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 <iostream>
#include <cmath>
#include <string>
#include <algorithm>
#include <set>
#include <vector>
#include <map>
#include <list>
#include <stack>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
#define mkp make_pair
#define pr(num) cout << num << "\n"
#define max(a, b) ((a)>(b) ? (a):(b))
#define min(a, b) ((a)<(b) ? (a):(b))
#define INF 1000000000000000000
#define MOD 1000000007LL
#define MAX 100100
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef map<ll, ll> Map;


int main(void) {
    ll N, A;
    cin >> N >> A;
    ll x[55];
    ll i, j, k;
    for(i=0; i<N; i++) {
        cin >> x[i];
    }
    
    
    ll dp[N+5][N*A+5][N+5];
    memset(dp, 0, sizeof(dp));
    
    dp[0][0][0]=1;
    for(i=1; i<=N; i++) {
        for(j=0; j<=N*A; j++) {
            for(k=1; k<=i; k++) {
                if(j+x[i-1]<=N*A)
                    dp[i][j+x[i-1]][k]=dp[i-1][j][k-1];
                dp[i][j][k-1]=dp[i-1][j][k-1];
            }
        }
    }
    
    
    ll ans=0;
    for(k=1; k<=N; k++) {
        ans+=dp[N][k*A][k];
    }
    
    pr(ans);
}