導語

各位小夥伴們週末快樂~

第六題終於結束了，最終有兩人挑戰成功！分別是風間仁和hotwinter，恭喜！

第六題作者以兩人攻破此題的成績，登上防守方第一名的位置。

防守方hotwinter 逆襲至第一名，風間仁在短暫的落後之後，再次重回第三名。

現在還剩三題，比賽愈發緊張。

究竟誰能奪取最後的勝利呢？我們翹首以望。各位選手，加油！

接下來讓我們一起來看看第六題的點評、出題思路和解析。

看雪評委netwind點評

作者採用了修改過的DES演算法對資料進行了加密。然後將加密結果轉換為大數，用miracl庫做大數計算。最後呼叫luajit指令碼執行大整數計算，把計算結果和記憶體資料進行對比驗證。破解此題要識別出正確的luajit版本號並反編譯luajit指令碼，識別作者實現的FFT大數乘法，識別DES函式的修改點並實現解密函式才能完成。

第六題作者簡介

loudy，2007年接觸看雪，熱衷程式設計、特別是逆向分析。當前從事工作出於保密規定暫不公開。愛好跑步、籃球、電影。在論壇多以潛水學習為主，也希望多認識有相同愛好的朋友。

第六題設計思路

程式設計流程清楚，未加殼，未做任何反逆向手段，但存在幾個坑點，以下一一說明：

1、輸入長度未固定，需要分析（16-24位）。

2、對輸入做DES加密，但該DES被修改。

（1）表

/*

char DES::Expand_Table[EXPAND_SIZE] = {

31, 0, 1, 2, 3, 4,

3,  4, 5, 6, 7, 8,

7,  8, 9,10,11,12,

11,12,13,14,15,16,

15,16,17,18,19,20,

19,20,21,22,23,24,

23,24,25,26,27,28,

27,28,29,30,31, 0};

*/

被修改為

char DES::Expand_Table[EXPAND_SIZE] = {

0,  0, 1, 2, 3, 3,

4,  4, 5, 6, 7, 7,

8,  8, 9,10,11,11,

12,12,13,14,15,15,

16,16,17,18,19,19,

20,20,21,22,23,23,

24,24,25,26,27,27,

28,28,29,30,31,31};

改變了加密結果，需要分析出來。

（2）塊加密時（8位元組），置換表的使用順序和子金鑰的使用順序做了改變。

IP_Transform(bitStr);

std::string halfBitStr;

halfBitStr.resize(bitStr.size() / 2);

std::string eBitStr;

eBitStr.resize(EXPAND_SIZE);

for(int i = SUBKEY_NUM-1; i >= 0; --i)

{

Expand_Transform(bitStr.substr(bitStr.size() / 2), eBitStr);

XOR(eBitStr, std::string(subKey[i],SUBKEY_LENGHT), SUBKEY_LENGHT);

SBox_Transform(eBitStr, halfBitStr);

Permute_Transform(halfBitStr);

XOR(bitStr, halfBitStr, halfBitStr.size());

if(i != 0)

LeftCycle(bitStr, 0, bitStr.size(), bitStr.size() / 2);

}

IP_1_Transform(bitStr);

（3）全文加密時，多加了一個異或操作，改變加密結果，需要分析出來。

for(size_t i = 1; i < plain.size() / 8; ++i)

{

block = plain.substr(i * 8, 8);

for(int j=0;j<8;j++)

{

block[j] = block[j]^tmpblock[7-j];

}

EncryptBlock(block, subKey);

result.append(block);

tmpblock.assign(block);

}

3、加密結果轉換為大數，用miracl庫做大數計算。

（1）呼叫函式multiply與173做乘法運算。

big bigx = mirvar(0);

bigx->len = 1;

bigx->w[0] = 173;

multiply(bigx,bigtmp1,bigx);

（2）呼叫函式fft_mult與0x719做乘法運算。

big bigy = mirvar(0);

bigy->len = 1;

bigy->w[0] = 0x719;

fft_mult(bigx,bigy,bigy);

但其中動了小手腳（在fft_mult最後做了一次乘方運算和一次減法運算，減1001）。

power(z,2,z,z);

decr(z,1001,z);

4、呼叫自己實現的FFT乘法，乘以317。

int* out1 = mull(desOut,"317");

5、呼叫luajit指令碼執行大整數計算。

--xut c程式匯入的引數

--func1加法

--func2減法

--func3乘法

local

yyy =

"1574592838300862641516215149137548264158058079230003764126382984039489925466995870724568174393389905601620735902909057604303543552180706761904"

function myst()

c = func1(xut,"1001")

d = 0

for i=1,100 do

for j=1,100 do

c = func1(c,"1001")

end

c = func1(c,"10101")

end

c = func3(c,"983751509373")

for i=1,13 do

c = func2(c,"1023")

for j=1,14 do

c = func2(c,"1203")

for k=1,15 do

c = func2(c,"1230")

for l=1,16 do

c = func2(c,"1231")

end

end

end

end

c = func1(c,"1")

c = func3(c,"2")

e = get(c)

--print(e)

lene = string.len(e)

leny = string.len(yyy)

if(lene==leny)

then

--print("same length")

--[ local variable definition --]

i = 1

d = 1

--[ while loop execution --]

while( i <= lene )

do

if(string.byte(e,i)~=string.byte(yyy,i))

then

--[ terminate the loop using break statement --]

d = 0

break

end

i = i+1

end

end

return d

end

計算結果和固定值“1574592838300862641516215149137548264158058079230003764126382984039489925466995870724568174393389905601620735902909057604303543552180706761904”比較，相等則註冊成功，不然失敗。但此處也有坑，我把luajit版本號改了，本來是luajit2.0.5，我把字串改為luajit2.1.0-beta3，反編譯難度加大。

破解思路

1、 識別正確luajit版本號，反編譯luajit指令碼，理清其中流程，通過固定值“1574592838300862641516215149137548264158058079230003764126382984039489925466995870724568174393389905601620735902909057604303543552180706761904”，得到luajit指令碼的正確輸入。

2、 識別自己實現的FFT大數乘法，反推乘法函式mull的輸入。

3、 識別miracl庫，並找到fft_mult函式的修改點（函式最後），反推fft_mult的輸入。

4、 反推multiply的輸入，並轉化為16進位制。

5、 識別DES函式的修改點，並自己實現解密函式，將上一步結果解密，得到最終的key(“KXCTF201710BYLoudy08”)，破解完成。

下面選取 hotwinter 的破解分析

這題真的是做的我想砸牆……前前後後花了不下24小時……各種坑……放棄HITCON一直在剛……

這題是一道數學 + 密碼學 + lua + C++ STL的極其噁心人的一道題……大概一共分為三大部分，其中摻雜了各種不常見的庫函式和更改過的庫函式……這題經驗就是善用工具……

首先慣例直接上strings

跟著引用找到main函式，如下

大概可以看出邏輯如下，我們的輸入會經過兩個大結構的變換，第一個是des_decrypt（之後我們會說怎麼識別他），第二個是MIRACL的大整數操作，我們輸入做了以下兩個操作以後，會扔進lua裡面進行第三次操作，最終進行比較，大致邏輯就是這樣

