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<cstdio>
#include<cstdlib>
#include<vector>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<string>
#include<cmath>
using namespace std;
typedef long long ll;
const int maxn = 5e5 + 5;
const int inf = 0x3f3f3f3f;
ll dp[55][55][3010],a[maxn];
int main() {
    ll n,ave;
    cin>>n>>ave;
    dp[0][0][0]=1;
    for(int i=1;i<=n;i++)
    cin>>a[i];
    for(int i=1;i<=n;i++)
    {
        for(int j=i;j>=0;j--)
        {
            for(int k=3000;k>=0;k--)
            {
                dp[i][j][k]+=dp[i-1][j][k];
                if(k>=a[i]&&j>=1) dp[i][j][k]+=dp[i-1][j-1][k-a[i]];
            }
        }
    }
    ll ans=0;
    for(int i=1;i<=n;i++)
    {
        ans+=dp[n][i][ave*i];
    }
    cout<<ans;
    return 0;
}