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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>
unsigned long long int func(unsigned long long int arg);
int main(){
    unsigned long long int n;
    scanf("%llu",&n);
    unsigned long long int ans;
    unsigned long long int fcount = 0;
    unsigned long long int count = -1;
    unsigned long long int fuga = 1;
    if(n % 2 != 0){
        printf("0");
        return 0;
    }
    while(1){
        if(n % 5 == 0){
            while(1){
                fuga *=5;
                if(n <= fuga){
                    break;
                }
                count++;
            }
            ans = n / 5;
            break;
        }else{
            n -= 2;
        }
    }
    if(ans % 2 == 0){
        ans /= 2;
    }else{
        ++ans;
        ans /= 2;
    }
    unsigned long long int atc = ans / 20;
    if(count != 0){   
        count = (atc - 1)  * 5;
    } 

    printf("%llu",ans + count);

    return 0;
}