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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>

int main(void){
    int n,a;
    long long ans=0;
    scanf("%d%d",&n,&a);
    int x[n],max=a;
    for (int i=0; i<n; i++){
        scanf("%d",&x[i]);
        if (max<x[i]){
            max=x[i];
        }
    }
    int j,k,l;
    long long dp[n+2][n+2][2*n*max];
    for (j=0; j<=n; j++){
        for (k=0; k<=j; k++){
            for (l=0; l<=k*max; l++){
                if (j==0 && k==0 && l==0){
                    dp[j][k][l]=1;

                } else if (j>=1 && l<x[j-1]) {
                    dp[j][k][l]=dp[j-1][k][l];
                    
                    
                } else if (j>=1 && k>=1 && l>=x[j-1]){
                    dp[j][k][l]=dp[j-1][k][l]+dp[j-1][k-1][l-x[j-1]];
                    
                } else {
                    dp[j][k][l]=0;
                }
                if (j==n && k>=1 && l==k*a){
                    ans+=dp[j][k][l];
                }
            }
        }

    }
    printf("%lld\n",ans);
	return 0 ;
}