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 (GCC 5.4.1) (judge verdict: Accepted)


  
 #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define INF 1e9+7
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define all(x) (x).begin(),(x).end()
ll min(ll a,ll b){
  if(a>b){
    return b;
    }
      else{
        return a;
      }
  }
  
  ll max(ll a,ll b){
    if(a>b){
      return a;
    }
    else{
      return b;
    }
  }

ll power(ll a,ll b){
 ll ans=1;
 for(ll i=1;i<=b;i++){
ans*=a;
  }
  return ans;
}

 ll gcd(ll m,ll n){
 	if(n==0){
       return m;
     }
   return gcd(n,m%n);
 }
ll lcm(ll a,ll b){
  ll g=gcd(a,b);
  return a/gcd(a,b)*b;
		

}
int main(){
ll N;
  cin>>N;
  
  vector<ll>A(N);
   rep(i,N){
    cin>>A[i];
  }
  
  double P=0;
  
  ll S=A[0];
 for(ll i=1;i<N;i++){
   S=lcm(S,A[i]);
   
  }
  
  
 
  
  rep(i,N){
    P+=S/A[i];
          }
    
  double ans=S*1.0/P;
  
  printf("%.14lf",ans);
 
}