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>

int main(void){
    int n, a;
    scanf("%d%d", &n, &a);           //カードの枚数と平均の目標値
    int x[n+1];
    for(int i=1;i<n+1;i++){
        scanf("%d", &x[i]);          //各カードの数字
    }
    
    
    int X=x[0];
    
    for(int i=1;i<n;i++){
        if(X<x[i]) X = x[i];          //xの最大値
    }
    
    int dp[n+1][n+1][n*X+1];          //i枚のカードからj枚選んで合計をkにする場合の数
    
    for(int i=0;i<n+1;i++){
        for(int j=0;j<n+1;j++){
            for(int k=0;k<(n*X+1);k++){
                if(i==0&&j==0&&k==0)dp[i][j][k] = 1;       //0枚引いて合計が0になる場合の数は当然1
                else if(i>0){
                    if(x[i]>k) dp[i][j][k] = dp[i-1][j][k];    //1枚以上引く場合でx[i]がkより大きい場合
                    else if(i>0&&j>0&&x[i]<=k)dp[i][j][k] = dp[i-1][j][k]+dp[i-1][j-1][k-x[i]];    //1枚以上引く場合でx[i]がk以下の場合
                }
                else dp[i][j][k]=0;
            }
        }
    }
    
    int sum=0;
    
    for(int i=1;i<n+1;i++){
        sum += dp[n][i][i*a];
    }
    
    printf("%d\n", sum);
    
    
    return 0;
}