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 Scala (2.11.7) (judge verdict: Wrong Answer)

import scala.io.StdIn

object Main {

  def main(args: Array[String]): Unit = {

    val Array(n, a) = readIntArray()
    val xs = readIntArray().map(_.toLong).zipWithIndex.toVector

    val alreadyComputed = collection.mutable.Set.empty[Seq[(Long, Int)]]

    def solve(ys: Seq[(Long, Int)]): Long = {
      ys match {
        case Nil =>
          0L
        case head +: Nil =>
          if (head._1 == a && !alreadyComputed(Seq(head))) 1L else 0L
        case head +: tail =>
//          println(s"--- head : ${head}")
          var count = 0
          (1 to tail.length).foreach { i =>
            tail.combinations(i).foreach { cb =>
//              println(s"${i} : ${cb}")
              if ((cb.map(_._1).sum + head._1).toDouble / (1 + i) == a && !alreadyComputed(head +: cb)) {
                count += 1
                alreadyComputed += (head +: cb)
              } else if (cb.map(_._1).sum.toDouble / i == a && !alreadyComputed(cb)) {
                count += 1
                alreadyComputed += cb
              }
            }
          }
//          println(s"::: count : ${count}")
          solve(tail) + count
      }
    }

    println(solve(xs))
  }

  private def readIntArray(): Array[Int] =
    StdIn.readLine().trim.split(' ').filter(_.nonEmpty).map(_.toInt)
}