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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long[] a = new long[n];
		for(int i = 0; i < n; i++){
			a[i] = sc.nextLong();
		}
		long num = lcm(a,n);
		//System.out.println(l);
		long deno = sum(a,n,num);
		//System.out.println(deno);
		double ans = num / (double)deno;
		System.out.println(ans);
	
	}
	
	public static long gcd(long a, long b){
		if(a == 0)
			return b;
		
		return gcd(b%a,a);
	}
	
	public static long lcm(long[] a, long n){
		long lc = 1;
		for(int i = 0; i < n; i++){
			lc =  (lc * a[i])/gcd(lc,a[i]);
		}
		return lc;
	}
	
	public static long sum(long[] a,int n,long num){
		long count = 0;
		for(int i = 0; i < n; i++){
			count += (num/a[i]);
		}
		return count;
	}
	
	
	
	}