Submission #1175627

---

Source Code Expand

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <cmath>

using namespace std;
 
#define FOR(i,a,b) for (int i = (a); i < (b); i++)
#define REP(i,n) FOR(i,0,n)

//最大公約数
long gcd(long a, long b){
	if (a%b==0){
		return b;
	}
	else{
		return gcd(b,a%b);
	}
}

//最小公倍数
long lcm(long a, long b){
	return (a*b) / gcd(a,b);
}

typedef long long ll;

ll a[5];

int main(){
	ll n;
	
	cin >> n;
	
	a[0] = 25;
	
	FOR(i,1,5){
		a[i] = a[i-1]*100 + 25;
	}
	
	int ans = 0;
	
	REP(i,n){
		REP(j,5){
			if (a[j] > i) break;
			if (i%a[j] == 0){
				ans++;
				break;
			}
		}
	}
	
	cout << ans << endl;
	
	return 0;
}

Submission Info

Judge Result