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: Wrong Answer)

#include <stdio.h>

int main(void){
	int n, a, num[50] = {0};
	long sum = 0;
	long dp[51][2501];
	// long dp[51][51][2501];

	scanf("%d%d", &n, &a);
	for(int i=0; i < n; i++) {
		scanf("%d", &num[i]);
	}

	// dp[0][0][0] = 1;
	// for(int i=1; i <= n; i++) {
	//  for(int j=0; j <= i; j++) {
	//      for(int k=0; k <= i*50; k++) {
	//          if(k < num[i-1]) {
	//              dp[i][j][k] = dp[i-1][j][k];
	//          }else if(j >= 1 && k >= num[i-1]) {
	//              dp[i][j][k] = dp[i-1][j][k] + dp[i-1][j-1][k-num[i-1]];
	//          }
	//      }
	//  }
	// }
	//
	// for(int i=1; i <= n; i++) {
	//  sum += dp[n][i][i*a];
	// }

	dp[0][0] = 1;
	for(int i=1; i <= n; i++) {
		for(int j=i; j > 0; j--) {
			for(int k=num[i-1]; k <= j*50; k++) {
				dp[j][k] += dp[j-1][k-num[i-1]];
			}
		}
	}

	for(int i=1; i <= n; i++) {
		sum += dp[i][i*a];
	}
	printf("%ld", sum);

	return 0;
}