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

B. Economy Game

1 second

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).

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, spending 1 234 567 + 123 456 + 1234 = 1 359 257 game-coins in total.

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题目大意：

给你一个数，问你能否仅由1234567，123456，1234组成

题目分析：

我是暴力莽一波，然后剪枝，剪当然是剪n大的啦。

#include<bits/stdc++.h>using namespace std;const int maxn=1e9+7;int flag;int main(){    int n;    scanf("%d",&n);        flag=0;        int n1=n/1234567;        int n2=n/123456;        for(int i=0;i<=n1;i++)            for(int j=0;j<=n2;j++)                {                    int tmp=n-1234567*i-123456*j;                    if(tmp==0){flag=1;break;}                    else                        if(tmp<1234)break;                    if((n-1234567*i-123456*j)%1234==0){flag=1;break;}                }        if(flag)puts("YES");        else puts("NO");}