HackerRank "Square Subsequences" !!!

Firt thought: an variation to LCS problem – but this one has many tricky detail.

I learnt the solution from this link:
https://github.com/wangyongliang/Algorithm/blob/master/hackerrank/Strings/Square%20Subsequences/main.cc

And here is his code with my comments..

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;long long lcs(string &a, string &b)
{
    int sizea = a.length();
    int sizeb = b.length();
    vector<vector<long long>> dp(sizea, vector<long long>(sizeb));    //  We only consider prefixes with a[0] matched
    //  to avoid duplicate counting
    for(int i = ; i < sizeb; i ++)
    {
        if(a[] == b[i]) dp[][i] = ;
        if(i) dp[][i] += dp[][i - ]; // we count accumulated
        dp[][i] %= ;
    }    for(int i = ; i < sizea; i ++)
    {
        dp[i][]  = dp[i - ][]; // all from dp[0][0]; part of init
        for(int j = ; j < sizeb; j ++)
        {
            dp[i][j] = dp[i - ][j] + dp[i][j - ]; // accumulated version of LCS.
            if(a[i] != b[j]) dp[i][j] -= dp[i-][j-]; // TODO: need 2nd thought on why
            dp[i][j] %= ;
        }
    }
    return dp.back().back();
}int main()
{
    int t; cin >> t;
    string str;
    while(t--)
    {
        cin >> str;
        long long ans = ;
        for(int i = ; i < str.length(); i ++)
        {
            string sb = str.substr(i, str.length()- i);
            string sa = str.substr(, i);
            ans += lcs(sb, sa);
            ans %= ;
        }
        cout << ans << endl;
    }    return ;
}