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 PyPy3 (2.4.0) (judge verdict: Runtime Error)

x,y = map(int,input().split())

#print(x,y)

def make_divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n//i)

    # divisors.sort()
    return divisors
x_li=(make_divisors(x))
y_li=(make_divisors(y))
z_li=(list(set(x_li)&set(y_li)))
#print(z_li)

import math
count=1
c=[]
"""
for n in z_li:
  if n == 1:
    count=count+1
    continue
  for k in range(2, int(math.sqrt(n)) + 1):
    print(n,k)
    if n % k == 0:
      break;
    else:
      c.append(n)
      count = count+1
print(c,count)
"""

def is_prime(n):
    if n == 1: return False

    for k in range(2, int(math.sqrt(n)) + 1):
        if n % k == 0:
            return False

    return True
  
  
for n in z_li:
  if(is_prime(n)==True):
    count=count+1
print(count)