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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int A = sc.nextInt();
    int[] x = new int[N+1];
    int sum = 0;
    for (int i = 1; i <= N; i++) {
      x[i] = sc.nextInt();
      sum += x[i];
    }
    
    int[][][] dp = new int[N+1][N+1][sum+1];
    dp[0][0][0] = 1;
    for (int j = 1; j <= N; j++) {
      for (int k = 0; k <= j; k++) {
        for (int s = 0; s <= sum; s++) {
          if (k > 0 && s - x[j] >= 0) {
            // x[j] can be used.
            // not use x[j] + use x[j]
            dp[j][k][s] = dp[j-1][k][s] + dp[j-1][k-1][s-x[j]];
          } else {
            // x[j] can't be used => not use x[j] case only
            dp[j][k][s] = dp[j-1][k][s];
          }
        }
      }
    }
    
    long ans = 0;
    for (int k = 1; k <= N; k++) {
      ans += dp[N][k][k*A];
    }

    System.out.println(ans);
  }
}