For this Question: Given an integer N, return the number of trailing zeroes in N!. 
Where N!=N*(N-1)*(N-2)....1Input contains a single integer N.

Constraints:-
1<= N <=1000000000000000Print the number of zeros in factorial(N)Sample Input
5

Sample Output
1

Explanation:-
5! = 120
number of zeroes = 1

Sample Input;
11

Sample output:
2, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;


long long  fd(long long  n)
{
    // Initialize result
    long long  count = 0;

    // Keep dividing n by powers of
    // 5 and update count
    for (long long  i = 5; n / i >= 1; i *= 5)
        count += n / i;

    return count;
}

// Driver code
int main()
{
    int t;
    t=1;
    while(t--){
 long long  n;
 cin>>n;
 cout<<fd(n)<<endl;
 }
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not?