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Add Project Euler problem 800 solution 1 (TheAlgorithms#8567)
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* [Sol1](project_euler/problem_587/sol1.py)
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* Problem 686
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* [Sol1](project_euler/problem_686/sol1.py)
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* Problem 800
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* [Sol1](project_euler/problem_800/sol1.py)
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## Quantum
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* [Bb84](quantum/bb84.py)

Diff for: project_euler/problem_800/__init__.py

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Diff for: project_euler/problem_800/sol1.py

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

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