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 <algorithm>
#include <iostream>

using namespace std;

constexpr int kMaxN = 51;
constexpr int kMaxX = 51;

int x[kMaxN] = {0};
long long dp[kMaxN][kMaxN][kMaxN * kMaxX] = {{{0}}};

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int N, A;
  cin >> N >> A;
  int X = 0;
  for (int i = 1; i <= N; ++i) {
    cin >> x[i];
    X = max(X, x[i]);
  }

  for (int j = 0; j <= N; ++j) {
    for (int k = 0; k <= N; ++k) {
      for (int s = 0; s <= N * X; ++s) {
        if (j == 0 && k == 0 && s == 0) {
          dp[j][k][s] = 1;
        } else if (1 <= j && s < x[j]) {
          dp[j][k][s] = dp[j - 1][k][s];
        } else if (1 <= j && 1 <= k && x[j] <= s) {
          dp[j][k][s] = dp[j - 1][k][s] + dp[j - 1][k - 1][s - x[j]];
        } else {
          dp[j][k][s] = 0;
        }
      }
    }
  }
  long long ans = 0;
  for (int k = 1; k <= N; ++k) ans += dp[N][k][k * A];
  cout << ans << endl;

  return 0;
}