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;

#define ll long long

int bin_pow(int a, int pow) {
    if (pow == 0) return 1;
    if (pow == 1) return a;
    if (pow % 2 == 0) {
        int tmp = bin_pow(a, pow / 2);
        return tmp * tmp;
    }
    return a * bin_pow(a, pow - 1);
}

int fact(int x) {
    int ans = 1;

    for (int i = 1; i <= x; i++) ans *= i;
    return ans;
}

int count(vector<int> a) {
    int ans = 0;
    unordered_set<int> was;
    for (int i = 0; i < a.size(); i++) {
        int count = a[i] - 1;
        for (int j = 1; j < a[i]; j++) {
            count -= was.count(j);
        }
        ans += count * fact(a.size() - i - 1);
        was.insert(a[i]);
    }

    return ans;
}

ll m = 1000000007;

bool isPrime(int x) {
    for (int i = 2; i * i <= x; i++) {
        if (x % i == 0) return false;
    }
    return true;
}

int main() {
    cin.tie();
    ios_base::sync_with_stdio(false);

    ll n;
    cin >> n;

    if (n % 2 == 1) {
        cout << 0;
        return 0;
    }

    ll ans = 0;

    for (ll i = 5; i <= n; i *= 5) {
        ans += n / i / 2;
    }

    cout << ans;
    return 0;
}