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 C++14 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <limits>
#include <string>
#include <algorithm>
#include <bitset>
#include <unordered_set>
#include <unordered_map>
#include <map>
#include "math.h"
#include "assert.h"

using namespace std;

#define ll long long

#define BUFF ios::sync_with_stdio(false)
#define epsilon 1e-07
#define pi 2 * acos(0.0);
#define mod 1000000007   // 1e9+7
#define cerr if(false)cerr
#define watch(x) cerr << #x << " = " << x << endl;

#define inf std::numeric_limits<int>().max();

int main() {
	int n;
	cin >> n;
	cout.precision(10);
	vector<double> v(n);
	for(int i=0; i<n; i++) {
		cin >> v[i];
	}
	for(int i=0; i<n; i++) {
		v[i] = (1 / v[i]);
	}
	double ans = 0;
	for(int i=0; i<n; i++) {
		ans += v[i];
	}
	cout << (1/ans) << endl;
}