§ 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;
}