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>
#define pb push_back
#define f first
#define s second
#define ind(a) scanf("%d", &a)
#define inlld(a) scanf("%lld", &a)
#define ind2(a, b) scanf("%d%d", &a, &b)
#define inlld2(a, b) scanf("%lld%lld", &a, &b)
#define ind3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define inlld3(a, b, c) scanf("%lld%lld%lld", &a, &b, &c)

using namespace std;

const int N=55;
const int MOD=1e9+7;

typedef long long ll;
typedef long double ld;

ll n, req, arr[N], dp[N][N][N*N];

ll solve(ll i, ll num_taken, ll sum_taken)
{
    if(i==n+1)
    {
        if(num_taken==0)
            return 0;
        if(sum_taken%num_taken)
            return 0;
        if(sum_taken/num_taken==req)
            return 1;
        return 0;
    }
    if(dp[i][num_taken][sum_taken]!=-1)
        return dp[i][num_taken][sum_taken];
    ll ret=0;
    // if(num_taken!=0 && sum_taken%num_taken==0 && sum_taken/num_taken==req)
    //     ret=1;
    ret+=solve(i+1, num_taken+1, sum_taken+arr[i]);
    ret+=solve(i+1, num_taken, sum_taken);
    return dp[i][num_taken][sum_taken]=ret;
}

int main() 
{
    inlld2(n, req);
    for(ll a=1; a<=n; a++)
        inlld(arr[a]);
    memset(dp, -1, sizeof(dp));
    printf("%lld\n", solve(1, 0, 0));
    return 0;
}