Counting ones

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

const int N = 2e5;
int T[N];

int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> T[i];
  }
  int a = 0, b = 0;
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < i; ++j) {
      int pc = __builtin_popcountll(ll(T[i])*T[j]);
      if (pc > ans) {
        ans = pc;
        a = T[j]; b = T[i];
      }
    }
  }
  cout << ans << '\n' << a << ' ' << b << '\n';
}

Test 1

Verdict: ACCEPTED

Test 2

Verdict: ACCEPTED

Test 3

Verdict: ACCEPTED

Test 4

Verdict: ACCEPTED

Test 5

Verdict: ACCEPTED

Test 6

Verdict: ACCEPTED

Test 7

Verdict: ACCEPTED

Test 8

Verdict: ACCEPTED

Test 9

Verdict: ACCEPTED

Test 10

Verdict: ACCEPTED