破解 Acronis Cyber Backup 12.5
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破解 Acronis Cyber Backup 12.5

大类
Crack
技术标签
逆向-Windows
开发-HookPatch-Windows
优先级
Critical
开始日期
Nov 25, 2020
状态
Finalized
Public
Public
最后更新
Sep 8, 2021

Round 1：逆向Windows验证思路（12.5.16327）

通过直接分析安装后文件夹的结构以及输错license Key时的日志，可以很轻松的得到调用关系：
  • Management Server → AccountServer → LicenseServer
将LicenseServer的pyc反编译得到源码，可以分析到在model/license_key.py中有：
from acrolic_check import check_license as parse_serial_number
 
对应acrolic_check搜索后可知在PyShell模块中，ida分析可知从函数入口开始，该模块与内置的14种key参数做验证，成功了哪个，哪个就是这个key对应的发行版。
Key的大概验证算法：
v45 = (A^x · k^y) % B
v44 = inv(v45, C) · y = y / v45 （mod C）
A、B、C固定，k可以任意指定，v45、v44已知，求x、y

MD5(v44[XXX] + v45[XXX]) == v45
 

Round 2：破解 Windows

破解思路

破解思路：hook libcrypto10使k变为0，同时令y = 0，则A^x可为任意数，剩下控制y即可实现等式
 

Keygen

def GenLicPayload(dist, timeLimit, lang, genDate, edition):
    LangTable = [
      "ru-RU",
      "en-US",
      "de-DE",
      "ko-KR",
      "ja-JP",
      "pl-PL",
      "nl-NL",
      "fr-FR",
      "cs-CZ",
      "ja-JP",
      "fr-FR",
      "pt-BR",
      "es-ES",
      "zh-CN",
      "zh-TW",
      "it-IT",
      "th-TH",
      "fr-FR",
      "sk-SK",
      "pt-PT",
      "gsw-CH",
      "fr-CA",
      "en-GB"
    ]
    import struct
    langID = LangTable.index(lang)
    licPayload = struct.pack('HBBH', (dist & 0xfff) | (timeLimit << 12), edition, langID, genDate)
    return licPayload

licPayload = GenLicPayload(dist, timeLimit, lang, genDate, edition)

#v45 = ((ExeHash3 ** InputVal2) * (Key ** InputVal1) % ExeHash1)  # input2 -> 0 --> v45 = 1
#v44 = inv(v45 % ExeHash2) * InputVal1 % ExeHash2
#v44[:14] = MD5(v44[14:] + v45)[:14]
#   (licPayload == v44[14:])

# If Key = 1, InputVal2 = 0
#    v45 = 1
#    v44 = InputVal1 % ExeHash2


# If Key = 1,
# v45 = ((ExeHash3 ** InputVal2) % ExeHash1)
# v44 = inv(v45 % ExeHash2) * InputVal1 % ExeHash2 --> InputVal1 = v44 * v45 % ExeHash2
# make InputVal2 = 1, v45 = ExeHash3 % ExeHash1


#ExeHash1_LE = '1903ADC443A8359DD274D921D5A42B2BB4D91CF266863C325457AC231398981C1C0FE7B9ED2C7B17A60E581C8CED6AFF0342355FA0407543462F6CBE7FEF27B67DB786E4565D4A775B22E5ACECB732DEBBA6009B313340D2A3E95643C1DEE565413ECB465977122969D8932B9289C0EDECE972D48D8184095CE8736059450DBD'
ExeHash1 = int('0x' + '1903ADC443A8359DD274D921D5A42B2BB4D91CF266863C325457AC231398981C1C0FE7B9ED2C7B17A60E581C8CED6AFF0342355FA0407543462F6CBE7FEF27B67DB786E4565D4A775B22E5ACECB732DEBBA6009B313340D2A3E95643C1DEE565413ECB465977122969D8932B9289C0EDECE972D48D8184095CE8736059450DBD'.decode('hex')[::-1].encode('hex'), 16)
#ExeHash2_LE = '37EB5D79D091D8782D57834A91139691361C09F3000000000000000000000000'
ExeHash2 = int('0x' + '37EB5D79D091D8782D57834A91139691361C09F3000000000000000000000000'.decode('hex')[::-1].encode('hex'), 16)
#ExeHash3_LE = '983951689797F3446B08BECCFE405582702966370110C0A31ECA0FEC96C06979C28685E241A9B18547B175CD5D96C3388820695AE662E424A6DABD3FF25302357D738034DFB7BE5FB7C014CBD2C52F82D5EF5C02E758C99FBA0CE8C50F1A7CB53D9028F8ACD16465114FCC87489B7B04062C8ED36FE55D4758ED7B26AD65DDB7'
ExeHash3 = int('0x' + '983951689797F3446B08BECCFE405582702966370110C0A31ECA0FEC96C06979C28685E241A9B18547B175CD5D96C3388820695AE662E424A6DABD3FF25302357D738034DFB7BE5FB7C014CBD2C52F82D5EF5C02E758C99FBA0CE8C50F1A7CB53D9028F8ACD16465114FCC87489B7B04062C8ED36FE55D4758ED7B26AD65DDB7'.decode('hex')[::-1].encode('hex'), 16)


