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;
using System.Linq;

namespace AtcoderCsharp
{
    class Program
    {
        static void Main(string[] args)
        {
            long[] NA = Console.ReadLine().Split().Select(long.Parse).ToArray();
            long[] cards = Console.ReadLine().Split().Select(long.Parse).ToArray();
            long[][][] dp = new long[NA[0]+1][][];
            for (long i = 0; i < NA[0] + 1; i++)
            {
                dp[i] = new long[NA[0]+1][];
                for(long j = 0;j< NA[0]+1;j++)
                dp[i][j] = new long[NA[0] * NA[1]+1];
            }
            dp[0][0][0] = 1;
            for (long i = 1; i < NA[0]+1; i++)
            {
                for (long j = 0; j < NA[0] + 1; j++)
                {
                    //i枚目まで見て、j枚使って、3番目の数を作る方法
                    for (long k = 0; k <= NA[0] * NA[1]; k++)
                    {
                        try
                        {
                            dp[i][j][k] += dp[i - 1][j][k];
                        }
                        catch { }
                        try
                        {
                            dp[i][j][k] += dp[i - 1][j - 1][k - cards[i - 1]];
                        }
                        catch { continue; }
                    }
                }
            }
            long output = 0;
            for (long i = 0; i <= NA[0]; i++)
            {
                output += dp[NA[0]][i][i * NA[1]];
            }
            Console.WriteLine(output - 1);
        }
    }
}