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 <bits/stdc++.h>

using namespace std;

long long n, s;

int main() {
    cin >> n >> s;
    for (int i = 2; 1LL * i * i <= n; ++i) {
        long long val = 0;
        long long x = n;
        while (x > 0) {
            val += x % i;
            x /= i;
        }
        if (val == s) {
            cout << i << endl;
            return 0;
        }
    }
    if (s > n) {
        cout << -1 << endl;
        return 0;
    }
    long long result = 1e18+7;
    for (int i = 1; 1LL * i * i <= n - s; ++i) {
        if ((n - s) % i == 0) {
            long long val;
            {
                val = i + 1;
                if (n / val < val && (val - 1) * (n / val) == n - s) {
                    result = min(result, val);
                }
            }
            {
                val = (n - s) / i + 1;
                if (n / val < val && (val - 1) * (n / val) == n - s) {
                    result = min(result, val);
                }
            }
        }
    }
    if (result <= n) {
        cout << result << endl;
        return 0;
    }
    cout << -1 << endl;
    return 0;
}