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 <iostream>
#include <string>
#include <map>
#include <queue>
#include <algorithm>
#include <vector>
#include <math.h>
#include <bitset>
#include <cmath>
#include <set>
#include <queue>

typedef long long ll;
#define rep(i,s,n)for(ll i = s;i<n;i++)
#define repe(i,s,n)for(ll i = s;i<=n;i++)
#define rep_r(i,s,n)for(ll i = n;i>s;i--)
#define rep_re(i,s,n)for(ll i = n;i>=s;i--)
using namespace std;
static const ll MOD = 1e9 + 7;
static const ll INF = 1e25;
static const ll MAX_N = 100001;

int main() {
	ll a; cin >> a;
	if (a == 1) {
		cout << 1 << endl;
		return 0;
	}
	double ans = 0;
	if (a & 1) {
		double p = a;
		ans = ((p - 1) / 2 + 1) / p;
		//ans = (double)((p / 2 + 1) / p);
	}
	else {
		double p = a;
		ans = (double)p / 2 / p;
	}
	printf("%0.8f", ans);
	return 0;
}