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 <string>
#include <string.h>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <stdio.h>
#include <vector>
#include <thread>
#include <atomic>
#include <mutex>
#include <future>
#define int long long
#define endre getchar();getchar();return 0
#define INF 1000000007
using namespace std;
bool prime(int n) {
	for (int i = 2; i <= sqrt(n); i++) {
		if (n%i == 0)return false;
	}
	return n != 1;
}
int gcd(int x, int y) {
	if (x < y)swap(x, y);
	if (y == 0)return x;
	return (y, x%y);
}
int lcm(int x, int y) {
	return x * y / gcd(x, y);
}
int kai(int n) {
	if (n == 0)return 1;
	return (kai(n - 1)*n) % INF;
}
int mod_pow(int x, int y, int mod) {
	int res = 1;
	while (y > 0) {
		if (y & 1) {
			res = res * x%mod;
		}
		x = x * x%mod;
		y >>= 1;
	}
	return res;
}
int n, a, x[55], dp[105][105][5005], cnt;
signed main() {
	cin >> n >> a;
	for (int i = 0; i < n; i++)cin >> x[i];
	dp[0][0][0] = 1;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j <= i; j++) {
			for (long long k = 0; k < 5000; k++) {
				dp[i + 1][j + 1][k + x[i]] += dp[i][j][k];
				dp[i + 1][j][k] += dp[i][j][k];
			}
		}
	}
	for (int i = 1; i <= n; i++)cnt += dp[n][i][a*i];
	cout << cnt << endl;
	endre;
}