Counting trailing 0s of n! It is not very hard to figure out how to count it - simply count how many 5s. Since even numbers are always more than 5s, we don't have to consider even numbers.

But counting 5, is tricky. Naive method is quite slow - I got 12+s for 1000000000. And after reading below article, I got 0.04s. The former counts individually, duplicated; but the latter counts smart:

#include 
      
       #include 
       
        using namespace std;int main() {    int cnt; cin >> cnt;        if(cnt == 0) return 0;        while(cnt --)    {        unsigned long long n; cin >> n;                unsigned cnt = 0;             for (int d = 5; d <= n; d *= 5) {            cnt += n / d;        }        cout << cnt << endl;    }    return 0;}