"""
Take the number 192 and multiply it by each of 1, 2, and 3:

    192 * 1 = 192
    192 * 2 = 384
    192 * 3 = 576

By concatenating each product we get the 1 to 9 pandigital, 192384576.
We will call 192384576 the concatenated product of 192 and (1,2,3).

The same can be achieved by starting with 9 and multiplying by 1,2,3,4, and 5,
giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).

What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product
of an integer with (1,2,...,n) where n > 1?
"""

import time


def timer(func):
    def wrapper(*args, **kwargs):
        start_time = time.time()
        result = func(*args, **kwargs)
        end_time = time.time()
        print(f"Execution time: {end_time - start_time:.6f} seconds")
        return result
    return wrapper


# def max_pandiMulti():
#     max_num = 0
#     key = [1,2,3,4,5]
#     res = (0, [])
#     ok = False
#     for i in range(9999, 9, -1):
#         mul_keys = [i * j for j in key]
#         for j in range(5) :
#             numx = "".join(map(str, mul_keys[:j+1]))
#             if len(numx) == 9 and set(numx) == set('123456789'):
#                 max_num = max(max_num, int(numx))
#                 if max_num == int(numx):
#                     res = (i, key[:j+1])
#                     ok = True
#                 break
#         if ok:
#             break
#     return max_num, res


def max_pandiMulti_better():
    max_num = 0
    res = (0, [])
    # 配置列表：(n, start, end)，其中 start > end，表示从大到小遍历
    configs = [
        (2, 9999, 5000),  # n=2: i ∈ [5000, 9999]
        (3, 333, 100),    # n=3: i ∈ [100, 333]
        (4, 33, 25),      # n=4: i ∈ [25, 33]
        (5, 9, 5)         # n=5: i ∈ [5, 9]
    ]

    for n, start, end in configs:
        for i in range(start, end - 1, -1):
            # 生成拼接字符串
            s = ''.join(str(i * k) for k in range(1, n + 1))
            # 检查是否为1-9 pandigital
            if set(s) == set('123456789'):
                num = int(s)
                if num > max_num:
                    max_num = num
                    res = (i, list(range(1, n + 1)))
                break  # 当前n中，i递减，第一个满足条件的即为最大，可终止内层循环

    return max_num, res


@timer
def main():
    maxnum, ress = max_pandiMulti_better()
    print(f"{maxnum} : {ress}")


if __name__ == "__main__":
    main()