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: Wrong Answer)

import java.util.Scanner;

public class Main {
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int A = sc.nextInt();
        int i = 0;
        int[] x = new int[N + 1];
        x[0] = 0;
        int sum = 0;
        while(i < N){
            x[i + 1] = sc.nextInt();
            sum += x[i + 1];
            i++;
        }
        int[][][] dp = new int[N + 1][N + 1][sum + 1];
        dp[0][0][0] = 1;

        for (int s = 0; s <= sum; s++) {
            for(int k = 0; k <= N; k++){
                for(int j = k; j <= N; j++){
                    if(s >= x[j] && j >= 1 && k >= 1){
                        dp[j][k][s] = dp[j - 1][k][s] + dp[j - 1][k - 1][s - x[j]];
                    }else if(s < x[j] && j >= 1){
                        dp[j][k][s] = dp[j - 1][k][s];
                    }else if(j != 0 || k != 0 || s != 0){
                        dp[j][k][s] = 0;
                    }
                }
            }
        }
//        for (int s = 0; s <= sum; s++) {
//            System.out.println("sum: " + s);
//            for(int j = 0; j <= N; j++) {
//                for (int k = 0; k <= N; k++) {
//                    System.out.print(dp[j][k][s] + " ");
//                }
//                System.out.println();
//            }
//            System.out.println("-----------------------------------");
//        }


        long ans = 0;
        for(int k = 1; k <= N; k++){
            for(int s = 0; s <= sum; s++){
                if(k * A == s){
                    ans += dp[N][k][s];
                }
            }
        }
        System.out.println(ans);
    }
}