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<map>
#include<iostream>
#include<deque>
#include<algorithm>
#include<string>
#include<cctype>
#include<iomanip>
#include<vector>
#include<queue>
#include<math.h>
#include<bitset>
#include<random>
#include<set>
#include <cstdint>
#include<tuple>
using namespace std;
using ll = __int64_t;
using lll = __int128_t;
typedef pair<ll,ll> P;
#define REP(i,b,e) for(ll i=(ll)b;i<(ll)e;i++)
#define rep0(i,n) REP(i,0ll,n)
#define rep1(i,n) REP(i,1ll,n+1)
const ll longinf=1ll<<60;
const int inf=(1<<30)-1;
const ll mod=1e9+7;

int main(void){
  double n;
  cin>>n;
  if((int)n%2==1){
    cout<<setprecision(10)<<((n+1.0)/2.0)/n<<endl;
  }
  else{
    cout<<setprecision(10)<<1.0/2.0<<endl;
  }
  
  return 0;
}