number of divisors (O(sqrt(n) time complex)

#include <iostream>
using namespace std;
int main() {
    int n;
    cin >> n;
    for (int q = 0; q < n; q++) {
        int x;
        int div_num = 0;
        cin >> x;
        for (int i = 1; i * i <= x; i++) {
            if (x % i == 0) {
                div_num += i * i == x ? 1 : 2;
            }
        }
        cout << div_num << '\n';
    }
}