三个for循环变为两个

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.

题目大意：a物品价格1234567    b物品价格123456    c价格 1234   问 n元钱随意组合三种物品是否正好能全部花完

解题思路：这题用三个for循环，简单粗暴省事，但是一定超时，所以要优化为2个for循环。那就一个for循环控制a物品的数量，一个for循环控制b物品的数量，剩下的都卖c物品就行了

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <list>
#define LL long long
#define INF 0x3f3f3f3f
#define mem(a, b) memset(a, b, sizeof(a))
#define PI 3.1415926
using namespace std;
const int N = 199999;
int main() {
    int n;
    while(scanf("%d", &n)!= EOF)
    {
        int t = 0;
        if(n % 1234567 == 0 || n % 123456 == 0 || n % 1234 == 0)
        {
            cout << "YES" << endl;
            break;
        }
        for(int i = 0; i < 100; i++)
        {
            int q = n - 1234567 * i;
            if(q > 0)     //这里一定要判断q是否大于0，否则负数也能可能除尽
            {
                for(int j = 0; j < 1000; j++)
                {
                    int w = q - 123456 * j;
                    if(w >= 0)
                    {
                    if(w == 0 || w % 1234 == 0)
                    {
                        cout << "YES" << endl;
                        t = 1;
                        break;
                    }
                    }
                }
            }
            if(t)
                break;
        }
        if(t == 0)
            cout << "NO" << endl;
    }
    return 0;
}