迷失丛林

根据关键字TryAgain定位到判断函数，确定输出位数，和字符输入格式

进入主函数，主函数分为两个部分

• 第一部分，根据前16位输入的字节填充入key中，生成256个510位的字数组
  
  随后添加256个256字数组，对应下标记录每次结果的254（2+4+8+0x10+0x20+0x40+0x80)至510（245+256）位对应值出现次数，响应位置不为0的次数要等于对应值才可以进入下一步判断
  
  这里直接用脚本先求出256位key中不存在的8个值，就是对应我们密码前16位字符
  key_number = [0x1e, 0x28, 0x4b, 0x6d, 0x8c, 0xa3, 0xd2, 0xfb]
  全排列爆破下对应顺序，40320次

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

  def key_enumeration(enum_key):   key = [0xA2, 0x9B, 0xF4, 0xDF, 0xAC, 0x7C, 0xA1, 0xC6, 0x16, 0xD0, 0x0F, 0xDD, 0xDC, 0x73, 0xC5, 0x6B, 0xD1, 0x96,          0x47, 0xC2, 0x26, 0x67, 0x4E, 0x41, 0x82, 0x20, 0x56, 0x9A, 0x6E, 0x33, 0x92, 0x88, 0x29, 0xB5, 0xB4, 0x71,          0xA9, 0xCE, 0xC3, 0x34, 0x50, 0x59, 0xBF, 0x2D, 0x57, 0x22, 0xA6, 0x30, 0x04, 0xB2, 0xCD, 0x36, 0xD5, 0x68,          0x4D, 0x5B, 0x45, 0x9E, 0x85, 0xCF, 0x9D, 0xCC, 0x61, 0x78, 0x32, 0x76, 0x31, 0xE3, 0x80, 0xAD, 0x39, 0x4F,          0xFA, 0x72, 0x83, 0x4C, 0x86, 0x60, 0xB7, 0xD7, 0x63, 0x0C, 0x44, 0x35, 0xB3, 0x7B, 0x19, 0xD4, 0x69, 0x08,          0x0B, 0x1F, 0x3D, 0x11, 0x79, 0xD3, 0xEE, 0x93, 0x42, 0xDE, 0x23, 0x3B, 0x5D, 0x8D, 0xA5, 0x77, 0x5F, 0x58,          0xDB, 0x97, 0xF6, 0x7A, 0x18, 0x52, 0x15, 0x74, 0x25, 0x62, 0x2C, 0x05, 0xE8, 0x0D, 0x98, 0x2A, 0x43, 0xE2,          0xEF, 0x48, 0x87, 0x49, 0x1C, 0xCA, 0x2B, 0xA7, 0x8A, 0x09, 0x81, 0xE7, 0x53, 0xAA, 0xFF, 0x6F, 0x8E, 0x91,          0xF1, 0xF0, 0xA4, 0x46, 0x3A, 0x7D, 0x54, 0xEB, 0x2F, 0xC1, 0xC0, 0x0E, 0xBD, 0xE1, 0x6C, 0x64, 0xBE, 0xE4,          0x02, 0x3C, 0x5A, 0xA8, 0x9F, 0x37, 0xAF, 0xA0, 0x13, 0xED, 0x1B, 0xEC, 0x8B, 0x3E, 0x7E, 0x27, 0x99, 0x75,          0xAB, 0xFE, 0xD9, 0x3F, 0xF3, 0xEA, 0x70, 0xF7, 0x95, 0xBA, 0x1D, 0x40, 0xB0, 0xF9, 0xE5, 0xF8, 0x06, 0xBC,          0xB6, 0x03, 0xC9, 0x10, 0x9C, 0x2E, 0x89, 0x5C, 0x7F, 0xB1, 0x1A, 0xD6, 0x90, 0xAE, 0xDA, 0xE6, 0x5E, 0xB9,          0x84, 0xE9, 0x55, 0xBB, 0xC7, 0x0A, 0xE0, 0x66, 0xF2, 0xD8, 0xCB, 0x00, 0x12, 0xB8, 0x17, 0x94, 0x6A, 0x4A,          0x01, 0x24, 0x14, 0x51, 0x07, 0x65, 0x21, 0xC8, 0x38, 0xFD, 0x8F, 0xC4, 0xF5, 0xFC]   unknown_key = enum_key   test_key = unknown_key + key     count = [2, 4, 8, 0x10, 0x20, 0x40, 0x80]   result = [0, 0, 0, 0]     bit_count = 1   pos = 0     for times in range(256):       mid_res = [0] * 512       mid_res2 = [0] * 256       mid_res[0] = test_key[bit_count - 1]       mid_res[1] = bit_count       res_count = 0       for i in count:           pos += i           start = pos           while i > 0:               mid_res[start] = test_key[mid_res[res_count]]               mid_res[start + 1] = (mid_res[res_count] + 1) % 256               start += 2               res_count += 1               i -= 1       for number in range(res_count, 510):           mid_res2[mid_res[number]] += 1       if mid_res2[0]:           result[0] += 1       if mid_res2[14]:           result[1] += 1       if mid_res2[40]:           result[2] += 1       if mid_res2[79]:           result[3] += 1       bit_count = (bit_count + 1) % 256       pos = 0   if result[0] == 169 and result[1] == 172 and result[2] == 167 and result[3] > 200:       print(result)       print(unknown_key)

  得到Key前八位分别为0x4B, 0x6D, 0x28, 0x8C, 0xFB, 0xD2, 0x1E, 0xA3，由于输入每两位做了一次翻转，所以输入的前十六位为B4D682C8BF2DE13A

• 第二部分，先调整了key值，这部分可以不用管，只要前十六位正确可以直接在内存0x404000处dump0x100字新key
  
  最后，根据输入的后十六位字符组成的8个字对新key的前八个字进行运算，且一一对应互不影响，生成的结果和GoodJob~0x47 0x6F 0x6F 0x64 0x4A 0x6F 0x62 0x7E进行匹配，所以只需要针对每个字节进行一次爆破即可：

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

  for i in range(8):     res.append(key_res[i]) target = [0x48, 0x6F, 0x6F, 0x64, 0x4A, 0x6F, 0x62, 0x7F] for pos in range(len(target)):     for number in range(0xFF):         source = number         for i in range(8):             if number & 1 != 0:                 res[pos] = (res[pos] + 1) % 256             else:                 res[pos] = key_res[res[pos]]             number = number >> 1         if res[pos] == target[pos]:             print(source)         res[pos] = key_res[pos]

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