首先先說一下庫函式都是怎麼識別出來的吧（DES是人工扒演算法和常數扒出來的……），這裡用到了bindiff這個工具https://www.zynamics.com/software.html,他會比較兩個idb之間每個函式的相似性，並列出最相似的函式，對於像MIRACL和LuaJIT這種開源程式來說很好用。我們需要做的就是下載對應版本的原始碼，編譯成exe/dll

（因為linux下calling convention不同，所以bindiff可能識別沒那麼準確（大概））以後開IDA執行一遍初始分析，然後儲存idb後退出。注意這裡每個版本都最好建debug build，這樣大部分的函式名稱會被保留，並生成pdb。bindiff目前還沒有支援IDA7.0……也就是我為什麼換回了IDA 6.95……

LuaJIT很好編譯……直接扔vs command line  tools裡面跑他的mvcsbuild就好了，這裡檢視build的bat指令碼後發現假如第一個引數是debug的話，他就會建成debug build（即 mvsbuild.exe

debug）非常方便，因此lua庫函式立馬就標完了（不過這裡實際執行的時候遇到了問題……luaji2.1.0-beta3使用的是2.0版的Opcode按理說並不能跑起來……即使改成1.0版也還是會出錯……不知道為什麼程式裡面可以用2.1.0-beta3這個build……）。lua庫函式所在的檔案是lua51.dll，不是luajit.exe，不過lua庫函式改動不大，所以bindiff識別準確率還是挺高的。所有lua的C API函式基本上都是bindiff識別的

MIRACL這個東西就dt了……我們知道用了MIRACL是因為strings裡面的這句話

MIRACL的build只有一個指令碼在lib/ms32doit.bat……而且寫的非常糟糕……手動加了一些編譯

flag以後終於是可以編譯了……生成檔案為miracl.lib

（用起來不是很方便，於是改編譯指令碼為dll），MIRACL的識別準確率香蕉LuaJIT來說就沒有那麼高了……不是很清楚為什麼，不過用的也不算太多，所以還好，這裡可以點開第一個mirsys_basic函式，然後照著原始碼把miracl的結構體給擼出來（因為有各種ifndef之類的……所以沒法直接看原始碼擼），這樣之後會方便很多

這裡我們注意到一點，MIRACL其實是有stack

backtrace的，他的識別關鍵在於miracl結構體中trace這個結構（這也是為什麼我們要照著mirsys_basic擼結構體的原因），找到這個以後我們MIRACL的識別就很簡單了。stack trace主要是靠MR_IN()組成，每個主要函式都有自己的數字，所以只要對照這個就能找到大部分的函式了，這裡以power為例子進行說明

紅框中標出的即為該函式的MR_IN的stack trace number，下面我們直接grep 原始碼目錄就可找到對應的定義

MIRACL幾乎沒有說明文件……所有東西都是靠我自己從原始碼裡面翻出來的……dt，這裡我還用到了cscope這個工具來在本地快速查詢C和C++ project的函式定義很引用（有時候grep的結果太多了），使用方法很簡單

find . -iname "*.c" -o -iname "*.cpp" -o -iname "*.hpp" -o -iname "*.h" > cscope.files （這裡可以加任何你想要的檔案字尾，只要find函式能找到就行）

cscope -bqk （建立索引）

cscope -d （避免重複建立索引）

個人感覺cscope只適合相對較小的project……如果是linux kernel那種……最好還是自己搭一個opengrok的伺服器比較好

基本庫函式標完……下面就進入正式的逆向工作……

一。des_cbc_decrypt

這個函式是des cbc的解密（加密？）函式，作者更改了很多的常量……使得我們不得不自己找一個des的庫更改後使用。des的識別是靠了這串陣列

以16進位制陣列搜尋（搜 0xe,0x4,0xd,0x1,0x2,0xf,0xb,0x8,0x3,0xa,0x6,0xc）可找到http://kodu.ut.ee/~lipmaa/teaching/crypto2000/des/des.C，檢視後，發現是DES的sbox，由於des並不是所有操作都需要用到table，找到一個和所給des相似的庫找了我很長時間（這裡也是剛開始覺得常量不會改……），最終選用了這個https://gist.githubusercontent.com/eigenein/1275094/raw/62d1fee5e9d19df44fab6950ecbcd9c82b9b9805/pyDes.py

根據這個演算法，對應的des的表改過來就好了（有個表只改了一個位元組……實在是很坑爹）。des的key

schedule用到的表也需要改（給的例子裡面bits是按MSB encode的這個也需要改）。同時des_decrypt裡面roundkey是從最後一輪開始的……（也就是為什麼我說他是decrypt），但cbc的演算法明顯是block

cipher的加密演算法……這裡還要翻一下iv。同時des_decrypt裡面對調Ininital

permutation和final permutation，調整後des的程式碼（3DES我懶得刪了）

#############################################################################

#                              Documentation                              #

#############################################################################

# Author:  Todd Whiteman

# Date:    16th March, 2009

# Verion:  2.0.0

# License:  Public Domain - free to do as you wish

# Homepage: http://twhiteman.netfirms.com/des.html

#

# This is a pure python implementation of the DES encryption algorithm.

# It's pure python to avoid portability issues, since most DES

# implementations are programmed in C (for performance reasons).

#

# Triple DES class is also implemented, utilising the DES base. Triple DES

# is either DES-EDE3 with a 24 byte key, or DES-EDE2 with a 16 byte key.

#

# See the README.txt that should come with this python module for the

# implementation methods used.

#

# Thanks to:

#  * David Broadwell for ideas, comments and suggestions.

#  * Mario Wolff for pointing out and debugging some triple des CBC errors.

#  * Santiago Palladino for providing the PKCS5 padding technique.

#  * Shaya for correcting the PAD_PKCS5 triple des CBC errors.

#

"""A pure python implementation of the DES and TRIPLE DES encryption algorithms.

Class initialization

--------------------

pyDes.des(key, [mode], [IV], [pad], [padmode])

pyDes.triple_des(key, [mode], [IV], [pad], [padmode])

key    -> Bytes containing the encryption key. 8 bytes for DES, 16 or 24 bytes

for Triple DES

mode    -> Optional argument for encryption type, can be either

pyDes.ECB (Electronic Code Book) or pyDes.CBC (Cypher Block Chaining)

IV      -> Optional Initial Value bytes, must be supplied if using CBC mode.

Length must be 8 bytes.

pad    -> Optional argument, set the pad character (PAD_NORMAL) to use during

all encrypt/decrpt operations done with this instance.

padmode -> Optional argument, set the padding mode (PAD_NORMAL or PAD_PKCS5)

to use during all encrypt/decrpt operations done with this instance.

I recommend to use PAD_PKCS5 padding, as then you never need to worry about any

padding issues, as the padding can be removed unambiguously upon decrypting

data that was encrypted using PAD_PKCS5 padmode.

Common methods

--------------

encrypt(data, [pad], [padmode])

decrypt(data, [pad], [padmode])

data    -> Bytes to be encrypted/decrypted

pad    -> Optional argument. Only when using padmode of PAD_NORMAL. For

encryption, adds this characters to the end of the data block when

data is not a multiple of 8 bytes. For decryption, will remove the

trailing characters that match this pad character from the last 8

bytes of the unencrypted data block.

padmode -> Optional argument, set the padding mode, must be one of PAD_NORMAL

or PAD_PKCS5). Defaults to PAD_NORMAL.

