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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

using ll = long long int;

const int MAX = (int)(1e5 + 5);
const ll INF = (ll)(1e10 + 5);

const int MAX_N = (int)(50 + 5);
const int MAX_X = (int)(50 + 5);

int n, a;
int x[MAX_N];

ll dp[MAX_N][MAX_N][MAX_N * MAX_X];

ll ans = 0;

int main(void) {
  // Here your code !
  scanf("%d %d", &n, &a);

  for (int i = 1; i <= n; ++i) {
    scanf("%d", &(x[i]));
  }

  dp[0][0][0] = 1LL;

  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= i; ++j) {
      for (int k = 0; k < n * MAX_X; ++k) {
        if (k - x[i] < 0) {
          dp[i][j][k] = dp[i - 1][j][k];
        }
        else {
          if (j > 0) dp[i][j][k] = dp[i - 1][j][k] + dp[i - 1][j - 1][k - x[i]];
        }
      }
    }
  }

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

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

  return 0;
}