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: Runtime Error)

#include <bits/stdc++.h>

#define all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const int MN = 50 + 5, LN = 17, MOD = 1e9 + 7, INF = 0x3f3f3f3f, BSZ = 320;
int N, A, a[MN];
ll dp[MN][MN][MN * MN];

int main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    cin >> N >> A;
    int sum = 0;
    for (int i = 1; i <= N; i++) cin >> a[i], sum += a[i];
    dp[0][0][0] = 1;
    for (int i = 1; i <= N; i++) {
        for (int j = 0; j <= i; j++) {
            for (int k = 0; k <= sum; k++) {
                dp[i][j][k] += dp[i - 1][j][k];
                if (j > 0 && k >= a[i]) {
                    dp[i][j][k] += dp[i - 1][j - 1][k - a[i]];
                }
            }
        }
    }
    ll ret = 0;
    for (int i = 1; i * A <= sum; i++) ret += dp[N][i][i * A];
    cout << ret << '\n';
    return 0;
}