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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

int isprime(long N){
    int m=N/2;
    for(int i = 2; i <= m; i++)
    {
        if(N % i == 0)
        {
            return 0;
        }
    }
    return 1;
}

int main()
{
    long num1,num2;
    int counter=0;
    cin>>num1>>num2;
    vector <long> cDivisors;

    long x = __gcd(num1,num2);
    for(int i=2; i<=x; i++){
        if((num1%i==0)&&(num2%i==0)&&isprime(i))
            cDivisors.push_back(i);
    }

    cout<<cDivisors.size()+1<<endl;
    /*for(int i=1; i<60; i++){
        if(60%i==0){
            cout<<i<<endl;
            cDivisors.push_back(i);
        }
    }
    for(int i=1; i<cDivisors.size()-1; i++){
        for(int j=i+1; j<cDivisors.size(); j++){
            //cout<<cDivisors[i]<<"\t"<<cDivisors[j]<<"\t"<<__gcd(cDivisors[i],cDivisors[j])<<endl;
            if((isprime(cDivisors[i])&&cDivisors[i]%cDivisors[j]!=0))
            counter++;
        }
    }
    cout<<counter<<endl;*/
}