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<iostream>
#include<cstdio>
#include<stack>
#include<queue>
#include<algorithm>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<math.h>
using namespace std; 
#define sim template < class c
#define ris return * this
#define dor > debug & operator <<
#define eni(x) sim > typename \
  enable_if<sizeof dud<c>(0) x 1, debug&>::type operator<<(c i) {
sim > struct rge { c b, e; };
sim > rge<c> range(c i, c j) { return rge<c>{i, j}; }
sim > auto dud(c* x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct debug {
#ifdef LOCAL
~debug() { cerr << endl; }
eni(!=) cerr << boolalpha << i; ris; }
eni(==) ris << range(begin(i), end(i)); }
sim, class b dor(pair < b, c > d) {
  ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d) {
  *this << "[";
  for (auto it = d.b; it != d.e; ++it)
    *this << ", " + 2 * (it == d.b) << *it;
  ris << "]";
}
#else
sim dor(const c&) { ris; }
#endif
};
#define pp(...) "\033[94m [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "
#define dg1(x) debug() << pp(x)
#define dg2(x,y) debug() << pp(x) pp(y)
#define dg3(x,y,z) debug() << pp(x) pp(y) pp(z)
#define dg4(w,x,y,z) debug() << pp(w) pp(x) pp(y) pp(z)
#define msg(x) debug() << x
#define int long long
const int xx = 2e5;
const int MOD = 1e9 + 7;
int lcm(int a, int b){
  return a*(b/__gcd(a,b));
}
int findTrailingZeros(int n) 
{ 
    int count = 0; 
  
    for (int i = 5; n / i >= 1; i *= 5) 
        count += n / i; 
  
    return count; 
}

signed main(){

  ios::sync_with_stdio(false);cin.tie(0);

  int n;
  cin >> n;
  int ans ;
  n = n/2;
  if(n%2 == 0){
    ans = findTrailingZeros(2*n - n);// - findTrailingZeros(n);
  }
  else
  {
   ans = 0;

  }
  cout << ans << '\n';
  
}