Compute the number of trailing zeros in N-factorial

Solution

#include <iostream>

using namespace std;

int main(int argc, char* argv[])

{

int n = 100;

int m = n;

int num;

num = 0;

while( n / 5 > 0){

num += n / 5;

n /= 5;

}

cout << "factoral (" << m << ")" << " has " << num << " zero trailings. " << endl;

return 0;

}