CodeForces - 681B Economy Game

本文介绍了一个游戏经济模拟问题,玩家需要判断初始游戏币是否能完全用于购买特定价格的房子、汽车和电脑。通过数学方法验证消费可能性,并提供了一个简单的算法实现。

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B. Economy Game
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Kolya is developing an economy simulator game. His most favourite part of the development process is in-game testing. Once he was entertained by the testing so much, that he found out his game-coin score become equal to 0.

Kolya remembers that at the beginning of the game his game-coin score was equal to n and that he have bought only some houses (for1 234 567 game-coins each), cars (for 123 456 game-coins each) and computers (for 1 234 game-coins each).

Kolya is now interested, whether he could have spent all of his initial n game-coins buying only houses, cars and computers or there is a bug in the game. Formally, is there a triple of non-negative integers ab and c such thata × 1 234 567 + b × 123 456 + c × 1 234 = n?

Please help Kolya answer this question.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 109) — Kolya's initial game-coin score.

Output

Print "YES" (without quotes) if it's possible that Kolya spent all of his initial n coins buying only houses, cars and computers. Otherwise print "NO" (without quotes).

Examples
input
1359257
output
YES
input
17851817
output
NO
Note

In the first sample, one of the possible solutions is to buy one house, one car and one computer, spending1 234 567 + 123 456 + 1234 = 1 359 257 game-coins in total.


找是否有a,b,c满足a × 1 234 567 + b × 123 456 + c × 1 234 = n



#include <iostream>
#include <cstdio>
#include <algorithm>
#define ll long long
using namespace std;
int main()
{
	int n;
	cin>>n;
	int n1,n2,f=0;
	n1=n/1234567;
	n2=n/123456;
	for(int i=0;i<=n1;i++)
	{
		for(int j=0;j<=n2;j++)
		{
			int d=n-i*1234567-j*123456;
			if(d>=0&&d%1234==0)
			{
				f=1;
				break;
			}
		}
		if(f)
		break;
	}
	if(f)
	printf("YES\n");
	else
	printf("NO\n");
}


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