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: Wrong Answer)

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

#if __has_include("print.hpp")
  #include "print.hpp"
#endif

#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define MOD 1000000007

typedef long long ll;
typedef pair<int, int> p;

// dp[i][j][k] = x1 - xi からj選んで和がkな数
int dp[60][60][10100];

int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, a;
  cin >> n >> a;
  vector<int> v(n);
  dp[0][0][0] = 1;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }

  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < 3000; k++) {
        if(dp[i][j][k]){
          dp[i+1][j][k] += dp[i][j][k];
          dp[i+1][j+1][k + v[i]] += dp[i][j][k];
        }
      }
    }
  }

  ll res = 0;
  for (int i = 1; i < n+1; i++) {
    res += dp[n][i][i * a];
  }
  cout << res << endl;
}