#include <queue>
#include <math.h>
#include <cmath>
#include <stack>
#include <stdio.h>
#include <iostream>
#include <vector>
#include <iomanip>
#include <string.h>
#include <algorithm>
#include <cstring>
#include <bits/stdc++.h>
#include <limits>
using namespace std;
long long n,wu;
int main(){
	cin>>n;
	for(long long i=5;i<=n;i*=5){
		wu+=n/i;
	}
	cout<<wu;
	return 0;
}

#include<bits/stdc++.h>
#define int long long
using namespace std;
	int n,sum;
signed main(){
	cin>>n;
	while(n!=0){
		n=n/5;
		sum=sum+n;
	}
	cout<<sum;
}

#include<iostream>
#include<sstream>
#include<iomanip>
#include<stdio.h>
#include<math.h>
#include<string>
#include<string.h>
using namespace std;
int main(){
	long long n,sum=0;
	cin>>n;
	while(n!=0){
		sum+=n/5;
		n/=5;
	}
	cout<<sum;
	return 0;
}

简单奥数，求其中因数5的个数就行（2*5=10，但2的数量一定比5多）

#include<iostream>
#include<sstream>
#include<iomanip>
#include<stdio.h>
#include<math.h>
#include<string>
#include<string.h>
using namespace std;
int main(){
	long long n,sum=0;
	cin>>n;
	while(n!=0){
		sum+=n/5;
		n/=5;
	}
	cout<<sum;
	return 0;
}