For this Question: You are given resistance value of N resistors. Find the net resistance of the system when all of these N resistors are connected in parallel.

If there are three resistors A1, A2, A3, when they are connected in parallel, the net resistance will be 1/((1/A1) + (1/A2) + (1/A3))

Since this number can also have a fraction part, you only have to print the floor of the result obtained.
For example, if value of 1/((1/A1) + (1/A2) + (1/A3)) if 7.54567, you only have to print 7.First line contains a single integer N denoting the number of resistors. 
Next line contains N space separated integers containing the value of different resistors.

Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^9Print the integral part or floor of the value obtained from the formula 1/((1/A1) + (1/A2) + ..... + (1/AN)).Input
2
10 30

Output
7

1/((1/10) + (1/30)) = 30/4 = 7.5 and floor of 7.5 is 7, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
	public static void main (String[] args)throws IOException {
                   BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
					int N = Integer.parseInt(br.readLine());
					double arr[]  = new double[N];
					String str[] = br.readLine().trim().split(" ");
					for(int i=0;i<N;i++)
					arr[i]=Integer.parseInt(str[i]);
					double resistance=0;
					int equResistance=0;
					for(int i=0;i<N;i++)
					arr[i]=Integer.parseInt(str[i]);
					for(int i=0;i<N;i++)
					{
						resistance=resistance+(1/arr[i]);
					}
					equResistance = (int)Math.floor((1/resistance));
					System.out.println(equResistance);
	}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?