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 <bits/stdc++.h>
using namespace std;
long long n,ave;
long long a[55];
long long dp[55][55][2501];
int main(void)
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin>>n>>ave;
    long long sum=0;
    for(long long i=1;i<=n;++i)
    {
        dp[i][0][0]=1;
        cin>>a[i];
        sum+=a[i];
    }
    dp[0][0][0]=1;
    for(long long i=1;i<=n;i++)
    {
        for(long long j=1;j<=i;j++)
        {
            for(long long k=1;k<=sum;k++)
            {
                if(k<a[i])
                {
                     dp[i][j][k]=dp[i-1][j][k];
                }
                else
                {
                    dp[i][j][k]=dp[i-1][j][k]+dp[i-1][j-1][k-a[i]];
                }
            }
        }
    }
    long long  ans=0;
    for(long long j=1;j<=n&&j*ave<=sum;j++)
    {
            ans+=dp[n][j][j*ave];
    }
    cout << ans <<endl;
    return 0;
}