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 <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <utility>
#include <string>

using namespace std;

int main() {
	int n, a;

	cin >> n >> a;
	vector<int> x(n);
	for (int i = 0; i < n; i++) {
		cin >> x[i];
	}

	// dp[使用枚数][合計]
	int max_sum = 50 * n;
	vector<vector<long long>> dp(n + 1, vector<long long>(max_sum + 100, 0));
	dp[0][0] = 1;

	for (int i = 0; i < n; i++) { // x[i]
		for (int j = n - 1; j >= 0; j--) {
			for (int k = 0; k < max_sum; k++) {
				// if (dp[j][k] > 0) {
				// 	printf("%d %d %d\n", j, k, dp[j][k]);
				// }
				// dp[i+1][k] += dp[i][k];
				dp[j+1][k + x[i]] += dp[j][k];
			}
		}
	}

	// for (int i = 0; i <= n; i++) {
	// 	printf("%d: ", i);
	// 	for (int j = 0; j < 40; j++) {
	// 		printf("%d ", dp[i][j]);
	// 	}
	// 	printf("\n");
	// }

	long long ans = 0;
	for (int i = 1; i <= n; i++) {
		ans += dp[i][i * a];
	}
	cout << ans << endl;
	return 0;
}