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 <string.h>
#include <ctype.h>
#define max(a, b)	(((a) > (b)) ? (a) : (b))		/* ２個の値の最大値 */
#define min(a, b)	(((a) < (b)) ? (a) : (b))		/* ２個の値の最小値 */
#define ENTER		printf("\n")					/* 改行プリント */
int DBG = 1;										/* デバッグプリント 提出時は0 */
/* main *************************************************************/
int main()
{
	
	long long		a,b,c,i,j,k,n,m,x,y,ans = 0;
	
	scanf("%lld", &n);
	a = 1;
	if (n & a == 1) {
		printf("0\n");
		return 0;
	}
	
	//偶数//
	
	ans  = n / 10;
	ans += n / 100;
	ans += n / 1000;
	ans += n / 10000;
	ans += n / 100000;
	ans += n / 1000000;
	ans += n / 10000000;
	ans += n / 100000000;
	ans += n / 1000000000;
	ans += n / 10000000000ll;
	ans += n / 100000000000ll;
	ans += n / 1000000000000ll;
	ans += n / 10000000000000ll;
	ans += n / 100000000000000ll;
	ans += n / 1000000000000000ll;
	ans += n / 10000000000000000ll;
	ans += n / 100000000000000000ll;
	ans += n / 1000000000000000000ll;

	ans += n / 50;
	ans -= n / 200;
	/*
	ans -= n / 500;
	ans -= n / 5000;
	ans -= n / 50000;
	ans -= n / 500000;
	ans -= n / 5000000;
	ans -= n / 50000000;
	ans -= n / 500000000;
	ans -= n / 5000000000ll;
	ans -= n / 50000000000ll;
	ans -= n / 500000000000ll;
	ans -= n / 5000000000000ll;
	ans -= n / 50000000000000ll;
	ans -= n / 500000000000000ll;
	ans -= n / 5000000000000000ll;
	ans -= n / 50000000000000000ll;
	ans -= n / 500000000000000000ll;
	*/
	
	printf("%lld\n",ans);
	
	return 0;
}