solution

#include <bits/stdc++.h>

using namespace std;

int isAnagram(string a, string b){
  int x[26] = {};
  int y[26] = {};
    for (int i=0; i<a.length(); i++){
        x[a[i]-97]++;
    }
    for (int i=0; i<b.length(); i++){
        y[b[i]-97]++;
    }
    for (int i=0; i<26; i++){
        if (x[i] != y[i]) return 0;
    }
    return 1;
}

// Complete the sherlockAndAnagrams function below.
int sherlockAndAnagrams(string s) {
    int n = s.length();
    vector<string> all;
    for (int len = 1; len <= n; len++) {
      for (int i = 0; i <= n - len; i++) {
        int j = i + len - 1;
        string v = "";
        for (int k = i; k <= j; k++){
            v += s[k];
        }
        all.push_back(v);
      }
    }

int answer = 0;
    for (int i=0; i<all.size(); i++){
        for (int j=0; j<all.size(); j++){
            if (i!= j && all[i].length() == all[j].length()){
                int a = isAnagram(all[i], all[j]);
                answer+= a;
            }
        }
    }
    return answer/2;
}

int main()
{
    ofstream fout(getenv("OUTPUT_PATH"));

    int q;
    cin >> q;
    cin.ignore(numeric_limits<streamsize>::max(), '\n');

    for (int q_itr = 0; q_itr < q; q_itr++) {
        string s;
        getline(cin, s);

        int result = sherlockAndAnagrams(s);

        fout << result << "\n";
    }

    fout.close();

    return 0;
}