Example

-------

from pyDes import *

data = "Please encrypt my data"

k = des("DESCRYPT", CBC, "\0\0\0\0\0\0\0\0", pad=None, padmode=PAD_PKCS5)

# For Python3, you'll need to use bytes, i.e.:

#  data = b"Please encrypt my data"

#  k = des(b"DESCRYPT", CBC, b"\0\0\0\0\0\0\0\0", pad=None, padmode=PAD_PKCS5)

d = k.encrypt(data)

print "Encrypted: %r" % d

print "Decrypted: %r" % k.decrypt(d)

assert k.decrypt(d, padmode=PAD_PKCS5) == data

See the module source (pyDes.py) for more examples of use.

You can also run the pyDes.py file without and arguments to see a simple test.

Note: This code was not written for high-end systems needing a fast

implementation, but rather a handy portable solution with small usage.

"""

import sys

# _pythonMajorVersion is used to handle Python2 and Python3 differences.

_pythonMajorVersion = sys.version_info[0]

# Modes of crypting / cyphering

ECB =  0

CBC =  1

# Modes of padding

PAD_NORMAL = 1

PAD_PKCS5 = 2

# PAD_PKCS5: is a method that will unambiguously remove all padding

#            characters after decryption, when originally encrypted with

#            this padding mode.

# For a good description of the PKCS5 padding technique, see:

# http://www.faqs.org/rfcs/rfc1423.html

# The base class shared by des and triple des.

class _baseDes(object):

def __init__(self, mode=ECB, IV=None, pad=None, padmode=PAD_NORMAL):

if IV:

IV = self._guardAgainstUnicode(IV)

if pad:

pad = self._guardAgainstUnicode(pad)

self.block_size = 8

# Sanity checking of arguments.

if pad and padmode == PAD_PKCS5:

raise ValueError("Cannot use a pad character with PAD_PKCS5")

if IV and len(IV) != self.block_size:

raise ValueError("Invalid Initial Value (IV), must be a multiple of " + str(self.block_size) + " bytes")

# Set the passed in variables

self._mode = mode

self._iv = IV

self._padding = pad

self._padmode = padmode

def getKey(self):

"""getKey() -> bytes"""

return self.__key

def setKey(self, key):

"""Will set the crypting key for this object."""

key = self._guardAgainstUnicode(key)

self.__key = key

def getMode(self):

"""getMode() -> pyDes.ECB or pyDes.CBC"""

return self._mode

def setMode(self, mode):

"""Sets the type of crypting mode, pyDes.ECB or pyDes.CBC"""

self._mode = mode

def getPadding(self):

"""getPadding() -> bytes of length 1. Padding character."""

return self._padding

def setPadding(self, pad):

"""setPadding() -> bytes of length 1. Padding character."""

if pad is not None:

pad = self._guardAgainstUnicode(pad)

self._padding = pad

def getPadMode(self):

"""getPadMode() -> pyDes.PAD_NORMAL or pyDes.PAD_PKCS5"""

return self._padmode

def setPadMode(self, mode):

"""Sets the type of padding mode, pyDes.PAD_NORMAL or pyDes.PAD_PKCS5"""

self._padmode = mode

def getIV(self):

"""getIV() -> bytes"""

return self._iv

def setIV(self, IV):

"""Will set the Initial Value, used in conjunction with CBC mode"""

if not IV or len(IV) != self.block_size:

raise ValueError("Invalid Initial Value (IV), must be a multiple of " + str(self.block_size) + " bytes")

IV = self._guardAgainstUnicode(IV)

self._iv = IV

def _padData(self, data, pad, padmode):

# Pad data depending on the mode

if padmode is None:

# Get the default padding mode.

padmode = self.getPadMode()

if pad and padmode == PAD_PKCS5:

raise ValueError("Cannot use a pad character with PAD_PKCS5")

if padmode == PAD_NORMAL:

if len(data) % self.block_size == 0:

# No padding required.

return data

if not pad:

# Get the default padding.

pad = self.getPadding()

if not pad:

raise ValueError("Data must be a multiple of " + str(self.block_size) + " bytes in length. Use padmode=PAD_PKCS5 or set the pad character.")

data += (self.block_size - (len(data) % self.block_size)) * pad

elif padmode == PAD_PKCS5:

pad_len = 8 - (len(data) % self.block_size)

if _pythonMajorVersion < 3:

data += pad_len * chr(pad_len)

else:

data += bytes([pad_len] * pad_len)

return data

def _unpadData(self, data, pad, padmode):

# Unpad data depending on the mode.

if not data:

return data

if pad and padmode == PAD_PKCS5:

raise ValueError("Cannot use a pad character with PAD_PKCS5")

if padmode is None:

# Get the default padding mode.

padmode = self.getPadMode()

if padmode == PAD_NORMAL:

if not pad:

# Get the default padding.

pad = self.getPadding()

if pad:

data = data[:-self.block_size] + \

data[-self.block_size:].rstrip(pad)

elif padmode == PAD_PKCS5:

if _pythonMajorVersion < 3:

pad_len = ord(data[-1])

else:

pad_len = data[-1]

data = data[:-pad_len]

return data

def _guardAgainstUnicode(self, data):

# Only accept byte strings or ascii unicode values, otherwise

# there is no way to correctly decode the data into bytes.

if _pythonMajorVersion < 3:

if isinstance(data, unicode):

raise ValueError("pyDes can only work with bytes, not Unicode strings.")

else:

if isinstance(data, str):

# Only accept ascii unicode values.

try:

return data.encode('ascii')

except UnicodeEncodeError:

pass

raise ValueError("pyDes can only work with encoded strings, not Unicode.")

return data

#############################################################################

#                                  DES                                    #

#############################################################################

class des(_baseDes):

"""DES encryption/decrytpion class

Supports ECB (Electronic Code Book) and CBC (Cypher Block Chaining) modes.

pyDes.des(key,[mode], [IV])

key  -> Bytes containing the encryption key, must be exactly 8 bytes

mode -> Optional argument for encryption type, can be either pyDes.ECB

(Electronic Code Book), pyDes.CBC (Cypher Block Chaining)

IV  -> Optional Initial Value bytes, must be supplied if using CBC mode.

Must be 8 bytes in length.

pad  -> Optional argument, set the pad character (PAD_NORMAL) to use

during all encrypt/decrpt operations done with this instance.

padmode -> Optional argument, set the padding mode (PAD_NORMAL or

PAD_PKCS5) to use during all encrypt/decrpt operations done

with this instance.

