Jug Hard
Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 1172 Solved: 180
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
an+bn=d;
证明等式的成立，就是扩展欧几里得的证明
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
int GCD(int a,int b)
{
return b==0?a:GCD(b,a%b);
}
int main()
{
int T;
int n ,m,d;
scanf("%d",&T);
while(T--)
{
scanf("%d %d %d",&n,&m,&d);
printf("%s
",d%GCD(n,m)==0?"Yes":"No");
}
return 0;
}