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

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 (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.

就是一个很简单的枚举数学题，当m小于0要break是因为可能负数取模等于0，要防止这种情况

#include
using namespace std;

typedef __int64 ll;
const ll num1=1234567;
const ll num2=123456;
const ll num3=1234;
const ll m1=811;
const ll m2=8101;

int main(){
    ll n,m;
    scanf("%I64d",&n);
    bool ans=true,go=true;
    for(int i=0;ans&&i<=m1;++i){
        for(int j=0;ans&&j<=m2;++j){
            m=n-(i*num1+j*num2);
            if(m<0){
                break;
            }
            if(m%num3==0){
                ans=false;
            }
        }
    }
    puts(ans?"NO":"YES");
    return 0;
}