dist = 0 # default
timeLimit = 0 # no time limit
lang = 'en-US'
genDate = 0x1fff # maximum, in fact future = 978350400 + 86400 * 0x1fff
edition=4 # VOLUME (no limit)


def convertToLEB(v, n):
    return hex(v & ((2 << (8*n)) - 1))[2:].rstrip('L').rjust(2*n,'0').decode('hex')[::-1]

from hashlib import md5
inputVal2 = 1
v45 = (ExeHash3 ** inputVal2) % ExeHash1
md5input = licPayload + convertToLEB(v45, 128)
h = md5(md5input).digest()
v44 = h[:14] + licPayload
v44 = int('0x' + v44[::-1].encode('hex'), 16)
inputVal1 = v44 * v45 % ExeHash2

fullPayload = convertToLEB(inputVal1, 20) + convertToLEB(inputVal2, 20)


ALPHA = '23456789ABCDEFGHJKLMNPQRSTUVWXYZ'

def decode_acronis_key(a):
    bins = ''
    for c in a:
        bins = bin(ALPHA.index(c))[2:].rjust(5, '0') + bins
    
    bins = bins.rjust(((len(bins) + 7) // 8) * 8, '0')
    
    ret = []
    for i in range(0, len(bins), 8):
        ret.append(int(bins[i:i+8], 2))
    
    return ''.join((map(chr, ret[::-1])))


def encode_acronis_key(raw):
    binstr = ''.join([bin(ord(c))[2:].rjust(8, '0') for c in raw[::-1]])
    binstr = binstr.rjust(((len(binstr) + 4) // 5) * 5, '0')
    ret = ''
    for i in range(len(binstr), 0, -5):
        ret += ALPHA[int(binstr[i-5:i], 2)]
    
    return ret

#INPUT = 'FW3E5WZS24UNA4QH7XXFFG5PQZPJ8B4T4ERA9QAYKGKY3H2TKK4LFLUS4MBKXZYU'
#ret = decode_acronis_key(INPUT)
#print ret.encode('hex')
#assert ret.encode('hex') == '8d0736f8c740688a847da5f7d6dca8f6576892c8825d742cf2d1451f1ec8310ad9a4c662a6d8bfd7'
#ret = encode_acronis_key(ret)
#assert ret == INPUT

licCodeRaw = encode_acronis_key(fullPayload)
licCode = []
for i in range(0, len(licCodeRaw), 8):
    licCode.append(licCodeRaw[i:i+8])

print '-'.join(licCode)
SampleKey: 23AAJGQQ-TRLP63MJ-9BUQUQZH-9KAMCVDH-32222222-22222222-22222222-22222222

破解实现

破解实现：使用https://github.com/kevinalmansa/DLL_Wrapper 做了一个DLL Forwarder，劫持BN_bin2bn函数，并在load对应bignum时替换为0
最后使用特殊的license： 23AAJGQQ-TRLP63MJ-9BUQUQZH-9KAMCVDH-32222222-22222222-22222222-22222222 即可

Round 3：破解Linux

Linux的acrolic_check同样位于PyShell，libcrypto10.dll变为liblibcrypto10.so。
破解实现：使用https://github.com/yugr/Implib.so 做了一个So Wrapper，同样劫持BN_bin2bn函数即可

打包破解docker

Agent端：直接抄的官方的Dockerfile
Server端：经总结必须要有systemd，否则就会各种出错。借鉴自https://github.com/silvenga-docker/acronis-backup
 
Copyright 2022 Misty