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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;

class Program
{
    static void Main(string[] args)
    {
        var na = Console.ReadLine().Split();
        var n = int.Parse(na[0]);
        var a = int.Parse(na[1]);

        var s = Console.ReadLine().Split();
        var x = new int[n + 1];
        for (int i = 0; i < n; i++)
        {
            x[i + 1] = int.Parse(s[i]);
        }

        var dp = new long[n + 1, n + 1, a * n + 1];
        dp[0, 0, 0] = 1;

        for (int i = 1; i < dp.GetLength(0); i++)
        {
            for (int j = 0; j < dp.GetLength(1); j++)
            {
                for (int k = 0; k < dp.GetLength(2); k++)
                {
                    if (0 < j && x[i] <= k)
                    {
                        dp[i, j, k] = dp[i, j, k] = dp[i - 1, j, k] + dp[i - 1, j - 1, k - x[i]];
                    }
                    else
                    {
                        dp[i, j, k] = dp[i, j, k] = dp[i - 1, j, k];
                    }
                }
            }
        }

        var ans = 0L;
        for (int j = 1; j <= n; j++)
        {
            ans += dp[n, j, a * j];
        }

        Console.WriteLine(ans);
    }
}