GDUT_寒假训练题解报告_专题I_L题个人题解报告_you are given two integers a a and b b. you may pe-优快云博客

本报告解析GDUT寒假训练题中的一道数学逻辑题，探讨通过特定操作使两个整数同时归零的可能性。文章详细分析了题目背后的数学原理，并提供了一个简洁的代码实现。

GDUT_寒假训练题解报告_专题I_L题个人题解报告

题目：
You are given two integers a and b. You may perform any number of operations on them (possibly zero).

During each operation you should choose any positive integer x and set a:=a−x, b:=b−2x or a:=a−2x, b:=b−x. Note that you may choose different values of x in different operations.

Is it possible to make a and b equal to 0 simultaneously?

Your program should answer t independent test cases.

Input
The first line contains one integer t (1≤t≤100) — the number of test cases.

Then the test cases follow, each test case is represented by one line containing two integers a and b for this test case (0≤a,b≤109).

Output
For each test case print the answer to it — YES if it is possible to make a and b equal to 0 simultaneously, and NO otherwise.

You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).

Example
Input
3
6 9
1 1
1 2
Output
YES
NO
YES
题目挺水的，笔纸写一下就知道有一个关系式

2x+y=a
x+2y=b
3(x+y)=a+b

那么首先ab都要>=3;其次就是只要ab为3的倍数，那么x，y都能取出来，而且是任意的，所以现在只要判断一下就好了
代码：

using namespace std;
int t,a,b;
int main()
{
	scanf("%d",&t);
	for(int time=0;time<t;time++)
	{
		scanf("%d%d",&a,&b);
		if(a>b)
		{
			int temp=a;
			a=b;b=temp;
		}
		if(b>2*a)printf("NO\n");
		else if(b==2*a)printf("YES\n");
		else
		{
			if((a+b)%3==0)printf("YES\n");
			else printf("NO\n");
		}
	}
	return 0;
}