Add Project Euler problem 800 solution 1 (#8567)

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project_euler/problem_800/sol1.py

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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)
+"""
+
+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() = }")