Problem:
Score : 300 points

Problem Statement

Tak has N cards. On the i-th (1 \leq i \leq N) card is written an integer x_i.
He is selecting one or more cards from these N cards, so that the average of the integers written on the selected cards is exactly A.
In how many ways can he make his selection?

Constraints

1 \leq N \leq 50

1 \leq A \leq 50

1 \leq x_i \leq 50

N,\,A,\,x_i are integers.

Partial Score

200 points will be awarded for passing the test set satisfying 1 \leq N \leq 16.

Input

The input is given from Standard Input in the following format:

N A
x_1 x_2 ... x_N

Output

Print the number of ways to select cards such that the average of the written integers is exactly A.

Sample Input 1

4 8
7 9 8 9

Sample Output 1

5

The following are the 5 ways to select cards such that the average is 8:

Select the 3-rd card.

Select the 1-st and 2-nd cards.

Select the 1-st and 4-th cards.

Select the 1-st, 2-nd and 3-rd cards.

Select the 1-st, 3-rd and 4-th cards.

Sample Input 2

3 8
6 6 9

Sample Output 2

0

Sample Input 3

8 5
3 6 2 8 7 6 5 9

Sample Output 3

19

Sample Input 4

33 3
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

Sample Output 4

8589934591

The answer may not fit into a 32-bit integer.

Solution in C (GCC 9.2.1) (judge verdict: Runtime Error)

#include<stdio.h>

int main(){
    int N;
    int A;
    scanf("%d%d",&N,&A);

    int l=1;
    for(int i=0;i<N;i++){
        l=l*2;
    }

    int a[l-1][N];
    for(int i=0;i<l-1;i++){
        for(int j=0;j<N;j++){
            a[i][j]=0;
        }
    }
    int s=1;
    int t=0;
    while(t<l-1){
      for(int i=0;s>0;i++){
        a[t][i]=s%2;
        s=s/2;
      }
      t++;
      s=t+1;
    } 

    int x[N];
    for(int i=0;i<N;i++){
        scanf("%d",&x[i]);
    }

    int y=0;
    int sum=0;
    int count=0;
    int ave=0;
    int rest=0;
    for(int i=0;i<l-1;i++){
        for(int j=0;j<N;j++){
            if(a[i][j]==1){
                sum+=x[j];
                count++;
            }
        }
        ave=sum/count;
        rest=sum%count;
        if(ave==A&&rest==0){
            y++;
        }
        sum=0;
        count=0;
        ave=0;
        rest=0;
    }

    printf("%d\n",y);
    return 0;
}