CF--1624B.Make AP

最新推荐文章于 2025-11-30 22:51:53 发布

原创最新推荐文章于 2025-11-30 22:51:53 发布 · 367 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#p2p #网络协议 #网络

本文探讨了一个关于等差数列的问题，即如何通过一次操作将三个正整数a, b, c变为等差数列。关键在于判断a+c是否等于2b以及是否能通过乘以某个正整数使它们成为等差数列。解决方案涉及到了整除性和条件判断，对于编程和数学思维有一定启发。

Polycarp has 33 positive integers aa, bb and cc. He can perform the following operation exactly once.

Choose a positive integer mm and multiply exactly one of the integers aa, bb or cc by mm.

Can Polycarp make it so that after performing the operation, the sequence of three numbers aa, bb, cc (in this order) forms an arithmetic progression? Note that you cannot change the order of aa, bb and cc.

Formally, a sequence x_1, x_2, \dots, x_nx1,x2,…,xn is called an arithmetic progression (AP) if there exists a number dd (called "common difference") such that x_{i+1}=x_i+dxi+1=xi+d for all ii from 11 to n-1n−1. In this problem, n=3n=3.

For example, the following sequences are AP: [5, 10, 15][5,10,15], [3, 2, 1][3,2,1], [1, 1, 1][1,1,1], and [13, 10, 7][13,10,7]. The following sequences are not AP: [1, 2, 4][1,2,4], [0, 1, 0][0,1,0] and [1, 3, 2][1,3,2].

You need to answer tt independent test cases.

Input

The first line contains the number tt (1 \le t \le 10^41≤t≤104) — the number of test cases.

Each of the following tt lines contains 33 integers aa, bb, cc (1 \le a, b, c \le 10^81≤a,b,c≤108).

Output

For each test case print "YES" (without quotes) if Polycarp can choose a positive integer mm and multiply exactly one of the integers aa, bb or cc by mm to make [a, b, c][a,b,c] be an arithmetic progression. Print "NO" (without quotes) otherwise.

You can print YES and NO in any (upper or lower) case (for example, the strings yEs, yes, Yes and YES will be recognized as a positive answer).

Sample 1

Inputcopy	Outputcopy
11 10 5 30 30 5 10 1 2 3 1 6 3 2 6 3 1 1 1 1 1 2 1 1 3 1 100000000 1 2 1 1 1 2 2	YES YES YES YES NO YES NO YES YES NO YES

Note

In the first and second test cases, you can choose the number m=4m=4 and multiply the second number (b=5b=5) by 44.

In the first test case the resulting sequence will be [10, 20, 30][10,20,30]. This is an AP with a difference d=10d=10.

In the second test case the resulting sequence will be [30, 20, 10][30,20,10]. This is an AP with a difference d=-10d=−10.

In the third test case, you can choose m=1m=1 and multiply any number by 11. The resulting sequence will be [1, 2, 3][1,2,3]. This is an AP with a difference d=1d=1.

In the fourth test case, you can choose m=9m=9 and multiply the first number (a=1a=1) by 99. The resulting sequence will be [9, 6, 3][9,6,3]. This is an AP with a difference d=-3d=−3.

In the fifth test case, it is impossible to make an AP.

a b c -->性质等差数列：a+c=2b

10 5 30 a+c>2b ：那么为了使两边一样大，需要增大b，使得a+c=2b 那么b是否有这个潜力呢？再写个判断 ((a+c)/2)%b==0? 来判断我们需要的那个b是否为现在的b的倍数，是的话就YES，不是就NO

30 5 10 同上

1 2 3 a+c=2b 直接输出YES

1 6 3 a+c<2b ：同理增大a或者是c 判断一下只要有一个可以构成a+c=2b就YES，否则NO

#include<bits/stdc++.h>
using namespace std;
int a,b,c;

int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		cin>>a>>b>>c;
		if((a+c)>(2*b))//a+c大了，那么就让b也变大 
		{
			if((a+c)%(2*b)==0)
			cout<<"YES"<<endl;
			else
			cout<<"NO"<<endl;
		}
		else if((a+c)<(2*b))//为了使两边相等，那么增大a或者是c 
		{
			if(((2*b)-a)%c==0)
			cout<<"YES"<<endl;
			else if(((2*b)-c)%a==0)
			cout<<"YES"<<endl;
			else 
			cout<<"NO"<<endl;
			 
		}
		else
		{
			cout<<"YES"<<endl;
		}
	}
	return 0;
}