Problem:
Score : 500 points

Problem Statement

For an integer n not less than 0, let us define f(n) as follows:

f(n) = 1 (if n < 2)

f(n) = n f(n-2) (if n \geq 2)

Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).

Constraints

0 \leq N \leq 10^{18}

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of trailing zeros in the decimal notation of f(N).

Sample Input 1

12

Sample Output 1

1

f(12) = 12 × 10 × 8 × 6 × 4 × 2 = 46080, which has one trailing zero.

Sample Input 2

5

Sample Output 2

0

f(5) = 5 × 3 × 1 = 15, which has no trailing zeros.

Sample Input 3

1000000000000000000

Sample Output 3

124999999999999995

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

 ll  GetDigit(ll num)
{
    return log10(num) + 1;
}

int main(){
    ll n;
    cin >> n;
    if(n%2 == 1){
        cout << "0";
        return 0;
    }
    ll ans = 0, cnt = 0;

    double dig = GetDigit(n);
    ll v = 10;
  
   while(v<=n)
   {
       cnt = floor(n / v);
       //long double p = powl(v, dig - 1);
       //ll k = floor(n / (ll)p);
       //if (k != 0)
           //ans += cnt * k;
        //else
            ans += cnt;

       v *= 5;
    }
    
    cout << ans;
}