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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Main main = new Main();
        main.solve();
    }
    private int ceil(int x, int y) {
        if (x % y == 0) {
            return x / y;
        }
        return x / y + 1;
    }
    public void solve() {
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int N = scan.nextInt();
        int A = scan.nextInt();
        int[] x = new int[N];
        int sum = 0;
        for (int i = 0; i < N; i++) {
            x[i] = scan.nextInt();
            sum += x[i];
        }
        int limit = ceil(sum, A) * A + 1;
        long[][][] dp = new long[N+1][N+1][limit];
        for (int i = 0; i <= N; i++) {
            for (int j = 0; j <= N; j++) {
                Arrays.fill(dp[i][j], 0);
            }
        }
        dp[0][0][0] = 1;
        for (int i = 1; i <= N; i++) {
            for (int j = 0; j <= N; j++) {
                for (int k = 0; k < limit; k++) {
                    // i枚目まででj枚を選択いて合計がk
                    if (x[i-1] <= k && 1 <= j) {
                        dp[i][j][k] = dp[i - 1][j][k] + dp[i - 1][j - 1][k - x[i - 1]];
                    } else {
                        dp[i][j][k] = dp[i-1][j][k];
                    }
                }
            }
        }
        long answer = 0;
        for (int j = 1; j <= N; j++) {
            answer += dp[N][j][j*A];
        }
        System.out.println(answer);
    }
}