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;
const int N=55;
long long dp[N][N*N];      //dp值會爆int
int main()
{
    int n,a;
    cin>>n>>a;
    dp[0][0]=1;
    for(int i=1;i<=n;i++)
    {
        int x;
        cin>>x;
        for(int j=i-1;j>=0;j--)    //每一次個數向下減少
            for(int k=0;k<=N*j;k++)    //優化時間  由於k最大為N*j次
                dp[j+1][k+x]+=dp[j][k];       //個數加1 則找到前面個數的值相加
    }
    long long ans=0;       
    for(int i=1;i<=n;i++)
            ans+=dp[i][i*a];    //累計所有個數會出現平均值滿足a的情況
    cout<<ans<<endl;
    return 0;
}