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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MM = 1000000000;
const int MOD = MM + 7;
const int MAX = 510000;
#define rep(i, n) for(ll i=0; i<n; i++)
#define Rep(i, j, n) for(ll i=j; i<n; i++)
#define all(vec) vec.begin(), vec.end()
template<class T> inline bool chmin(T& a, T b) {if(a > b) {a = b; return true;} return false;}
template<class T> inline bool chmax(T& a, T b) {if(a < b) {a = b; return true;} return false;}
const ll INF = 1LL << 60;
const double pi = acos(-1.0);

ll dp[55][55][3000];

int main() {
    ll n, a; cin >> n >> a;
    rep(i, 55) rep(j, 55) rep(k, 3000) dp[i][j][k] = 0;
    dp[0][0][0] = 1;
    rep(i, n) {
        ll x; cin >> x;
        rep(j, n+1) {
            rep(k, n*a) {
                dp[i+1][j][k] += dp[i][j][k];
                dp[i+1][j+1][k+x] += dp[i][j][k];
            }
        }
    }
    ll ans = 0;
    Rep(i, 1, n+1) ans += dp[n][i][i*a];
    cout << ans << endl;
}