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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdbool.h>
#include <math.h>

int main(void){

    long long N;
    long long A;

    scanf("%lld%lld", &N, &A);

    long long x[N+1];

    long long sum = 0;

    for(long long i = 1; i <= N; i++){
        scanf("%lld", &x[i]);
        sum += x[i];
    }

    long long dp[N+1][N+1][sum+1];

    for(long long j = 0; j <= N; j++){
        for(long long k = 0; k <= N; k++){
            for(long long s = 0; s <= sum; s++){
                if(j >= 1 && s < x[j]){
                    dp[j][k][s] = dp[j-1][k][s];
                }else if(j >= 1 && k >= 1 && s >= x[j]){
                    dp[j][k][s] = dp[j-1][k][s] + dp[j-1][k-1][s-x[j]];
                }else if(j == 0 && k == 0 && s == 0){
                    dp[j][k][s] = 1;
                }else{
                    dp[j][k][s] = 0;
                }
            }
        }
    }

    long long ans = 0;

    long long max;

    if(sum/A < N){
        max = sum/A;
    }else{
        max = N;
    }

    for(long long k = 1; k <= max; k++){
        ans += dp[N][k][k*A];
    }

    printf("%lld\n", ans);

    return 0;
}