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: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll=long long;
#define rep(i,N) for(int i=0; i<N; i++){    ;
int main() {
ll N;
cin>>N;
ll h=N/10;
int count=0;
long long answer=0;
  if(N%2==0){
for(int i=0; i<39; i++){
if(h>4){
h=h/5;
count++;
}
else
{
    break;
}
}
for(int i=0; i<count+2; i++){
    long long Q=10;
    if(i!=0){
    for(int j=0; j<i; j++){
        Q=Q*5;
    }
    }
answer+=N/Q;
}
  }
  cout<<answer<<endl;
 }