Description

Link.

$ n $ distinct integers $ x_1,x_2,\ldots,x_n $ are written on the board. Nezzar can perform the following operation multiple times.

• Select two integers $ x,y $ (not necessarily distinct) on the board, and write down $ 2x-y $ . Note that you don't remove selected numbers.

Now, Nezzar wonders if it is possible to have his favorite number $ k $ on the board after applying above operation multiple times.

Solution

Let us onsider how to express all the number we can express.
Since \(2x-y=x+(x-y)\), the expression is \(x_{i}+\sum_{j,k}x_{j}-x{k}\).
Since \(x_{i}=x_{1}+(x_{1}-(x_{1}+(x_{1}-x_{i})))\), the expression could be written as \(x_{1}+\sum_{i}x_{i}-x_{1}\), which means the answer exists where \(\sum_{i}x_{i}-x_{1}=K-x_{1}\) has solutions.
Beout's identity works.

#include<bits/stdc++.h>
typedef long long ll;
#define sf(x) scanf("%d",&x)
#define ssf(x) scanf("%lld",&x)
int main()
{
	int T,n;
	ll k;
	sf(T);
	while(T-->0)
	{
		sf(n),ssf(k);
		ll g=0,x1,x;
		ssf(x1);
		for(int i=2;i<=n;++i)	ssf(x),g=std::__gcd(x-x1,g);
		puts((k-x1)%g==0?"YES":"NO");
	}
	return 0;
}