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>
#define ll long long
#define rep(X,N) for(ll X = 0; X < (N); X++)
#define PI (acos(-1.0))
#define sorta(V) sort(V.begin(), V.end())
#define pback push_back
#define mpair make_pair
#define MODN 1000000007
#define ALL(V) (V).begin(),(V).end() 

using namespace std;

int main(){

    int n,a;
    cin >> n >> a;

    vector<int> x;

    rep(i,n){
        int tmp;
        cin >> tmp;
        x.pback(tmp);
    }

    ll dp[51][51][2501] = {};

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

    for(int i = 1; i <= n; i++){
        for(int j = 0; j <= i; j++){
            for(int k = 0; k <= n * 50; k++){
                //xiを使わなかった場合
                assert(dp[i][j][k] == 0);
                dp[i][j][k] = dp[i][j][k] + dp[i - 1][j][k];

                //xiを使う場合
                if(k - x[i - 1] >= 0) dp[i][j][k] = dp[i][j][k] + dp[i - 1][j - 1][k - x[i - 1]];
            }
        }
    }

    ll ans = 0;

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

    cout << ans << endl;

    return 0;
}