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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.io.*;
class Main {
    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int N = Integer.parseInt(st.nextToken());
        int A = Integer.parseInt(st.nextToken());
        st = new StringTokenizer(br.readLine());
        int[] cards = new int[N];
        for(int i=0;i<N;i++){
            cards[i]=Integer.parseInt(st.nextToken());
        }

        int[][][] dp = new int[52][52][2502];
        dp[0][0][0]=1;
        for(int i=1; i<=N; i++){
            for(int j=0; j<=i; j++){
                for(int k=0; k<=2501; k++){
                    dp[i][j][k] += dp[i-1][j][k];
                    if(j-1>=0 && k-cards[i-1]>=0)
                        dp[i][j][k] += dp[i-1][j-1][k-cards[i-1]];
                }
            }
        }
        long ans = 0;
        for(int j=1; j<=N; j++){
            ans+=dp[N][j][A*j];
        }
        System.out.println(ans);
    }
}