"""
The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9
in some order, but it also has a rather interesting sub-string divisibility property.

Let d_1 be the 1st digit, d_2 be the 2nd digit, and so on. In this way, we note the following:

    d_2d_3d_4 is divisible by 2
    d_3d_4d_5 is divisible by 3
    d_4d_5d_6 is divisible by 5
    d_5d_6d_7 is divisible by 7
    d_6d_7d_8 is divisible by 11
    d_7d_8d_9 is divisible by 13
    d_8d_9d_10 is divisible by 17

Find the sum of all 0 to 9 pandigital numbers with this property.
"""

import time
from itertools import permutations
from tkinter.constants import TOP


def timer(func):
    def wrapper(*args, **kwargs):
        start_time = time.perf_counter()
        result = func(*args, **kwargs)
        end_time = time.perf_counter()
        print(f"Execution time: {end_time - start_time} seconds")
        return result

    return wrapper


def get_all_numbers() -> list[int]:
    ori = list(permutations(list(range(10))))
    res = []
    for x in ori:
        if x[3] % 2 != 0:
            continue
        if x[5] not in [0, 5]:
            continue
        for i, d in enumerate([2, 3, 5, 7, 11, 13, 17]):
            if int("".join(map(str, x[i + 1 : i + 4]))) % d != 0:
                break
        else:
            res.append(int("".join(map(str, x))))
    return res


def optimized_solve() -> tuple[int, list[int]]:
    total = 0

    # 从最后三位开始，这些必须能被17整除
    candidates = []
    for n in range(102, 1000, 17):
        s = str(n).zfill(3)
        if len(set(s)) == 3:  # 三位数字必须不同
            candidates.append(s)

    # 逐步向前添加数字，检查约束
    for i, prime in enumerate([13, 11, 7, 5, 3, 2]):
        new_candidates = []
        for cand in candidates:
            # 获取当前已使用的数字
            used_digits = set(cand)

            # 尝试添加一个新数字在最前面
            for d in "0123456789":
                if d not in used_digits:
                    # 新形成的三位数
                    new_triplet = d + cand[:2]
                    # 检查是否能被对应的质数整除
                    if int(new_triplet) % prime == 0:
                        new_candidates.append(d + cand)

        candidates = new_candidates

    # 现在candidates中包含的是d_2到d_10（9位数字）
    # 我们需要添加第一个数字d_1
    final_numbers = []
    for cand in candidates:
        used_digits = set(cand)
        # 找出缺失的那个数字
        for d in "0123456789":
            if d not in used_digits and d != "0":  # d_1不能为0
                num = d + cand
                # 验证这是一个0-9的pandigital数
                if len(set(num)) == 10:
                    final_numbers.append(num)

    # 计算总和
    total = sum(int(num) for num in final_numbers)
    return total, [int(num) for num in final_numbers]


@timer
def main():
    numbers = get_all_numbers()
    print(sum(numbers))


@timer
def main_op():
    total, _ = optimized_solve()
    print(total)


if __name__ == "__main__":
    main()
    main_op()