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.*;

public class Main {
	static long[][][] dp;
	static int[] arr;
	public static void main (String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int a = sc.nextInt();
		arr = new int[n];
		for (int i = 0; i < n; i++) {
			arr[i] = sc.nextInt();
		}
		dp = new long[n + 1][n + 1][a * n + 1];
		for (long[][] first : dp) {
			for (long[] second : first) {
				Arrays.fill(second, -1);
			}
		}
		long total = 0;
		for (int i = 1; i <= n; i++) {
			total += dfs(n, i, i * a);
		}
		System.out.println(total);
	}
	
	static long dfs(int x, int y, int total) {
		if (x < 0 || y < 0 || total < 0) {
			return 0;
		}
		if (dp[x][y][total] != -1) {
			return dp[x][y][total];
		}
		if (x == 0) {
			if (y == 0 && total == 0) {
				dp[x][y][total] = 1;
				return 1;
			} else {
				dp[x][y][total] = 0;
				return 0;
			}
		}
		long ans = dfs(x - 1, y, total) + dfs(x - 1, y - 1, total - arr[x - 1]);
		dp[x][y][total] = ans;
		return ans;
	}
}