B. Make AP

Problem - B - Codeforces

一开始是想要暴力枚举abc的

然后不想写那么麻烦，回想起了等差数列的性质 a+c == 2*b

然后呢 既然能被乘到满足条件 那就意味着能被整除

于是有了这样的代码

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;
#define ms(x, y)	memset(x, y, sizeof x)
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;

int main()
{
	ios::sync_with_stdio(false);cin.tie(0);
	int n;
	cin >> n;
	int a, b, c;
	while(n--)
	{
		cin >> a >> b >> c;
		if(a+c == 2*b)	cout << "Yes";

		else
		{
			if(2*b < a + c)	
			{
				if( (a+c) % (2*b) == 0 )	cout << "Yes";
				else	cout << "No";
			}
			else//(2*b > a+c)
			{
				if( (2*b - c) % a == 0 || (2*b - a) % c == 0 )	cout << "Yes";
				else	cout << "No";
			}
		}
		cout << endl;
	}
	return 0;
}