来点gcd

题意:是否存在一个子集使得子集中所有元素的$gcdgcd$为$xx\backslash \backslash$ 有一个需要注意的事情是：如果给定$xx$是2，那么那个子集中的所有数都应该是2的倍数,有了这个思想，我们就可以利用调和级数On的遍历x取1~n时存在的gcd。

vis[i]:标记$ii$是否是给定的$nn$个数内，是为$truetrue$，否则为$falsefalse\backslash \backslash$ res[i]:标记$ii$是否能由一个子集构造使得$gcd=igcd\; =\; i$

#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
# include<iostream>
# include<iomanip>
# include<algorithm>
# include<cmath>
# include<cstdio>
# include<set>
# include<stack>
# include<queue> 
# include<map>
# include<string>
# include<cstring> 


# define eps 1e-9
# define fi first
# define se second
# define ll long long
# define int ll
// cout<<fixed<<setprecision(n) 
using namespace std;
   
typedef unsigned long long ull;
typedef pair<int,int > PII; 
const int mod=1e9+7;
const int MAX=3e6+10;
const int Time=86400;
const int X=131;
const int inf=0x3f3f3f3f;
const double PI = 1e-4;
double pai = 3.14159265358979323846; 

int T,n,m;
bool vis[MAX],res[MAX];
int a[MAX];


void solve(){
	  cin >> n >> m;
	  for(int i = 1 ; i <= n ; i ++ ) vis[i] = false , res[i] = false; //初始化
	  for(int i = 1 ; i <= n ; i ++ ){
	  	 int x;
	  	 cin >> x;
	  	 vis[x] = true; //存在x
	  }
	  for(int i = 1 ; i <= n ; i ++ ){
	  	    int g = 0;
	  	    for(int j = i ; j <= n ; j += i) //gcd=i,构造的集合一定是i的倍数
	  	      if(vis[j]) g = __gcd(g,j); //输入的数据存在j，就求gcd
	  	    
	  	  if(g == i) res[i] = true; //如果能构造出来i，res[i]就为true
	  }
	  while(m--){
	  	  int x;
	  	  cin >> x;
	  	  if(res[x])  cout<<"YES\n";
	  	   else cout<<"NO\n";
	  }
}
signed main(){  
    std::ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    cin >> T;
	while(T--){
		solve();
	} 
    return 0; 
}
/*
2 2 6 6 2 1 5
vis 1 2 3 4 5 6
i 1 ~ 7
j i ~ 7
i = 1  
j: 1~7 
*/