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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

long long memo[50][50][50];
int x[51]={};

long long dp(int, int, int);

int main(){
    int N, A;
    long long sum=0;
    
    scanf("%d %d", &N, &A);
    for (int i=1; i<=N; i++) scanf("%d", &x[i]);
    
    for (int i=0; i<50; i++){
        for (int j=0; j<50; j++){
            for (int k=0; k<50; k++){
                memo[i][j][k] = -1;
            }
        }
    }
    
    for (int k=1; k<=N; k++){
        sum += dp(N, k, k*A);
    }
    
    printf("%lld\n", sum);
    
    return 0;
}

long long dp(int j, int k, int s){
    if (memo[j][k][s] != -1) return memo[j][k][s];
    long long result;
    if (j==0 && k==0 && s==0) result = 1;
    else if (j>=1 && x[j]>s) result = dp(j-1, k, s);
    else if (j>=1 && k>=1 && x[j]<=s) result = dp(j-1, k, s)+dp(j-1, k-1, s-x[j]);
    else result = 0;
    memo[j][k][s] = result;
    return result;
}