"""
Consider quadratic Diophantine equations of the form:
    x^2 - D y^2 = 1
For example, when D = 13 , the minimal solution in x is $649^2 - 13 * 180^2 = 1$ .
It can be assumed that there are no solutions in positive integers when D is square.
By finding minimal solutions in x for D={2,3,5,6,7} , we obtain the following:
    3^2 - 2 * 2^2 = 1
    2^2 - 3 * 2^2 = 1
    9^2 - 5 * 4^2 = 1
    5^2 - 6 * 2^2 = 1
    8^2 - 7 * 3^2 = 1
Hence, by considering minimal solutions in x for D <= 7 ,
the largest x is obtained when D = 5 .
Find the value of D <= 1000 in minimal solutions of x for
which the largest value of x is obtained.
"""

import time
from functools import wraps
from math import isqrt
from typing import Any, Callable, TypeVar

F = TypeVar("F", bound=Callable[..., Any])


def benchmark(repeat: int = 1) -> Callable[[F], F]:
    if repeat < 1:
        raise ValueError("repeat >= 1")

    def decorator(func: F) -> F:
        @wraps(func)
        def wrapper(*args: Any, **kwargs: Any) -> Any:
            total = 0.0
            result = None

            for _ in range(repeat):
                start = time.perf_counter()
                result = func(*args, **kwargs)
                end = time.perf_counter()
                total += end - start

            wrapper.avg_time = total / repeat  # type: ignore[attr-defined]
            wrapper.total_time = total  # type: ignore[attr-defined]

            print(
                f"[Benchmark] {func.__name__} | repeated {repeat} times | "
                f"average: {wrapper.avg_time:.6f}s | total: {wrapper.total_time:.6f}s"  # type: ignore[attr-defined]
            )
            return result

        wrapper.avg_time = 0.0  # type: ignore[attr-defined]
        wrapper.total_time = 0.0  # type: ignore[attr-defined]
        return wrapper  # type: ignore[return-value]

    return decorator


def diophantine_eq(D: int) -> int | None:
    a = [int(isqrt(D))]
    if a[0] * a[0] == D:
        return None
    p = [0, 1, int(isqrt(D))]
    m = 0
    d = 1
    while True:
        m = d * a[-1] - m
        d = (D - m * m) // d
        a.append((a[0] + m) // d)
        p.append(p[-1] * a[-1] + p[-2])
        if a[-1] == 2 * a[0]:
            if len(a) % 2 == 1:
                return p[-2]
            else:
                for ai in a[1:-1]:
                    p.append(p[-1] * ai + p[-2])
                return p[-1]


@benchmark(repeat=10)
def find_minimal_solution(limit: int) -> int:
    min_x: dict[int, int] = {}
    for D in range(2, limit + 1):
        x = diophantine_eq(D)
        if x is not None:
            min_x[D] = x
    return max(min_x, key=min_x.__getitem__)


if __name__ == "__main__":
    max_D = 1000
    dd = find_minimal_solution(max_D)
    print(f"D <= {max_D} |-> D of max x = {dd}")