Codeforces Round #357 (Div. 2) B. Economy Game 水题

B. Economy Game

题目连接：

http://www.codeforces.com/contest/681/problem/B

Description

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 (for 1 234 567 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 a, b and c such that a × 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).

Sample Input

1359257

Sample Output

YES

Hint

题意

给你n，让你判断是否存在非负整数解a,b,c

满足1234567a+123456b+1234c = n

题解：

暴力枚举a和b，然后o1判断c就好了

代码

#include<bits/stdc++.h>
using namespace std;

int main()
{
    long long n;
    cin>>n;
    long long num1 = 1234567;
    long long num2 = 123456;
    long long num3 = 1234;
    for(long long i=0;i<=100;i++)
        for(long long j=0;j<=10000;j++)
            if(n>=i*num1+j*num2)
            {
                long long p = n - i*num1-j*num2;
                if(p%num3==0)
                {
                    cout<<"YES"<<endl;
                    return 0;
                }
            }

    cout<<"NO"<<endl;
}