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>
#include <iostream>
#include <string>
#include <vector>
#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <math.h>

#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define repP(i, a, b) for(int i = a; i < b; i++)
#define repM(i, a, b) for(int i = a - 1; i >= b; i--)

using namespace std;

int main()
{
  long a = 0 , b = 0, c = 0, d = 0, e = 0, f = 0, i = 0, j = 0;
  long sum = 0, num = 5, count = 0, ans = 0, min = 0, max = 0;
  double d1 = 0, d2 = 0, d3 = 0, d4 = 0, d5 = 0, d6 = 0;
  int a1[999999], a2[999999], a3[999999];
  //char s1[200000], s2[200000], s3[200000];
  //string s1, s2, s3;
  cin >> a;
  
  if(a==0){
  cout<<1<<endl;
  }else if(a%2==1){
  cout<<0<<endl;
  }else if(a%2==0){
  num = 10;
  while(num<a){
  b+=a/num;
  num*=5;
  }
  cout<<b<<endl;
  }else{
  num = 5;
  while(num < a){
    b+= a/num;
    num*=5;
  }
  cout<<b<<endl;
  }
}