湖南中医药大学2017年集训队第四场选拔赛-Problem D: Jug Hard

Problem D: Jug Hard

Time Limit: 5 Sec Memory Limit: 128 MB
Submit: 192 Solved: 83
[Submit][Status][Web Board]
Description

You have two empty jugs and tap that may be used to fill a jug. When filling a jug from the tap, you can only fill it completely (i.e., you cannot partially fill it to a desired level, since there are no volume measurements on the jug).

You may empty either jug at any point.

You may transfer water between the jugs: if transferring water from a larger jug to a smaller jug, the smaller jug will be full and there will be water left behind in the larger jug.

Given the volumes of the two jugs, is it possible to have one jug with some specific volume of water?

Input

The first line contains T, the number of test cases (1 ≤ T 100 000). Each test case is composed of three integers: a b d where a and b (1 ≤ a, b ≤ 10 000 000) are the volumes of the two jugs, and d is the desired volume of water to be generated. You can assume that d ≤ max(a,b).
Output

For each of the T test cases, output either Yes or No, depending on whether the specific volume of water can be placed in one of the two jugs.
Sample Input

3
8 1 5
4 4 3
5 3 4
Sample Output

Yes
No
Yes

[分析]
这题曾经做过类似的。
数论。拓展欧几里得。（算法竞赛入门经典310-313）
设va为第一个桶的容量，vb为第二个桶的容量，v为所要求的容量。
所以存在x和y满足va*x+vb*y=v;
所有根据欧几里得得出va*x+vb*y=gcd（va,vb）;
因为题目不要求算出x和y，所有不用完全的拓展欧几里得。
只要连立上两式子可得v=gcd（va,vb）;
没完。
因为x和y存在多解。
所有最终得：
当v%gcd==0时，满足条件。
[代码]

#include<cstdio>

int gcd(int a, int b)
{
    return b ? gcd(b, a%b) : a;
}

int main()
{
    int va, vb, v;
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d%d%d", &va, &vb,&v);
        if (v%gcd(va, vb) == 0)printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}