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>
using namespace std;
typedef long long ll;
#define rep(i,n) for(int i=0;i<(n);i++)

int x[55];
ll dp[55][55][2550];

int main(void){
    int n, a; cin >> n >> a;
    rep(i, n){
        cin >> x[i]; 
    }

    rep(i, 55)rep(j, 55)rep(k, 2550) dp[i][j][k] = 0;
    dp[0][1][x[0]] = 1; dp[0][0][0] = 1;
    for (int i = 0; i < n - 1; ++i){
        for (int j = 0; j <= n; ++j){
            for (int k = 0; k <= 2500; ++k){
                if(dp[i][j][k] == 0) continue;
                //i + 1番目を選ぶ
                dp[i + 1][j + 1][k + x[i + 1]] += dp[i][j][k];
                //i + 1選ばない
                dp[i + 1][j][k] += dp[i][j][k];
            }
        }
    }

    ll ans = 0;
    for (int j = 1; j <= n; ++j){
        int aim = j * a;
        if(aim <= 2500 && dp[n - 1][j][aim] != 0){
            ans += dp[n - 1][j][aim];
        }
    }
    printf("%lld\n", ans);
    return 0;
}