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: Wrong Answer)

#include <iostream>
#include <stdio.h>

#define maxn 51
#define maxsum 5000
using namespace std;
long long N, A;
// dp[i][j][k] = j carti din primele i carti care dau suma k
// dp[1][0][0] = 1
// dp[i][j][k] = dp[i - 1][j][k] + dp[i - 1][j - 1][k - v[i]]
long long dp[maxn][maxn][maxsum];
int v[maxn];
int main()
{
    int mini = 100;
    long long rez = 0;
    scanf("%d %d", &N, &A);
    for (int i = 1; i <= N; ++i) {
        scanf("%d", &v[i]);
        mini = min(v[i], mini);
        // dp[i][1][v[i]] = 1;
        dp[i][0][0] = 1;
    }
    dp[0][0][0] = 1;
    for (int i = 1; i <= N; ++i) {
        for (int j = 1; j <= i; ++j) {
            for (int k = mini; k <= maxsum; ++k) {
                dp[i][j][k] = dp[i - 1][j][k] + dp[i - 1][j - 1][k - v[i]];
            }
        }
    }
    for (long long i = 1; i <= N; ++i) {
        rez += dp[N][i][i * A];
    }
    printf("%lld", rez);
    return 0;
}