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: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
 
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
 
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;

signed main() {
	IOS;
	int n;	cin >> n;
	double s = 0;
	for(int i = 1; i <= n; i++){
		double p;	cin >> p;
		s = s + (1/p);
	}	
	s = 1/s;
	cout << floor(s);
	return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?