Python/project_euler/problem_800/sol1.py
Maxim Smolskiy d66e1e8732
Add Project Euler problem 800 solution 1 ()
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2023-04-01 17:48:13 +05:30

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Python

"""
Project Euler Problem 800: https://projecteuler.net/problem=800
An integer of the form p^q q^p with prime numbers p != q is called a hybrid-integer.
For example, 800 = 2^5 5^2 is a hybrid-integer.
We define C(n) to be the number of hybrid-integers less than or equal to n.
You are given C(800) = 2 and C(800^800) = 10790
Find C(800800^800800)
"""
from math import isqrt, log2
def calculate_prime_numbers(max_number: int) -> list[int]:
"""
Returns prime numbers below max_number
>>> calculate_prime_numbers(10)
[2, 3, 5, 7]
"""
is_prime = [True] * max_number
for i in range(2, isqrt(max_number - 1) + 1):
if is_prime[i]:
for j in range(i**2, max_number, i):
is_prime[j] = False
return [i for i in range(2, max_number) if is_prime[i]]
def solution(base: int = 800800, degree: int = 800800) -> int:
"""
Returns the number of hybrid-integers less than or equal to base^degree
>>> solution(800, 1)
2
>>> solution(800, 800)
10790
"""
upper_bound = degree * log2(base)
max_prime = int(upper_bound)
prime_numbers = calculate_prime_numbers(max_prime)
hybrid_integers_count = 0
left = 0
right = len(prime_numbers) - 1
while left < right:
while (
prime_numbers[right] * log2(prime_numbers[left])
+ prime_numbers[left] * log2(prime_numbers[right])
> upper_bound
):
right -= 1
hybrid_integers_count += right - left
left += 1
return hybrid_integers_count
if __name__ == "__main__":
print(f"{solution() = }")