2019_GDUT_新生专题I选集 L Codeforces-1260B

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≤10e9).’
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

做法:数学题。每次操作的x都是相等的,也就是每次总共都减去了3个x,所以a+b应当是3的倍数。假设全部选择操作a:=a−2x, b:=b−x,会发现当a=2*b 的时候刚好可以为0,当a>2b的时候无论如何也没办法完成操作,b>2a同理。

代码:

#include<iostream>
#include<cstdio>
using namespace std;

int main()
{
	int T;
	long long a,b;
	cin>>T;
	while (T--)
	{
		cin>>a>>b;
		if ((a+b)%3==0 && (b<=2*a && a<=2*b))
		cout<<"YES"<<endl;
		else cout<<"NO"<<endl;
	}
	return 0;
}
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转载自blog.csdn.net/qq_39581539/article/details/103965038