2015浙江理工校赛G Jug Hard （数论orBFS）

题解

两个杯子倒水问题…一看10万个样例就犹豫了一下要不要BFS做，其实直接判断一下d能不能除尽a和b的gcd就好了..当然BFS也能做，只是慢了一点

代码

#include<cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define MAX 100000
#define LL long long
int cas=1,T;
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
int main()
{
    scanf("%d",&T);
    while (T--)
    {
       int a,b,d;
       scanf("%d%d%d",&a,&b,&d);
       if (b==d || a == d)
       {
           printf("Yes\n");
           continue;
       }
       if (d%gcd(a,b))
           printf("No\n");
       else
           printf("Yes\n");
    }

    return 0;
}

题目

Jug Hard

Time Limit: 10 Sec Memory Limit: 128 MB

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