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>
#include <boost/multiprecision/cpp_int.hpp>

#define repd(i, a, b) for (ll i = (a); i < (b); i++)
#define rep(i, n) repd(i, 0, n)

using namespace std;

namespace mp = boost::multiprecision;
using Bint = mp::cpp_int;

using ll = long long;
ll mod = pow(10, 9) + 7;

template <typename T> void output(T, ll);
ll gcd(ll, ll);
ll lcm(ll, ll);
Bint comb(Bint, Bint);
void prime_map(map<ll, ll> &, ll);

int main() {
    // source
    int n;
    cin >> n;
    int a[n];
    rep(i, n) { cin >> a[i]; }
    double mother = 0;
    rep(i, n) { mother += (double)1 / a[i]; }

    double ans = (double)1 / mother;

    cout << ans << endl;
    return 0;
}

Bint comb(Bint a, Bint b) {
    Bint ans = 1;

    if (a < b) {
        return 0;
    }

    rep(i, b) {
        ans *= a - i;
        ans /= i + 1;
    }

    return ans;
}

template <typename T> void output(T a, int precision) {
    if (precision > 0) {
        cout << setprecision(precision) << a << "\n";
    } else {
        cout << a << "\n";
    }
}

template <typename T> T min(T a, T b) {

    if (a < b) {
        return a;
    } else {
        return b;
    }
}

ll gcd(ll a, ll b) {
    if (b == 0) {
        return a;
    }
    return gcd(b, a % b);
}

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

void prime_map(map<ll, ll> &m, ll num) {
    ll cache = num;
    repd(i, 2, sqrt(num) + 2) {
        if (num % i == 0) {
            if (m[i] == 0) {
                m[i] = 1;
            } else {
                m[i]++;
            }

            num /= i;
            break;
        }
    }

    if (cache == num) {
        if (m[num] == 0) {
            m[num] = 1;
        } else {
            m[num]++;
        }
        return;
    } else {
        prime_map(m, num);
    }
}