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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

# C - 高橋君とカード / Tak and Cards

#import numpy as np
import math

N, A = map(int, input().split())
#x = np.array(list(map(int, input().split())))
x = list(map(int, input().split()))

# 数列aと整数Aを与えると、aからいくつか取って合計Aにするパターンが何通りあるかを返す関数
def partial_sum(a,A):
    n = len(a)
    dp = [[0]*(A+2) for _ in range(n+2)] # dp[i+1][j] a[i]までの数からいくつか選んで総和をjにするパターンの数
    dp[0][0] = 1
    
    for i in range(n):
        for j in range(A+1):
            if j >= a[i]:
                dp[i+1][j] = dp[i][j-a[i]] + dp[i][j]
            else:
                dp[i+1][j] = dp[i][j]
                
    return dp[n][A]

#x_diff = x - np.array(A) # ここから合計0になるように取る。負の数列についてpartial_sumが動かないことに注意
x_diff = [a - A for a in x]
x_diff_negative = [-a for a in x_diff if a < 0]
x_diff_positive = [a for a in x_diff if a > 0]

ans = 0
for i in range(1, sum(x)):
    ans += partial_sum(x_diff_negative,i) * partial_sum(x_diff_positive,i) 

# 0の選び方
n_pattern_0 = 0
#n_0 = np.count_nonzero(x_diff == 0)
n_0 = x_diff.count(0)
if n_0 > 0:
    for i in range(1, n_0+1):
        n_pattern_0 += math.factorial(n_0) // (math.factorial(n_0-i) * math.factorial(i))

# ここまで数えた全てのパターンそれぞれについて、x_diff中の0をいくつか含めた別パターンがある
# 0単体で選ぶ方法もある
ans += (ans+1) * n_pattern_0
    
print(ans)