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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>

int gcd(int a, int b) {
  int v0 = a, v1 = b, v2 = a % b;
  while (v2 != 0) {
    v0 = v1;
    v1 = v2;
    v2 = v0 % v1;
  };
  return v1;
}

int lcm(int a, int b) {
  return a * b / gcd(a, b);
}


int main(void){
    char str[11];
    int a[100];
    int value;
    float v1;

    int n;
    scanf("%d",&n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
    }
    
    value = a[0];
    for (int i = 1; i < n; i++) {
      value = lcm(value, a[i]);
    }
    // printf("%d", value);
    int sum = 0;
    for (int i = 0; i < n; i++) {
        sum = sum + value / a[i];
    } 
    v1 = (float)((float)value / (float)sum);
    printf("%e", v1);    
    return 0;
}