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>
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define RFOR(i,a,b) for(int i=(b) - 1;i>=(a);i--)
#define REP(i,n) for(int i=0;i<(n);i++)
#define RREP(i,n) for(int i=n-1;i>=0;i--)

#define PB push_back
#define MP make_pair
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define CLR(a) memset(a,0,sizeof(a))
#define SET(a,c) memset(a,c,sizeof(a))

#define DEBUG(x) cout<< #x <<": "<<x<<endl

using namespace std;

typedef long long int ll;
typedef vector<int> vi;
typedef vector<ll> vl;

const ll INF = INT_MAX/3;
const ll MOD = 1000000007;
const double EPS = 1e-14;
const int dx[] = {1,0,-1,0} , dy[] = {0,1,0,-1};

ll dp[51][51][2601];
int main(){
    int N,A,NX,x[51];
    cin >> N >> A;
    REP(i,N) cin >> x[i];
    NX = N * 50;
    dp[0][0][0] = 1;

    REP(i,N)REP(j,N)REP(k,NX){
        dp[i+1][j+1][k+x[i]] += dp[i][j][k];
        dp[i+1][j][k] += dp[i][j][k];
    }

    ll ans = 0;
    FOR(i,1,N+1) ans += dp[N][i][A*i];
    cout << ans << endl;

    return 0;
}