the number of 0s at the end

Input a positive integer n, how many 0s are at the end of n! (ie factorial)? For example: n = 10; n! = 3628800, so the answer is 2 

Enter description:

Input is one line, n(1 ≤ n ≤ 1000)

Output description:

Output an integer, that is, what the question requires

Example 1

enter

10

output

2

#include <iostream>
using namespace std;

intmain()
{
    int n, k = 0;
    cin >> n;
    
    while(n)
    {
      k = n / 5 + k;
      n = n / 5;
    }
    
    cout << k;
    return 0;
}

Explanation: For example, 100/5=20, that is, there are 20 numbers that contain factor 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85,90,95,100. Among these 20 numbers, there are 20/5=4 containing factors 5: 25, 50, 75, 100 (the first 20 numbers can be regarded as 5*1, 5*2, 5*3, 5*4, 5* 5, 5*6, 5*7, 5*8, 5*9, 5*10, 5*11, 5,12, 5*13, 5*14, 5*15, 5*16, 5*17 , 5*18, 5*19, 5*20), and then the factor 5 in these 4 numbers has been calculated again.