"""

# Permutation and translation tables for DES

__pc1 = [56, 48, 40, 32, 24, 16,  8,

0, 57, 49, 41, 33, 25, 17,

9,  1, 58, 50, 42, 34, 26,

18, 10,  2, 59, 51, 43, 35,

62, 54, 46, 38, 30, 22, 14,

6, 61, 53, 45, 37, 29, 21,

13,  5, 60, 52, 44, 36, 28,

20, 12,  4, 27, 19, 11,  3

]

# number left rotations of pc1

# changed!

__left_rotations = [

0x17, 0xa, 2, 5, 9, 2, 3, 2, 3, 2, 5, 7, 2, 9, 2, 7

]

# permuted choice key (table 2)

# changed!

__pc2 = [

13, 16, 10, 23,  0,  4,

2, 27, 14,  5, 20,  9,

22, 18, 11,  3, 25,  7,

15,  6, 26, 19, 12,  1,

40, 51, 30, 36, 46, 54,

29, 39, 50, 44, 32, 46,

43, 48, 38, 55, 33, 52,

45, 41, 49, 35, 28, 31

]

# initial permutation IP

__ip = [57, 49, 41, 33, 25, 17, 9,  1,

59, 51, 43, 35, 27, 19, 11, 3,

61, 53, 45, 37, 29, 21, 13, 5,

63, 55, 47, 39, 31, 23, 15, 7,

56, 48, 40, 32, 24, 16, 8,  0,

58, 50, 42, 34, 26, 18, 10, 2,

60, 52, 44, 36, 28, 20, 12, 4,

62, 54, 46, 38, 30, 22, 14, 6

]

# Expansion table for turning 32 bit blocks into 48 bits

__expansion_table = [

0,  0,  1,  2,  3,  3,

4,  4,  5,  6,  7,  7,

8,  8,  9, 10, 11, 11,

12, 12, 13, 14, 15, 15,

16, 16, 17, 18, 19, 19,

20, 20, 21, 22, 23, 23,

24, 24, 25, 26, 27, 27,

28, 28, 29, 30, 31, 31

]

# The (in)famous S-boxes

__sbox = [

# S1

[14, 4, 13, 1, 2, 15, 11, 8, 3, 10, 6, 12, 5, 9, 0, 7,

0, 15, 7, 4, 14, 2, 13, 1, 10, 6, 12, 11, 9, 5, 3, 8,

4, 1, 14, 8, 13, 6, 2, 11, 15, 12, 9, 7, 3, 10, 5, 0,

15, 12, 8, 2, 4, 9, 1, 7, 5, 11, 3, 14, 10, 0, 6, 13],

# S2

[15, 1, 8, 14, 6, 11, 3, 4, 9, 7, 2, 13, 12, 0, 5, 10,

3, 13, 4, 7, 15, 2, 8, 14, 12, 0, 1, 10, 6, 9, 11, 5,

0, 14, 7, 11, 10, 4, 13, 1, 5, 8, 12, 6, 9, 3, 2, 15,

13, 8, 10, 1, 3, 15, 4, 2, 11, 6, 7, 12, 0, 5, 14, 9],

# S3

[10, 0, 9, 14, 6, 3, 15, 5, 1, 13, 12, 7, 11, 4, 2, 8,

13, 7, 0, 9, 3, 4, 6, 10, 2, 8, 5, 14, 12, 11, 15, 1,

13, 6, 4, 9, 8, 15, 3, 0, 11, 1, 2, 12, 5, 10, 14, 7,

1, 10, 13, 0, 6, 9, 8, 7, 4, 15, 14, 3, 11, 5, 2, 12],

# S4

[7, 13, 14, 3, 0, 6, 9, 10, 1, 2, 8, 5, 11, 12, 4, 15,

13, 8, 11, 5, 6, 15, 0, 3, 4, 7, 2, 12, 1, 10, 14, 9,

10, 6, 9, 0, 12, 11, 7, 13, 15, 1, 3, 14, 5, 2, 8, 4,

3, 15, 0, 6, 10, 1, 13, 8, 9, 4, 5, 11, 12, 7, 2, 14],

# S5

[2, 12, 4, 1, 7, 10, 11, 6, 8, 5, 3, 15, 13, 0, 14, 9,

14, 11, 2, 12, 4, 7, 13, 1, 5, 0, 15, 10, 3, 9, 8, 6,

4, 2, 1, 11, 10, 13, 7, 8, 15, 9, 12, 5, 6, 3, 0, 14,

11, 8, 12, 7, 1, 14, 2, 13, 6, 15, 0, 9, 10, 4, 5, 3],

# S6

[12, 1, 10, 15, 9, 2, 6, 8, 0, 13, 3, 4, 14, 7, 5, 11,

10, 15, 4, 2, 7, 12, 9, 5, 6, 1, 13, 14, 0, 11, 3, 8,

9, 14, 15, 5, 2, 8, 12, 3, 7, 0, 4, 10, 1, 13, 11, 6,

4, 3, 2, 12, 9, 5, 15, 10, 11, 14, 1, 7, 6, 0, 8, 13],

# S7

[4, 11, 2, 14, 15, 0, 8, 13, 3, 12, 9, 7, 5, 10, 6, 1,

13, 0, 11, 7, 4, 9, 1, 10, 14, 3, 5, 12, 2, 15, 8, 6,

1, 4, 11, 13, 12, 3, 7, 14, 10, 15, 6, 8, 0, 5, 9, 2,

6, 11, 13, 8, 1, 4, 10, 7, 9, 5, 0, 15, 14, 2, 3, 12],

# S8

[13, 2, 8, 4, 6, 15, 11, 1, 10, 9, 3, 14, 5, 0, 12, 7,

1, 15, 13, 8, 10, 3, 7, 4, 12, 5, 6, 11, 0, 14, 9, 2,

7, 11, 4, 1, 9, 12, 14, 2, 0, 6, 10, 13, 15, 3, 5, 8,

2, 1, 14, 7, 4, 10, 8, 13, 15, 12, 9, 0, 3, 5, 6, 11],

]

# 32-bit permutation function P used on the output of the S-boxes

__p = [

15, 6, 19, 20, 28, 11,

27, 16, 0, 14, 22, 25,

4, 17, 30, 9, 1, 7,

23,13, 31, 26, 2, 8,

18, 12, 29, 5, 21, 10,

3, 24

]

# final permutation IP^-1

__fp = [

39,  7, 47, 15, 55, 23, 63, 31,

38,  6, 46, 14, 54, 22, 62, 30,

37,  5, 45, 13, 53, 21, 61, 29,

36,  4, 44, 12, 52, 20, 60, 28,

35,  3, 43, 11, 51, 19, 59, 27,

34,  2, 42, 10, 50, 18, 58, 26,

33,  1, 41,  9, 49, 17, 57, 25,

32,  0, 40,  8, 48, 16, 56, 24

]

# Type of crypting being done

ENCRYPT =      0x00

DECRYPT =      0x01

# Initialisation

def __init__(self, key, mode=ECB, IV=None, pad=None, padmode=PAD_NORMAL):

# Sanity checking of arguments.

if len(key) != 8:

raise ValueError("Invalid DES key size. Key must be exactly 8 bytes long.")

_baseDes.__init__(self, mode, IV, pad, padmode)

self.key_size = 8

self.L = []

self.R = []

self.Kn = [ [0] * 48 ] * 16    # 16 48-bit keys (K1 - K16)

self.final = []

self.setKey(key)

def setKey(self, key):

"""Will set the crypting key for this object. Must be 8 bytes."""

