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<algorithm>
#include<vector>
#include<stack>
#include <cmath>
#include <limits>
#include<set>
#include <iomanip>
#include <queue>
#pragma GCC optimize("Ofast")
using namespace std;
const long long INF=20000000000;
//setprecision(2) <<

const int MOD = 1000000007;

long long max(long long a,long long b){
	if(a<b)return b;
	else return a;	
}
long long min(long long a,long long b){
	if(a>b)return b;
	else return a;	
}

int A[100006];
int N;

int f(double n){
	int ret=0;
	bool a = (n<A[0]);
	for(int i=0;i<N-1;i++){
		if((A[i]-n)*(A[i+1]-n)<0){
			if(a){
				ret++;
				a=false;
			}
			else a=true;
		}
	}
	if(a)ret++;
	//cout<<n<<" "<<ret<<endl;
	return ret;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int N;
	cin>>N;
	double ans=(double)((int)N/2)/N;
	if(N%2==0)cout<<0.5<<endl;
	else cout<<(double)(1-ans)<<endl;
}