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: Accepted)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

class Main {
	@SuppressWarnings("resource")
	public static void main(String args[]) {
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int a = scanner.nextInt();
		List<Integer> x = new ArrayList<>();

		for (int i = 0; i < n; i++) {
			x.add(scanner.nextInt());
		}

		long ans = 0;
		long[][][] dp = new long[n + 1][n + 1][n * a + 1];
		dp[0][0][0] = 1;
		for (int i = 1; i <= n; i++) {
			for (int j = 0; j <= n; j++) {
				for (int k = 0; k <= n * a; k++) {
					if (j - 1 >= 0 && k - x.get(i - 1) >= 0) {
						dp[i][j][k] = dp[i - 1][j][k] + dp[i - 1][j - 1][k - x.get(i - 1)];
					} else {
						dp[i][j][k] = dp[i - 1][j][k];
					}

				}
			}
		}

		for (int j = 1; j <= n; j++) {
			ans += dp[n][j][a * j];

		}

		System.out.println(ans);
	}
}