Problem:
Score : 200 points

Problem Statement

Given is a sequence of N integers A_1, \ldots, A_N.

Find the (multiplicative) inverse of the sum of the inverses of these numbers, \frac{1}{\frac{1}{A_1} + \ldots + \frac{1}{A_N}}.

Constraints

1 \leq N \leq 100

1 \leq A_i \leq 1000

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \ldots A_N

Output

Print a decimal number (or an integer) representing the value of \frac{1}{\frac{1}{A_1} + \ldots + \frac{1}{A_N}}.

Your output will be judged correct when its absolute or relative error from the judge's output is at most 10^{-5}.

Sample Input 1

2
10 30

Sample Output 1

7.5

\frac{1}{\frac{1}{10} + \frac{1}{30}} = \frac{1}{\frac{4}{30}} = \frac{30}{4} = 7.5.

Printing 7.50001, 7.49999, and so on will also be accepted.

Sample Input 2

3
200 200 200

Sample Output 2

66.66666666666667

\frac{1}{\frac{1}{200} + \frac{1}{200} + \frac{1}{200}} = \frac{1}{\frac{3}{200}} = \frac{200}{3} = 66.6666....

Printing 6.66666e+1 and so on will also be accepted.

Sample Input 3

1
1000

Sample Output 3

1000

\frac{1}{\frac{1}{1000}} = 1000.

Printing +1000.0 and so on will also be accepted.

Solution in Scala (2.11.7) (judge verdict: Accepted)


import java.util.Scanner

import scala.collection.immutable


object Main {

  def read(): (Int, IndexedSeq[Int]) = {
    val sc: Scanner = new Scanner(System.in)
    val n = sc.nextInt
    val arr = (1 to n).map { _ =>
      sc.nextInt
    }
    sc.close()
    (n, arr)
  }


  def solve(n: Int, arr: IndexedSeq[Int]): Double = {
    val reversedNums = for(i <- arr) yield {
      1.toDouble / i
    }
    1 / reversedNums.sum
  }


  def main(args: Array[String]): Unit = {
    val (a, s) = read()
    println(solve(a, s))
  }

}