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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using vi = vector<int>;
using pi = pair<int,int>;

int MOD = 1e9+7;

// ll po(ll b, ll p){ return !p?1:po(b*b%MOD,p/2)*(p&1?b:1)%MOD;}
// ll pomd(ll a, ll k, ll md) { ll p; if (k == 0){ return 1; } p = pomd(a, k / 2, md); p = p * p % md; if (k % 2){ p = p * a % md; } return p; }
// vi arr(N); for(int i{1}; i<N; ++i) for(int j=i; j<N; j+=i) d[j]++; // counting divisors
//bool isPrime(ll n) { if (n <= 1) return false; if (n <= 3) return true; if (n % 2 == 0 || n % 3 == 0) return false; for (ll i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true; }

#define mp make_pair
#define pb push_back
#define rsz resize
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define trav(a, x) for(auto& a : x)
#define rot(s, k) rotate(s.begin(), s.begin()+ s.size()-(k%s.size()), s.end()); //+ve = right, -ve = left

void setIO(string name = ""){
    freopen((name+".in").c_str(), "r", stdin); 
    freopen((name+".out").c_str(), "w", stdout);
}	

int main(){
 	ios_base::sync_with_stdio(0); cin.tie(0); 
	// setIO();
	int n; cin >> n;
	cout << (n%2==0?double((n/2))/n:double(((n/2))+1)/n) << "\n";

}