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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

long long int dp[60][60][2510] = {{{1}}};	// i個からj個選んで和をsにする選び方の総数

// O(N^3X)の解法
int main(void)
{
	int i, j, A, N, s, x;
	long long int ans = 0;

	scanf("%d%d", &N, &A);
	dp[0][0][0] = 1;
	for (i = 1; i <= N; i++) {
		scanf("%d", &x);
		for (j = 0; j <= N; j++) {
			for (s = 0; s <= 2500; s++) {
				if (s < x) {
					dp[i][j][s] = dp[i - 1][j][s];
				} else if (j >= 1 && s >= x) {
					dp[i][j][s] = dp[i - 1][j - 1][s - x]	// xを選択した場合
								+ dp[i - 1][j][s];			// xを選択しなかった場合
				} 
			}
		}
	}

	for (i = 1; i <= N; i++) {
		ans += dp[N][i][i * A];
	}

	printf("%lld\n", ans);

	return 0;
}