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;
typedef long long ll;
typedef pair<int, int> P;
constexpr int di[] = { 0, 1, 0, -1 }, dj[] = { 1, 0, -1, 0 };
#define rep(i, n) for (int i = 0; i < n; i++)
#define repr(i, n) for (int i = n; i >= 0; i--)
#define SORT(v) sort((v).begin(), (v).end())
#define SORTR(v) sort((v).rbegin(), (v).rend())
#define all(v) (v).begin(), (v).end()
#define vmax 10000
constexpr ll inf = 1e18+7;
/*cout<<fixed<<setprecision(20);cin.tie(0);ios::sync_with_stdio(false);*/
ll n,a,ans=0;
int dp[51][51][2501];

int main(){
  cin>>n>>a;
  vector<int> x(n);
  rep(i,n){
    cin>>x[i];
  }
  rep(i,2501){
    dp[0][0][i]=0;
  }

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

  rep(i,n){
    rep(k,n){
      rep(j,2501){
      if(j-x[i]>=0){
        dp[i+1][k+1][j]=dp[i][k+1][j-x[i]];
        dp[i+1][k][j]=dp[i][k][j];
      }else{
        dp[i+1][k][j]=dp[i][k][j];
      }
      }
    }
  }
  rep(i,n+1){
    rep(k,n){
    rep(j,2501){
      if(j==k*a){
        ans+=dp[i][k][j];
      }
    }
    }
  }
  cout<<ans<<endl;
  return 0;
}