_baseDes.setKey(self, key)

self.__create_sub_keys()

def __String_to_BitList(self, data):

"""Turn the string data, into a list of bits (1, 0)'s"""

if _pythonMajorVersion < 3:

# Turn the strings into integers. Python 3 uses a bytes

# class, which already has this behaviour.

data = [ord(c) for c in data]

l = len(data) * 8

result = [0] * l

pos = 0

for ch in data:

i = 0

while i <= 7:

result[pos] = (ch >> i) & 1

pos += 1

i += 1

return result

def __BitList_to_String(self, data):

"""Turn the list of bits -> data, into a string"""

result = []

pos = 0

c = 0

while pos < len(data):

c += data[pos] << (pos % 8)

if (pos % 8) == 7:

result.append(c)

c = 0

pos += 1

if _pythonMajorVersion < 3:

return ''.join([ chr(c) for c in result ])

else:

return bytes(result)

def __permutate(self, table, block):

"""Permutate this block with the specified table"""

return list(map(lambda x: block[x], table))

# Transform the secret key, so that it is ready for data processing

# Create the 16 subkeys, K[1] - K[16]

def __create_sub_keys(self):

"""Create the 16 subkeys K[1] to K[16] from the given key"""

key = self.__permutate(des.__pc1, self.__String_to_BitList(self.getKey()))

i = 0

# Split into Left and Right sections

self.L = key[:28]

self.R = key[28:]

while i < 16:

j = 0

# Perform circular left shifts

while j < des.__left_rotations[i]:

self.L.append(self.L[0])

del self.L[0]

self.R.append(self.R[0])

del self.R[0]

j += 1

# Create one of the 16 subkeys through pc2 permutation

self.Kn[i] = self.__permutate(des.__pc2, self.L + self.R)

i += 1

# Main part of the encryption algorithm, the number cruncher :)

def __des_crypt(self, block, crypt_type):

"""Crypt the block of data through DES bit-manipulation"""

block = self.__permutate(des.__fp, block)

self.L = block[:32]

self.R = block[32:]

# Encryption starts from Kn[1] through to Kn[16]

if crypt_type == des.ENCRYPT:

iteration = 0

iteration_adjustment = 1

# Decryption starts from Kn[16] down to Kn[1]

else:

iteration = 15

iteration_adjustment = -1

i = 0

while i < 16:

# Make a copy of R[i-1], this will later become L[i]

tempR = self.R[:]

# Permutate R[i - 1] to start creating R[i]

self.R = self.__permutate(des.__expansion_table, self.R)

# Exclusive or R[i - 1] with K[i], create B[1] to B[8] whilst here

self.R = list(map(lambda x, y: x ^ y, self.R, self.Kn[iteration]))

B = [self.R[:6], self.R[6:12], self.R[12:18], self.R[18:24], self.R[24:30], self.R[30:36], self.R[36:42], self.R[42:]]

# Optimization: Replaced below commented code with above

#j = 0

#B = []

#while j < len(self.R):

#      self.R[j] = self.R[j] ^ self.Kn[iteration][j]

#      j += 1

#      if j % 6 == 0:

#              B.append(self.R[j-6:j])

# Permutate B[1] to B[8] using the S-Boxes

j = 0

Bn = [0] * 32

pos = 0

while j < 8:

# Work out the offsets

m = (B[j][0] << 1) + B[j][5]

n = (B[j][1] << 3) + (B[j][2] << 2) + (B[j][3] << 1) + B[j][4]

# Find the permutation value

v = des.__sbox[j][(m << 4) + n]

#print hex(j * 48 + (m << 4 + n))

# Turn value into bits, add it to result: Bn

Bn[pos] = (v & 8) >> 3

Bn[pos + 1] = (v & 4) >> 2

Bn[pos + 2] = (v & 2) >> 1

Bn[pos + 3] = v & 1

pos += 4

j += 1

# Permutate the concatination of B[1] to B[8] (Bn)

self.R = self.__permutate(des.__p, Bn)

# Xor with L[i - 1]

self.R = list(map(lambda x, y: x ^ y, self.R, self.L))

# Optimization: This now replaces the below commented code

#j = 0

#while j < len(self.R):

#      self.R[j] = self.R[j] ^ self.L[j]

#      j += 1

# L[i] becomes R[i - 1]

self.L = tempR

i += 1

iteration += iteration_adjustment

# Final permutation of R[16]L[16]

self.final = self.__permutate(des.__ip, self.R + self.L)

return self.final

# Data to be encrypted/decrypted

def crypt(self, data, crypt_type):

"""Crypt the data in blocks, running it through des_crypt()"""

# Error check the data

if not data:

return ''

if len(data) % self.block_size != 0:

if crypt_type == des.DECRYPT: # Decryption must work on 8 byte blocks

raise ValueError("Invalid data length, data must be a multiple of " + str(self.block_size) + " bytes\n.")

if not self.getPadding():

raise ValueError("Invalid data length, data must be a multiple of " + str(self.block_size) + " bytes\n. Try setting the optional padding character")

else:

data += (self.block_size - (len(data) % self.block_size)) * self.getPadding()

# print "Len of data: %f" % (len(data) / self.block_size)

if self.getMode() == CBC:

if self.getIV():

iv = self.__String_to_BitList(self.getIV())

else:

raise ValueError("For CBC mode, you must supply the Initial Value (IV) for ciphering")

# Split the data into blocks, crypting each one seperately

i = 0

dict = {}

result = []

#cached = 0

#lines = 0

while i < len(data):

# Test code for caching encryption results

#lines += 1

#if dict.has_key(data[i:i+8]):

#print "Cached result for: %s" % data[i:i+8]

#      cached += 1

#      result.append(dict[data[i:i+8]])

#      i += 8

#      continue

block = self.__String_to_BitList(data[i:i+8])

# Xor with IV if using CBC mode

if self.getMode() == CBC:

if crypt_type == des.ENCRYPT:

block = list(map(lambda x, y: x ^ y, block, iv))

#j = 0

#while j < len(block):

#      block[j] = block[j] ^ iv[j]

#      j += 1

processed_block = self.__des_crypt(block, crypt_type)

if crypt_type == des.DECRYPT:

processed_block = list(map(lambda x, y: x ^ y, processed_block, iv))

#j = 0

#while j < len(processed_block):

#      processed_block[j] = processed_block[j] ^ iv[j]

#      j += 1

iv = block

else:

iv = processed_block

else:

processed_block = self.__des_crypt(block, crypt_type)

# Add the resulting crypted block to our list

#d = self.__BitList_to_String(processed_block)

#result.append(d)

result.append(self.__BitList_to_String(processed_block))

#dict[data[i:i+8]] = d

i += 8

# print "Lines: %d, cached: %d" % (lines, cached)

# Return the full crypted string

if _pythonMajorVersion < 3:

return ''.join(result)

else:

return bytes.fromhex('').join(result)

def encrypt(self, data, pad=None, padmode=None):

"""encrypt(data, [pad], [padmode]) -> bytes

data : Bytes to be encrypted

pad  : Optional argument for encryption padding. Must only be one byte

padmode : Optional argument for overriding the padding mode.

The data must be a multiple of 8 bytes and will be encrypted

with the already specified key. Data does not have to be a

multiple of 8 bytes if the padding character is supplied, or

the padmode is set to PAD_PKCS5, as bytes will then added to

ensure the be padded data is a multiple of 8 bytes.

