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[原创]KCTF2021 迷失丛林write up
2021-11-17 20:14 16040

[原创]KCTF2021 迷失丛林write up

2021-11-17 20:14
16040

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

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    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:
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    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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最后于 2021-11-19 10:33 被hacktu编辑 ,原因: 添加标题
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