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>
#define rep(i, n) for(int i = 0; i < (n); i++)
#define MAX 510000
using namespace std;
typedef long long ll;
const int mod = 998244353;
#define V vector<ll>
#define VV vector<V>
#define VVV vector<VV>

signed main(void)
{
	int N, A, X;
	cin >> N >> A;
	int x[N + 1];

	X = A;
	x[0] = 0;
	for (int i = 1; i <= (N); i++)
	{
		cin >> x[i];
		X = max(x[i], X);
	}
	VVV dp(N + 1, VV(N + 1, V(N * X + 1, 0)));
	dp[0][0][0] = 1;
	for (int j = 0; j <= N; j++)
	{
		for (int k = 0; k <= N; k++)
		{
			for (int s = 0; s <= N * X; s++)
			{
				if (s < x[j] && j >= 1)
					dp[j][k][s] = dp[j - 1][k][s];
				else if (k >= 1 && j >= 1)
					dp[j][k][s] = dp[j - 1][k][s] + dp[j - 1][k - 1][s - x[j]];
			}
		}
	}
	ll ans = 0;
	for(int k = 1; k <= N; k++)
		ans += dp[N][k][k * A];
	cout << ans << endl;
	return 0;
}