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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
 
#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ
 
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
 
int main(void){
    int N,A;
    scanf("%d%d",&N,&A);
 
    int* x;
    NEW(x,N);
    for(int i=0;i<N;i++){
        scanf("%d",&x[i]);
    }
 
    //DP[i][j][k]...x[0]~x[i]までで、j枚使って計kになる場合の数
    long DP[N][N+1][2501];
    for(int i=0;i<N;i++){
        for(int j=0;j<=N;j++){
            for(int k=0;k<=500;k++) DP[i][j][k]=0;
        }
    }
    //初期値
    DP[0][0][0]=1;
    DP[0][1][x[0]]=1;
 
    //DPを埋める
    for(int i=1;i<N;i++){
        for(int j=0;j<=N;j++){
            for(int k=0;k<=50*i;k++){
                DP[i][j][k]+=DP[i-1][j][k];
                DP[i][j+1][k+x[i]]+=DP[i-1][j][k];
            }
        }
    }
 
    long ans=0;
    for(int i=1;i<=N;i++){
        ans+=DP[N-1][i][A*i];
    }
    printf("%ld\n",ans);
 
    return 0;
}