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 <stdlib.h>
#include <math.h>
#include <string.h>

// long long int型の整数の桁数を返す。入力は19桁まで
long long int digit(long long int num) {
    long long int nine = 0;
    long long int cnt = 0;
    while (1) {
        nine = nine*10 + 9;
        cnt++;
        if (cnt == 20) return -1;
        if (num - nine < 0) return cnt;
    }
}

int main(void) {
    long long int n, ans=0, keta;
    scanf("%lld", &n);
    if (n%2==1) {
        ans=0;
    } else {
        keta = digit(n);
        long long int i, lower, upper;
        /*for (i = 1; i < 10; i++) {
            lower = (long long int)pow(10, keta-1)*i;
            upper = (long long int)pow(10, keta-1)*(i+1);
            if (lower <= n < upper) {
                break;
            }
        }*/
        long long int ten=1, five=10;
        /*while (1) {
            ten*=10;
            if (n < ten) {
                break;
            }
            ans += (n - n%ten)/ten;
        }*/
        while (1) {
            five*=5;
            if (n < five) {
                break;
            }
            ans += (n - n%five)/five;
        }
    }
    printf("%lld\n", ans);
    return 0;
}