"""

data = self._guardAgainstUnicode(data)

if pad is not None:

pad = self._guardAgainstUnicode(pad)

data = self._padData(data, pad, padmode)

return self.crypt(data, des.ENCRYPT)

def decrypt(self, data, pad=None, padmode=None):

"""decrypt(data, [pad], [padmode]) -> bytes

data : Bytes to be encrypted

pad  : Optional argument for decryption padding. Must only be one byte

padmode : Optional argument for overriding the padding mode.

The data must be a multiple of 8 bytes and will be decrypted

with the already specified key. In PAD_NORMAL mode, if the

optional padding character is supplied, then the un-encrypted

data will have the padding characters removed from the end of

the bytes. This pad removal only occurs on the last 8 bytes of

the data (last data block). In PAD_PKCS5 mode, the special

padding end markers will be removed from the data after decrypting.

"""

data = self._guardAgainstUnicode(data)

if pad is not None:

pad = self._guardAgainstUnicode(pad)

data = self.crypt(data, des.DECRYPT)

return self._unpadData(data, pad, padmode)

#############################################################################

#                              Triple DES                                  #

#############################################################################

class triple_des(_baseDes):

"""Triple DES encryption/decrytpion class

This algorithm uses the DES-EDE3 (when a 24 byte key is supplied) or

the DES-EDE2 (when a 16 byte key is supplied) encryption methods.

Supports ECB (Electronic Code Book) and CBC (Cypher Block Chaining) modes.

pyDes.des(key, [mode], [IV])

key  -> Bytes containing the encryption key, must be either 16 or

24 bytes long

mode -> Optional argument for encryption type, can be either pyDes.ECB

(Electronic Code Book), pyDes.CBC (Cypher Block Chaining)

IV  -> Optional Initial Value bytes, must be supplied if using CBC mode.

Must be 8 bytes in length.

pad  -> Optional argument, set the pad character (PAD_NORMAL) to use

during all encrypt/decrpt operations done with this instance.

padmode -> Optional argument, set the padding mode (PAD_NORMAL or

PAD_PKCS5) to use during all encrypt/decrpt operations done

with this instance.

"""

def __init__(self, key, mode=ECB, IV=None, pad=None, padmode=PAD_NORMAL):

_baseDes.__init__(self, mode, IV, pad, padmode)

self.setKey(key)

def setKey(self, key):

"""Will set the crypting key for this object. Either 16 or 24 bytes long."""

self.key_size = 24  # Use DES-EDE3 mode

if len(key) != self.key_size:

if len(key) == 16: # Use DES-EDE2 mode

self.key_size = 16

else:

raise ValueError("Invalid triple DES key size. Key must be either 16 or 24 bytes long")

if self.getMode() == CBC:

if not self.getIV():

# Use the first 8 bytes of the key

self._iv = key[:self.block_size]

if len(self.getIV()) != self.block_size:

raise ValueError("Invalid IV, must be 8 bytes in length")

self.__key1 = des(key[:8], self._mode, self._iv,

self._padding, self._padmode)

self.__key2 = des(key[8:16], self._mode, self._iv,

self._padding, self._padmode)

if self.key_size == 16:

self.__key3 = self.__key1

else:

self.__key3 = des(key[16:], self._mode, self._iv,

self._padding, self._padmode)

_baseDes.setKey(self, key)

# Override setter methods to work on all 3 keys.

def setMode(self, mode):

"""Sets the type of crypting mode, pyDes.ECB or pyDes.CBC"""

_baseDes.setMode(self, mode)

for key in (self.__key1, self.__key2, self.__key3):

key.setMode(mode)

def setPadding(self, pad):

"""setPadding() -> bytes of length 1. Padding character."""

_baseDes.setPadding(self, pad)

for key in (self.__key1, self.__key2, self.__key3):

key.setPadding(pad)

def setPadMode(self, mode):

"""Sets the type of padding mode, pyDes.PAD_NORMAL or pyDes.PAD_PKCS5"""

_baseDes.setPadMode(self, mode)

for key in (self.__key1, self.__key2, self.__key3):

key.setPadMode(mode)

def setIV(self, IV):

"""Will set the Initial Value, used in conjunction with CBC mode"""

_baseDes.setIV(self, IV)

for key in (self.__key1, self.__key2, self.__key3):

key.setIV(IV)

def encrypt(self, data, pad=None, padmode=None):

"""encrypt(data, [pad], [padmode]) -> bytes

data : bytes to be encrypted

pad  : Optional argument for encryption padding. Must only be one byte

padmode : Optional argument for overriding the padding mode.

The data must be a multiple of 8 bytes and will be encrypted

with the already specified key. Data does not have to be a

multiple of 8 bytes if the padding character is supplied, or

the padmode is set to PAD_PKCS5, as bytes will then added to

ensure the be padded data is a multiple of 8 bytes.

"""

ENCRYPT = des.ENCRYPT

DECRYPT = des.DECRYPT

data = self._guardAgainstUnicode(data)

if pad is not None:

pad = self._guardAgainstUnicode(pad)

# Pad the data accordingly.

data = self._padData(data, pad, padmode)

if self.getMode() == CBC:

self.__key1.setIV(self.getIV())

self.__key2.setIV(self.getIV())

self.__key3.setIV(self.getIV())

i = 0

result = []

while i < len(data):

block = self.__key1.crypt(data[i:i+8], ENCRYPT)

block = self.__key2.crypt(block, DECRYPT)

block = self.__key3.crypt(block, ENCRYPT)

self.__key1.setIV(block)

self.__key2.setIV(block)

self.__key3.setIV(block)

result.append(block)

i += 8

if _pythonMajorVersion < 3:

return ''.join(result)

else:

return bytes.fromhex('').join(result)

else:

data = self.__key1.crypt(data, ENCRYPT)

data = self.__key2.crypt(data, DECRYPT)

return self.__key3.crypt(data, ENCRYPT)

def decrypt(self, data, pad=None, padmode=None):

"""decrypt(data, [pad], [padmode]) -> bytes

data : bytes to be encrypted

pad  : Optional argument for decryption padding. Must only be one byte

padmode : Optional argument for overriding the padding mode.

The data must be a multiple of 8 bytes and will be decrypted

with the already specified key. In PAD_NORMAL mode, if the

optional padding character is supplied, then the un-encrypted

data will have the padding characters removed from the end of

the bytes. This pad removal only occurs on the last 8 bytes of

the data (last data block). In PAD_PKCS5 mode, the special

padding end markers will be removed from the data after

decrypting, no pad character is required for PAD_PKCS5.

