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 IOS ios::sync_with_stdio(false); cin.tie(0);
#define FOR(i,s,n) for(int i = s; i < (n); i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(n) (n).begin(), (n).end()
#define RALL(n) (n).rbegin(), (n).rend()
#define ATYN(n) cout << ( (n) ? "Yes":"No") << endl;
#define CFYN(n) cout << ( (n) ? "YES":"NO") << endl;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int,int>;

int main(void)
{
    IOS
    int n; cin >> n;
    double cnt = 0;
    FOR(i,1,n+1) if(i % 2 == 1) cnt += 1;
    double ans = cnt / n;
    cout << fixed << setprecision(10) << ans << endl;

    return 0;
}