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Add Project Euler problem 800 solution 1 (TheAlgorithms#8567)
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""" | ||
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) | ||
""" | ||
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from math import isqrt, log2 | ||
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def calculate_prime_numbers(max_number: int) -> list[int]: | ||
""" | ||
Returns prime numbers below max_number | ||
>>> calculate_prime_numbers(10) | ||
[2, 3, 5, 7] | ||
""" | ||
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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 | ||
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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: | ||
""" | ||
Returns the number of hybrid-integers less than or equal to base^degree | ||
>>> solution(800, 1) | ||
2 | ||
>>> solution(800, 800) | ||
10790 | ||
""" | ||
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upper_bound = degree * log2(base) | ||
max_prime = int(upper_bound) | ||
prime_numbers = calculate_prime_numbers(max_prime) | ||
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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 | ||
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return hybrid_integers_count | ||
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if __name__ == "__main__": | ||
print(f"{solution() = }") |