[Python随笔]暴力解解决“崩铁”的引航罗盘解密

暴力解决游戏《崩坏：星穹铁道》中的引航罗盘解密。由于复杂度较低，所以暴力解具有可行性。

图片介绍参数设定：

import sys# 获取自然数余数def get_nature_remain(dividends: int, divisors: int):    remain = dividends % divisors    while remain < 0:        remain += divisors    return remain# 计算符号位，返回至3*3符号矩阵def get_sign(method, directions: tuple):    if (len(directions) != 1 and method < 3) or (len(directions) != 2 and method > 2):        print('输入格式不正确')        sys.exit(-2)    if method == 0:  # 外圈        return [(0, 0, directions[0]), ]    elif method == 1:  # 中圈        return [(1, 0, directions[0]), ]    elif method == 2:  # 内圈        return [(2, 0, directions[0]), ]    elif method == 3:  # 外中        return [(0, 1, directions[0]), (1, 1, directions[1])]    elif method == 4:  # 外内        return [(0, 2, directions[0]), (2, 1, directions[1])]    else:  # 中内        return [(1, 2, directions[0]), (2, 2, directions[1])]# 打印结果def result_print(index, rotate):    if index == 0:        print('最外圈转', rotate)    elif index == 1:        print('中间圈转', rotate)    elif index == 2:        print('最内圈转', rotate)    elif index == 3:        print('外中联动圈转', rotate)    elif index == 4:        print('外内联动圈转', rotate)    else:        print('中内联动圈转', rotate)# 解法为暴力解def star_rail_LuoPan(points: tuple, rotates: tuple, methods: tuple):    # 判断输入是否正确    if min(points) < 0 or min(rotates) < 0 or min(methods[0]) < 0 or max(points) > 5 or max(rotates) > 6 or max(methods[0]) > 6 or len(points) != 3 or len(rotates) != 3 or len(methods[0]) != 3 or len(set(methods[0])) != 3:        print('输入格式不正确')        sys.exit(-1)    # 常量参数初始化    p1, p2, p3 = points    r1, r2, r3 = rotates    # 变量参数矩阵    f = [0 for _ in range(6)]    x = methods[0][0] - 1    y = methods[0][1] - 1    z = methods[0][2] - 1    # 公式正负号，顺逆转    sign = [[1, 1, 1],            [1, 1, 1],            [1, 1, 1]]    for i in range(3):        for row, col, t_sign in get_sign(method=methods[0][i] - 1, directions=methods[1][i]):            sign[row][col] *= t_sign    flag = False  # 是否存在解，默认不存在解    # 6为一轮，超过6相当于从0开始（数论）    for i in range(6):        f[x] = i        for j in range(6):            f[y] = j            for k in range(6):                f[z] = k                # 计算公式                a1 = get_nature_remain(dividends=p1 + (sign[0][0]*f[0] + sign[0][1]*f[3] + sign[0][2]*f[4]) * r1, divisors=6)                a2 = get_nature_remain(dividends=p2 + (sign[1][0]*f[1] + sign[1][1]*f[3] + sign[1][2]*f[5]) * r2, divisors=6)                a3 = get_nature_remain(dividends=p3 + (sign[2][0]*f[2] + sign[2][1]*f[4] + sign[2][2]*f[5]) * r3, divisors=6)                # 当前参数是否满足正确解                if a1 + a2 + a3 == 0:                    print('*' * 60)                    result_print(x, i)                    result_print(y, j)                    result_print(z, k)                    flag = True  # 存在解    return flagif __name__ == '__main__':	# 程序设计有很多参数设定，都是程序猿，相信你能看懂[doge]    # 注意，假设取最终汇聚方向处为0°点。下面为一个例子：    p = (4, 4, 2)  # 设置三个圈点的初始值 < 6，例子的解释为最外圈起始点240°（4*60°，以下同理），中圈240°，内圈120°    r = (4, 4, 4)  # 设置一次转动的角度 <= 6，例子的解释为最外圈一次转240°，中圈240°，内圈240°    # m[0]，1为外圈单转，2为中圈单转，3为内圈单转，4为外中圈联动同转，5为外内圈联动同转，6为中外圈联动同转    # m[1]，转动方向顺逆，单转要求tuple大小为1，联动转要求tuple大小为2，与m[0]一一对应，1为顺时针，-1为逆时针    m = ((2, 3, 5), ((1,), (-1,), (-1, -1)),)    if not star_rail_LuoPan(points=p, rotates=r, methods=m):        print('不存在正确解')