strong> each new character to the <strong>front</strong> of the string. The strings are labeled from <code>1</code> to <code>n</code>, where the string with length <code>i</code> is labeled <code>s<sub>i</sub></code>.</p>

<ul>
	<li>For example, for <code>s = &quot;abaca&quot;</code>, <code>s<sub>1</sub> == &quot;a&quot;</code>, <code>s<sub>2</sub> == &quot;ca&quot;</code>, <code>s<sub>3</sub> == &quot;aca&quot;</code>, etc.</li>
</ul>

<p>The <strong>score</strong> of <code>s<sub>i</sub></code> is the length of the <strong>longest common prefix</strong> between <code>s<sub>i</sub></code> and <code>s<sub>n</sub></code> (Note that <code>s == s<sub>n</sub></code>).</p>

<p>Given the final string <code>s</code>, return<em> the <strong>sum</strong> of the <strong>score</strong> of every </em><code>s<sub>i</sub></code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;babab&quot;
<strong>Output:</strong> 9
<strong>Explanation:</strong>
For s<sub>1</sub> == &quot;b&quot;, the longest common prefix is &quot;b&quot; which has a score of 1.
For s<sub>2</sub> == &quot;ab&quot;, there is no common prefix so the score is 0.
For s<sub>3</sub> == &quot;bab&quot;, the longest common prefix is &quot;bab&quot; which has a score of 3.
For s<sub>4</sub> == &quot;abab&quot;, there is no common prefix so the score is 0.
For s<sub>5</sub> == &quot;babab&quot;, the longest common prefix is &quot;babab&quot; which has a score of 5.
The sum of the scores is 1 + 0 + 3 + 0 + 5 = 9, so we return 9.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;azbazbzaz&quot;
<strong>Output:</strong> 14
<strong>Explanation:</strong> 
For s<sub>2</sub> == &quot;az&quot;, the longest common prefix is &quot;az&quot; which has a score of 2.
For s<sub>6</sub> == &quot;azbzaz&quot;, the longest common prefix is &quot;azb&quot; which has a score of 3.
For s<sub>9</sub> == &quot;azbazbzaz&quot;, the longest common prefix is &quot;azbazbzaz&quot; which has a score of 9.
For all other s<sub>i</sub>, the score is 0.
The sum of the scores is 2 + 3 + 9 = 14, so we return 14.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

#define ll long long
#define pll pair<ll,ll>

struct hasher {
    ll n, _k, _p;
    vector<ll> base, kpow;
    
    void init(string s, ll k, ll p) {
        _p = p;
        n = s.length();
        base.resize(n+1);
        kpow.resize(n+1);
        
        base[0] = 0;
        kpow[0] = 1;
        for(ll i=0; i<n; i++) {
            ll num = (s[i]-'a'+1) % p;
            base[i+1] = (base[i]*k + num) % p;
            kpow[i+1] = (kpow[i]*k) % p;
        }
    }
    
    ll getHash(ll l, ll r) {
        ll ans = (base[r+1] - base[l]*kpow[(r-l+1)]) % _p;
        return (ans + _p) % _p;
    }
};

struct double_hash {
    hasher a, b;
    void init(string s) {
        a.init(s, 31, 1000000021);
        b.init(s, 37, 1000000009);
    }
    
    pair<ll,ll> getHash(ll l, ll r) {
        return {a.getHash(l, r), b.getHash(l, r)};
    }
};
double_hash myHash;

class Solution {
public:
    ll sumScores(string s) {
        ll n = s.length();
        myHash.init(s);
        
        ll sum = n;
        for(ll i=1; i<n; i++) {
            ll lo = 0, hi = n - i, cnt = 0;
            while(lo <= hi) {
                ll mid = (lo+hi)/2;
                if(myHash.getHash(0, mid - 1) == myHash.getHash(i, i + mid - 1)) {
                    cnt = mid;  
                    lo = mid + 1;  
                } 
                else {
                    hi = mid - 1; 
                }
            }
            sum += cnt;
        }
        return sum;
    }
};












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