Problem: The number 105 is quite special - it is odd but still it has eight divisors. Now, your task is this: how many odd numbers with exactly eight positive divisors are there between 1 and N (inclusive)?

Submission attempts (in order):
import math

n = int(input())

cnt = 0
def search_divisor_num_1(num):
    if num < 0:
        return None
    elif num == 1:
        return 1
    else:
        num_sqrt = math.floor(math.sqrt(num))
        prime_list = make_prime_list_2(num_sqrt)

        divisor_num = 1
        for prime in prime_list:
            count = 1
            while num % prime == 0:
                num //= prime
                count += 1
            divisor_num *= count

        if num != 1:
            divisor_num *= 2

        return divisor_num

for i in range(n+1):
    if i%2==0 and search_divisor_num_1(i) == 8:
        cnt += 1

print(cnt)

n = int(input())

cnt = 0
def search_divisor_num_1(num):
    if num < 0:
        return None
    elif num == 1:
        return 1
    else:
        num_sqrt = math.floor(math.sqrt(num))
        prime_list = make_prime_list_2(num_sqrt)

        divisor_num = 1
        for prime in prime_list:
            count = 1
            while num % prime == 0:
                num //= prime
                count += 1
            divisor_num *= count

        if num != 1:
            divisor_num *= 2

        return divisor_num

for i in range(n):
    if (i+1)%2==0 and search_divisor_num_1(i+1) == 8:
        cnt += 1

print(cnt)

n = int(input())

cnt = 0
def search_divisor_num_1(num):
    if num < 0:
        return None
    elif num == 1:
        return 1
    else:
        num_sqrt = math.floor(math.sqrt(num))
        prime_list = make_prime_list_2(num_sqrt)

        divisor_num = 1
        for prime in prime_list:
            count = 1
            while num % prime == 0:
                num //= prime
                count += 1
            divisor_num *= count

        if num != 1:
            divisor_num *= 2

        return divisor_num

for i in range(n+1):
    if i%2==0 and search_divisor_num_1(i) == 8:
        cnt += 1

print(cnt)

import math

n = int(input())

cnt = 0

def make_prime_list_2(num):
    if num < 2:
        return []

    
    prime_list = [i for i in range(num + 1)]
    prime_list[1] = 0 
    num_sqrt = math.sqrt(num)

    for prime in prime_list:
        if prime == 0:
            continue
        if prime > num_sqrt:
            break

        for non_prime in range(2 * prime, num, prime):
            prime_list[non_prime] = 0

    return [prime for prime in prime_list if prime != 0]

def search_divisor_num_1(num):
    if num < 0:
        return None
    elif num == 1:
        return 1
    else:
        num_sqrt = math.floor(math.sqrt(num))
        prime_list = make_prime_list_2(num_sqrt)

        divisor_num = 1
        for prime in prime_list:
            count = 1
            while num % prime == 0:
                num //= prime
                count += 1
            divisor_num *= count

        if num != 1:
            divisor_num *= 2

        return divisor_num

for i in range(n+1):
    if i%2==1 and search_divisor_num_1(i) == 8:
        cnt += 1

print(cnt)

Verdicts: Runtime Error

Runtime Error

Runtime Error

Accepted