Problem:
Score : 100 points

Problem Statement

Given is an integer N.

Takahashi chooses an integer a from the positive integers not greater than N with equal probability.

Find the probability that a is odd.

Constraints

1 \leq N \leq 100

Input

Input is given from Standard Input in the following format:

N

Output

Print the probability that a is odd.
Your output will be considered correct when its absolute or relative error from the judge's output is at most 10^{-6}.

Sample Input 1

4

Sample Output 1

0.5000000000

There are four positive integers not greater than 4: 1, 2, 3, and 4. Among them, we have two odd numbers: 1 and 3. Thus, the answer is \frac{2}{4} = 0.5.

Sample Input 2

5

Sample Output 2

0.6000000000

Sample Input 3

1

Sample Output 3

1.0000000000

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define pb push_back
#define ll long long
#define MX 1000009

bool marked[MX+1];

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

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

void sieve()
{
    marked[0] = true;
    marked[1] = true;
    for(int i = 4; i <= MX; i += 2)
        marked[i] = true;
    for(ll i = 3; i * i <= MX; i += 2)
    {
        for(ll j = i * i; j <= MX; j += i)
             marked[j] = true;
    }
}

int main()
{
    //cout << "Hello world!!" << endl;
 int n;
 cin >> n;
 int tot = n;
 int odd;
 if( n % 2 == 0)odd = n / 2;
 else odd = ( n / 2) + 1;
 double ans = (odd * 1.0) / tot;
 //cout << ans << endl;
printf("%0.10f\n", ans);

    return 0;
}