"""

ENCRYPT = des.ENCRYPT

DECRYPT = des.DECRYPT

data = self._guardAgainstUnicode(data)

if pad is not None:

pad = self._guardAgainstUnicode(pad)

if self.getMode() == CBC:

self.__key1.setIV(self.getIV())

self.__key2.setIV(self.getIV())

self.__key3.setIV(self.getIV())

i = 0

result = []

while i < len(data):

iv = data[i:i+8]

block = self.__key3.crypt(iv,    DECRYPT)

block = self.__key2.crypt(block, ENCRYPT)

block = self.__key1.crypt(block, DECRYPT)

self.__key1.setIV(iv)

self.__key2.setIV(iv)

self.__key3.setIV(iv)

result.append(block)

i += 8

if _pythonMajorVersion < 3:

data = ''.join(result)

else:

data = bytes.fromhex('').join(result)

else:

data = self.__key3.crypt(data, DECRYPT)

data = self.__key2.crypt(data, ENCRYPT)

data = self.__key1.crypt(data, DECRYPT)

return self._unpadData(data, pad, padmode)

CBC那裡庫函式好像有點問題……於是直接呼叫的ECB模式，然後自己寫的CBC，如下，同時可寫出逆運算。des_cbc_decrypt之後的hex encode也是先編碼低四位，後編碼高四位，所以需要更改一下變成16進位制的演算法

import pyDes

from pwn import xor

from gmpy2 import isqrt_rem

def hexc(inp):

inp = inp.encode("hex")

return "".join(inp[i + 1] + inp[i] for i in range(0, len(inp), 2))

def ihexc(inp):

inp = format(inp, 'x')

inp = '0' * (len(inp) % 2) + inp

return "".join(inp[i + 1] + inp[i] for i in range(0, len(inp), 2)).decode("hex")

def compute(inp):

des = pyDes.des("*2017*10")

rest = len(inp) % 8

if rest:

inp += "\x00" * (8 - rest)

inp = [inp[i:i+8] for i in range(0, len(inp), 8)]

iv = "\x00" * 8

out = [iv]

for i in range(len(inp)):

out.append(des.decrypt(xor(inp[i], out[i][::-1])))

return hexc("".join(out[1:]))

def inv_compute(inp):

des = pyDes.des("*2017*10")

inp = ihexc(inp)

ap = len(inp) % 8

if ap:

inp = "\x00" * (8 - ap) + inp

inp = [inp[i:i+8] for i in range(0, len(inp), 8)]

iv = "\x00" * 8

out = []

for i in range(len(inp)):

out.append(xor(des.encrypt(inp[i]), iv[::-1]))

iv = inp[i]

return "".join(out).rstrip("\x00")

DES 這部分花的時間大概最長……因為C++的STL也不是很好除錯……所以只能一輪一輪的對……看是不是改對了……

二. LuaJIT

我剛開始沒有直接看MIRACL，因為感覺太煩了……（看見了類似傅立葉變換的東西），所以從最後的lua開始繼續逆向，lua位元組碼和0x5A異或了，這裡直接idapython dump出來異或一下就好了，然後得到的是LuaJIT的bytecode（注意不是lua），位元組碼的enum定義我沒有找到……所以主要用了這兩個工具。由於之前從來沒研究過lua……所以還是花了不少時間的

https://github.com/viruscamp/luadec

看起來好像大家都在用的一個反編譯器，可得到原始碼，缺點：由於是JIT的緣故，所以得不到在函式中使用全域性變數引用，導致某些地方看起來很詭異……

而且好像必須要在同一個目錄下將檔案命名為test.lua才可以執行……（貌似可以改AutoIt指令碼，不過懶……）

得到的反編譯程式碼大概長這個德行……搜尋xyByInt，可以找到這個http://blog.csdn.net/xianyun2009/article/details/44796823，貌似是lua的某個大整數實現

經過精簡後，得到的程式碼大概是這樣的

function show(a)

print(get(a))

end

function get(a)

s = {a[#a]}

for i=#a-1, 1, -1 do

table.insert(s, string.format("%04d", a[i]))

end

return table.concat(s, "")

end

function create(s)

if s["xyBitInt"] == true then return s end

n, t, a = math.floor(#s/4), 1, {}

a["xyBitInt"] = true

if #s%4 ~= 0 then a[n + 1], t = tonumber(string.sub(s, 1, #s%4), 10), #s%4 + 1 end

for i = n, 1, -1 do a[i], t= tonumber(string.sub(s, t, t + 3), 10), t + 4 end

return a

end

function add(a,b)

local var_3_2 = create(a)

var_3_3 = create(b)

c = create("0")

t =  0

local var_3_1 = var_3_3

local var_3_0 = var_3_2

for var_3_5 =  1 , math.max(c, var_3_5), 1  do --location 0022, loop ends at 0048-1

if not unknown2 then

--jump to 0028 (if previous if statement is false)

--location 0028

if not unknown3 then

--jump to 0033 (if previous if statement is false)

--location 0033

t =  (  t +  0  +  0  )

local var_3_7 = t % uget_3_0

local var_3_8 = math.floor( t / uget_3_0 )

t = var_3_8

c[var_3_5] = var_3_7

end --location 0047, loops back to 0023-1

if t ~= 0 then

--jump to 0068 (if previous if statement is false)

repeat

var_3_4 = t % uget_3_0

local var_3_5 = math.floor( t / uget_3_0 )

t = var_3_5

c[ c +  1 ] = var_3_4

--jump to 0048 (if previous if statement is false)

--location 0068

until false or (previous if statement is true) --location 0068

return c

end

function bsub(a, b)

a, b, c, t = create(a), create(b), create("0"), 0

for i = 1, #a do

c[i] = a[i] - t - (b[i] or 0)

if c[i] < 0 then t, c[i] = 1, c[i] + mod  else t = 0 end

end

return c

end

function by(a, b)

a, b, c, t = create(a), create(b), create("0"), 0

for i = 1, #a do

for j = 1, #b do

t = t + (c[i + j - 1] or 0) + a[i] * b[j]

c[i + j - 1], t = t%mod, math.floor(t / mod)

end

if t ~= 0 then c[i + #b], t = t + (c[i + #b] or 0), 0 end

end

return c

end

function myst(xut)

c = add(xut, "1001")

d =  0

for i =  1 , 100 , 1  do --location 0011, loop ends at 0028-1

for j =  1 , 100 , 1  do --location 0015, loop ends at 0022-1

c = add(c, "1001")

end --location 0021, loops back to 0016-1

c = add(c, "10101")

end --location 0027, loops back to 0012-1

c = by(c, "983751509373")

for i =  1 , 13 , 1  do --location 0036, loop ends at 0073-1

c = bsub(c, "1023")

for j =  1 , 14 , 1  do --location 0045, loop ends at 0072-1

c = bsub(c, "1203")

for k =  1 , 15 , 1  do --location 0054, loop ends at 0071-1

c = bsub(c, "1230")

for l =  1 , 16 , 1  do --location 0063, loop ends at 0070-1

c = bsub(c, "1231")

--until false or (previous if statement is true) --location 0068

end --location 0069, loops back to 0064-1

end --location 0070, loops back to 0055-1

end --location 0071, loops back to 0046-1

end --location 0072, loops back to 0037-1

c = add(c, "1")

c = by(c, "2")

e = get(c)

lene = string.len(e)

leny = string.len(uget_6_0)

if lene == leny then

--jump to 0129 (if previous if statement is false)

i =  1

d =  1

if i <= lene then

--jump to 0129 (if previous if statement is false)

repeat

var_6_0 = string.byte(e, i)

local var_6_1 = string.byte(uget_6_0, i)

if var_6_0 ~= var_6_1 then

--jump to 0125 (if previous if statement is false)

d =  0

--jump to 0129 (if previous if statement is false)

i =  i +  1

--jump to 0105 (if previous if statement is false)

until false or (previous if statement is true) --location 0129

return d

end

function add(INPUT_VAR_0_)

return  INPUT_VAR_0_ + INPUT_VAR_1_

end

function sub(INPUT_VAR_0_)

return  INPUT_VAR_0_ - INPUT_VAR_1_

end

function someFunc9()

local var_9_0 = 10000 --var_9_0 NUMBER-NUMBER

local var_9_1 = "1574592838300862641516215149137548264158058079230003764126382984039489925466995870724568174393389905601620735902909057604303543552180706761904" --strings longer than 40 characters get cut

end

還有一些不重要的部分沒有化簡，不過可以大概看出lua的邏輯了，執行多次運算後，將其化為2進位制的字串和var_9_1做比較。這裡剛開始判斷錯了add，以為是sub，所以寫出來的程式碼不對……然後手動的去更改luaJIT的位元組碼，在get那裡直接return，然後用Lua的C API寫了個小的wrapper來測試

