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 Rust (1.15.1) (judge verdict: Accepted)

fn read<T: std::str::FromStr>() -> Vec<T> {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).ok();
    buf.split_whitespace()
        .map(|x| x.parse().ok().unwrap())
        .collect()
}

fn main() {
    let v: Vec<usize> = read();
    let n = v[0];
    let a = v[1] as i64;
    let x: Vec<i64> = read();

    let sum = x.iter().fold(0, |s, &t| s + t) as usize;
    let mut dp = vec![vec![0_usize; sum * 2 + 1]; n + 1];
    dp[0][sum] = 1;
    for i in 1..(n + 1) {
        for j in 0..(sum * 2 + 1) {
            dp[i][j] = dp[i - 1][j];
            let tmp = j as i64 - (a - x[i - 1]);
            if tmp >= 0 && tmp <= (sum * 2) as i64 {
                dp[i][j] += dp[i - 1][tmp as usize];
            }
        }
    }
    println!("{}", dp[n][sum] - 1);
}