7-1 understand everything

7-1 Know everything.

author  DAI, Longao

Company   Hangzhou Baiteng Education Technology Co., Ltd

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  Mi​  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  Mi​ (1 ≤ Mi​ ≤ 200), Represents the number of characteristic data of the new graph . And then there was  Mi​  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

Code length limit

16 KB

The time limit

400 ms

Memory limit

64 MB

Explain ：

#include <iostream>
#include <set>
using namespace std;

int main()
{
    set <int> pj;
    int n,k, m[201], tz[201][201];
    cin >> n >> k;
    int ns[n];
    for(int i=0; i<n; i++)  cin >> ns[i];
    for(int a=0; a<n-3; a++)
        for(int b=a+1; b<n-2; b++)
            for(int c=b+1; c<n-1; c++)
                for(int d=c+1; d<n; d++)
                    pj.insert((ns[a]+ns[b]+ns[c]+ns[d]));

    for(int i=0; i<k; i++)
    {
        cin >> m[i];
        for(int j=0; j<m[i]; j++)  cin >> tz[i][j];
    }
    
    for(int i=0; i<k; i++)
    {
        int cnt = 0;
        for(int j=0; j<m[i]; j++)
            for(auto p=pj.begin(); p!=pj.end(); p++)  if(tz[i][j]*4 == *p)  cnt++;

        if(cnt == m[i])  cout << "Yes" << endl;
        else  cout << "No" << endl;
    }
    

    return 0;
}