#include

#include "lua.h"

#include "lualib.h"

#include "lauxlib.h"

/* Convenience stuff */

static void close_state(lua_State **L) { lua_close(*L); }

#define cleanup(x) __attribute__((cleanup(x)))

#define auto_lclose cleanup(close_state)

int main(int argc, char *argv[])

{

/* Create VM state */

auto_lclose lua_State *L = luaL_newstate();

if (!L)

return 1;

luaL_openlibs(L); /* Open standard libraries */

/* Load config file */

if (argc > 1) {

luaL_loadfile(L, argv[1]); /* (1) */

//luaJIT_setmode(L, 0, 0);

int ret = lua_pcall(L, 0, 0, 0);

if (ret != 0) {

fprintf(stderr, "%s\n", lua_tostring(L, -1));

return 1;

}

}

lua_pushstring(L, argv[2]);

lua_setfield(L, -10002, "xut");

lua_getfield(L, -10002, "myst");

lua_pcall(L, 0, 1, 0);

printf("%s\n", lua_tolstring(L, -1, 0));

lua_settop(L, 0); /* (4) */

return 0;

}

這裡簡單說一下patch luaJIT位元組碼的過程，不知道位元組碼的enum很頭疼……於是上網搜尋的途中找到了這個有用的工具https://github.com/franko/luajit-lang-toolkit, 其中run.lua的-bx功能可以檢視lua位元組碼對應的16進位制位元組是什麼……讓更改位元組碼變得簡單了許多，效果大概是這個樣子

如果要增加一個kgc的話，需要更改的地方有kgc的長度和這段proto的總長度，其中proto總長度是U128LE編碼，然後大概學習了一下怎麼寫luaJIT位元組碼……後來發現自己把add看成了sub以後解決問題就沒啥用了……總的來說lua位元組碼還是挺難除錯的……不知道有沒有好的方法……之後可以考慮寫個hook的工具之類的……

第一次逆向luaJIT，感覺還是挺有趣的，學到了不少東西，luaJIT程式碼逆向出來後的逆運算如下

def lua(target):

target = int(target) / 2 - 1

for i in range(13):

for j in range(14):

for k in range(15):

for l in range(16):

target += 1231

target += 1230

target += 1203

target += 1023

target /= 983751509373

for i in range(100):

target -= 10101

for j in range(100):

target -= 1001

target -= 1001

return target

可以看出並不是很困難……大概是當時比較困……所以出了各種差錯……

三. MIRACL

攻破了luaJIT和des以後，就只剩下我們最後的MIRACL了……首先我們來看看他都做了什麼

首先把我們des_cbc_decrypt之後的結果16進位制化以後變成了一個mirvar的大整數，然後乘了個173，之後做了fft_mult（使用傅立葉變換進行快速大數乘法）1817，以後做了某個計算（其實是除法，這個之後我們說怎麼識別出來的）

看起來很簡單對不對……結果怎麼算怎麼不對……fft_mult那裡的輸出和輸入永遠對不上……（輸入的變換也不是線性的……所以只能逆向了）於是考慮可能fft_mult被更改過了……檢視fft_mult

之前都一樣……最後居然多出了兩行！

然後通過olly根據輸入輸出猜測，得到第一個是平方的演算法，第二個根據bindiff得到是decr（減法，不過內建了第二個減數為1001）

於是fft_mult的實際演算法是 (inp * 1817)^2 - 1001……嗯……怪不得怎麼都算不對

最後還差divide那個函式，那個函式長這樣……

嗯……看了一眼就不想看了……遂祭出olly的連蒙帶猜的的方法……輸入了幾組資料，觀察輸出

輸入:1

輸出：713

輸入：2

輸出：436

輸入：3

輸出：159

嗯……這個好像第二個反過來再除以輸入就是317啊……驗證了一下以後還真是

至此，我們已將所有程式逆向完畢……完整指令碼如下

import pyDes

from pwn import xor

from gmpy2 import isqrt_rem

def hexc(inp):

inp = inp.encode("hex")

return "".join(inp[i + 1] + inp[i] for i in range(0, len(inp), 2))

def ihexc(inp):

inp = format(inp, 'x')

inp = '0' * (len(inp) % 2) + inp

return "".join(inp[i + 1] + inp[i] for i in range(0, len(inp), 2)).decode("hex")

def compute(inp):

des = pyDes.des("*2017*10")

rest = len(inp) % 8

if rest:

inp += "\x00" * (8 - rest)

inp = [inp[i:i+8] for i in range(0, len(inp), 8)]

iv = "\x00" * 8

out = [iv]

for i in range(len(inp)):

out.append(des.decrypt(xor(inp[i], out[i][::-1])))

return hexc("".join(out[1:]))

def inv_compute(inp):

des = pyDes.des("*2017*10")

inp = ihexc(inp)

ap = len(inp) % 8

if ap:

inp = "\x00" * (8 - ap) + inp

inp = [inp[i:i+8] for i in range(0, len(inp), 8)]

iv = "\x00" * 8

out = []

for i in range(len(inp)):

out.append(xor(des.encrypt(inp[i]), iv[::-1]))

iv = inp[i]

return "".join(out).rstrip("\x00")

def lua(target):

target = int(target) / 2 - 1

for i in range(13):

for j in range(14):

for k in range(15):

for l in range(16):

target += 1231

target += 1230

target += 1203

target += 1023

target /= 983751509373

for i in range(100):

target -= 10101

for j in range(100):

target -= 1001

target -= 1001

return target

assert(compute("1122334455667788").upper() == "BEC20560FEB6FCEFF93B5F94B3CF259D")

assert(compute("1234567890abcdef").upper() == "C7BF2111CD9A94A30E31F260D9BF9432")

n1 = 0xad

n2 = 0x719

target = "1574592838300862641516215149137548264158058079230003764126382984039489925466995870724568174393389905601620735902909057604303543552180706761904"

target = lua(target)

target = int(str(target)[::-1])

assert(target % 317 == 0)

target /= 317

target += 1001

res, rem = isqrt_rem(target)

assert(int(rem) == 0)

prod = (n2 * n1)

assert(res % prod == 0)

res /= prod

print inv_compute(res)

"""

inp = "0011223344556677"

num1 = int(compute(inp), 16)

num1 *= n1

print hex(num1 * n2)

"""

總結

這題說難也不難……就是很繁瑣……而且考察的知識點比較雜……附上我的IDA6.95 idb和解題用的全部程式碼……感興趣的人可以看一眼……

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