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;

#ifdef LOCAL
    #define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
        #define eprintf(...) 42
#endif

#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define repi(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define all(x) (x).begin(),(x).end()
#define foreach(u,v) for(auto (u) : (v))
#define pb push_back
#define mp make_pair
#define mt make_tuple

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<ll, ll> pll;
typedef vector<ll> vl;

const int inf = 1e9;
const ll linf = 1LL<<60;
const ll mod = 1e9 + 7;
const double eps = 1e-9;

/*
*/

ll dp[51][51][2501];

int main()
{
  int n, a;
  cin >> n >> a;
  vi x(n);
  rep(i, n) cin >> x[i];

  dp[0][0][0] = 1;
  rep(i, n) rep(j, n) rep(k, 2501){
    dp[i+1][j][k] += dp[i][j][k];
    if(k + x[i] <= 2500){
      dp[i+1][j+1][k+x[i]] += dp[i][j][k]; 
    }
  }

  ll ans = 0;
  repi(i, 1, n+1) rep(k, 2501){
    if(a*i == k){
      ans += dp[n][i][k];
    }
  }
  
  cout << ans << endl;

  return 0;
}