7-1 understand everything

7-1 Know everything.
fraction 20
author DAI, Longao
Company Hangzhou Baiteng Education Technology Co., Ltd
b3ceb051352ac65c29767cc3ecf2b21192138add.jpg

as everyone knows , There are many words on the Internet that are not easy to say directly , However, some vague pictures can still make netizens understand what you are talking about . However, we must still give a heavy blow to this kind of speech , So please implement a simple matching algorithm .

Now we have collected some characteristic data of the original image , from N Less than 255 Composed of nonnegative integers , Suppose that for a given number of sheets M
i
​
Two are also less than 255 The characteristic data of the new graph composed of nonnegative integers , Each data can be calculated from the average of any four different data in the original figure , The new picture is called a similar picture of the original picture . For the given data , Please judge whether it's a similar picture .

Be careful , Different data does not mean different values of data , Instead, you can't take the same data multiple times . For two data with the same value , If you give it twice , You can take it twice .

Input format :
The first line of input is two integers N,K (1 ≤ N ≤ 50, 1 ≤ K ≤ 200), Indicates the number of characteristic data of the collected original drawing and the number of new drawings .

The next action N Less than 255 Non-negative integer , Represent the characteristic data of the original drawing .

final K That's ok , The first number in each line is M
i
​
(1 ≤ M
i
​
≤ 200), Represents the number of characteristic data of the new graph . And then there was M
i
​
Less than 255 Non-negative integer , Represent the characteristic data of the new graph .

Output format :
For each new picture , If it is a similar picture , Output in one line Yes, Otherwise output No.

sample input :
5 3
4 8 12 20 40
3 11 16 19
3 12 16 19
10 11 11 11 11 11 11 11 11 11 11
sample output :
Yes
No
Yes

#include <bits/stdc++.h> 

using namespace std;

int a[100],b[4];
bool st[1200];
int n,k;
void dfs(int t,int x)
{
    
	if(t==4)
	{
    
		int s=0;
		for(int i=0;i<4;++i)s+=b[i];
		//s/=4;
		st[s]=true;
		return ;
	}
	for(int i=x;i<n;++i){
    
		b[t]=a[i];
		dfs(t+1,i+1);
	}
}
int main()
{
    
	cin>>n>>k;
	for(int i=0;i<n;++i)cin>>a[i];
	dfs(0,0);
	while(k--)
	{
    
		int t;cin>>t;
		bool flag=true;
		for(int i=0;i<t;++i)
		{
    
			int x;cin>>x;
			x*=4;
			if(st[x]==false){
    
				flag=false;
			}
		}
		if(flag)cout<<"Yes"<<endl;
		else cout<<"No"<<endl;
	}
	return 0;
}