Problem p03618 (C++)

Buggy submission (status: Time Limit Exceeded):
//  Created by conan1024hao in 2019.
//  Copyright © 2019 conan1024hao. All rights reserved.
//  専用ライブラリです、自由にコピーして構いません。
//  感谢看我的代码！Wechat:conan1024hao QQ:810396815
#pragma GCC optimize("O3")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <iomanip>
#include <iostream>
#include <istream>
#include <iterator>
#include <list>
#include <map>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>
#define INF 10e16
#define MOD 1000000007
#define rep(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define mmax(x, y) (x > y ? x : y)
#define mmin(x, y) (x < y ? x : y)
#define chmax(x, y) x = mmax(x, y)
#define chmin(x, y) x = mmin(x, y)
#define all(x) (x).begin(), (x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
using namespace std;
// using Int=int_fast64_t;
typedef long long int ll;
typedef pair<ll, ll> Pll;
typedef pair<int, int> Pin;
long long GCD(long long a, long long b) { return b ? GCD(b, a % b) : a; }
long long LCM(long long a, long long b) { return a / GCD(a, b) * b; }
int dx[8] = {-1, 0, 1, 0};
int dy[8] = {0, -1, 0, 1};
char dir[4] = {'u', 'l', 'd', 'r'};
ll cmp1(pair<ll, string> a, pair<ll, string> b) {
  if (a.fi != b.fi)
    return a.fi < b.fi;
  else
    return a.se < b.se;
}
//--------------------------------------------------------------------------

//---------------------------------------------------------------------------
int main() { // 問題をちゃんと見ろ!!!!!!!!!!!!!!!!!      llか??????????
             // memset()!!!!!!!!!!!!
  cin.tie(0);
  ios::sync_with_stdio(false);
  //-------------------------------
  string s;
  cin >> s;
  ll n = s.length();
  ll ans = n * (n - 1) / 2;
  for (int i = 0; i < n - 1; i++) {
    for (int j = i + 1; j < n; j++) {
      if (s[i] == s[j])
        ans--;
    }
  }
  cout << ans + 1 << endl;
  //-------------------------------
  return 0;
}
//---------------------------------------------------------------------------


Fixed submission:
//  Created by conan1024hao in 2019.
//  Copyright © 2019 conan1024hao. All rights reserved.
//  専用ライブラリです、自由にコピーして構いません。
//  感谢看我的代码！Wechat:conan1024hao QQ:810396815
#pragma GCC optimize("O3")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <iomanip>
#include <iostream>
#include <istream>
#include <iterator>
#include <list>
#include <map>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>
#define INF 10e16
#define MOD 1000000007
#define rep(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define mmax(x, y) (x > y ? x : y)
#define mmin(x, y) (x < y ? x : y)
#define chmax(x, y) x = mmax(x, y)
#define chmin(x, y) x = mmin(x, y)
#define all(x) (x).begin(), (x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
using namespace std;
// using Int=int_fast64_t;
typedef long long int ll;
typedef pair<ll, ll> Pll;
typedef pair<int, int> Pin;
long long GCD(long long a, long long b) { return b ? GCD(b, a % b) : a; }
long long LCM(long long a, long long b) { return a / GCD(a, b) * b; }
int dx[8] = {-1, 0, 1, 0};
int dy[8] = {0, -1, 0, 1};
char dir[4] = {'u', 'l', 'd', 'r'};
ll cmp1(pair<ll, string> a, pair<ll, string> b) {
  if (a.fi != b.fi)
    return a.fi < b.fi;
  else
    return a.se < b.se;
}
//--------------------------------------------------------------------------

//---------------------------------------------------------------------------
int main() { // 問題をちゃんと見ろ!!!!!!!!!!!!!!!!!      llか??????????
             // memset()!!!!!!!!!!!!
  cin.tie(0);
  ios::sync_with_stdio(false);
  //-------------------------------
  string s;
  cin >> s;
  ll n = s.length();
  ll ans = n * (n - 1) / 2;
  map<char, ll> m;
  for (int i = 0; i < n; i++) {
    m[s[i]]++;
  }
  for (char i = 'a'; i <= 'z'; i++) {
    ans -= m[i] * (m[i] - 1) / 2;
  }
  cout << ans + 1 << endl;
  //-------------------------------
  return 0;
}
//---------------------------------------------------------------------------