CodeForces 681B

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=121183#problem/G

G - Economy Game

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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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

Input

1359257

Output

YES

Input

17851817

Output

NO

题意:给出一个游戏币数求买买房车电脑能不能恰好花完

不思考的暴力  超时

一个技巧减少一层循环

#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
int main(){
    ll n;
    while(cin>>n){

        int a=n/1234567;
        int b=n/123456;
        int c=n/1234;

        int flag=0;
        for(int i=0;i<=a+1;++i){
            for(int j=0;j<=b+1;++j){
                if(((ll)n-(ll)i*1234567-(ll)j*123456)>=0&&((ll)n-(ll)i*1234567-(ll)j*123456)%1234==0){flag=1;goto aaa;}//@1

            }
        }
        aaa:;
        if(!flag)cout<<"NO\n";
        else cout<<"YES\n";
    }
    return 0;
}

@1 如果此处再把c循环一遍一定超时