[Rev赛题复现]DASCTF Apr X FATE 2022

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[Rev赛题复现]DASCTF Apr X FATE 2022

本文为看雪论坛精华文章
看雪论坛作者ID:t0hka1


总共4题,贴了3题,还有一道go逆向的直接看这位师傅的吧。(https://bbs.pediy.com/thread-273162.htm


Crackme

 
几个关键点:mfc逆向,win32 加密api的识别,ZwSetInformationThread反调试。

看程序图标是个mfc的程序,先打开看看,随便输入一点东西,看到弹窗弹出:
[Rev赛题复现]DASCTF Apr X FATE 2022
直接拖进ida搜索Wrong!!!字符串,借此通过查看引用跳转到主函数。
 
先简单的修复一下变量名:
[Rev赛题复现]DASCTF Apr X FATE 2022
大概可以看到key的前4位经过一次sub_403510,后4位也经过一次sub_403510,整个key8位又经过一次sub_403510。
 
随后进入sub_403510函数看一看,看到有一堆为win32 加密api的函数。
[Rev赛题复现]DASCTF Apr X FATE 2022
那就对照着MSDN一个个函数查阅一下:
 
CryptAcquireContext 函数用于获取特定加密服务提供程序 (CSP) 中特定密钥容器的句柄。此返回的句柄用于调用使用所选 CSP 的 CryptoAPI 函数。
BOOL CryptAcquireContextA([out] HCRYPTPROV *phProv,[in]  LPCSTR     szContainer,[in]  LPCSTR     szProvider,[in]  DWORD      dwProvType,[in]  DWORD      dwFlags);


CryptCreateHash 函数启动数据流的哈希。它创建加密服务提供程序 (CSP) 哈希对象的句柄并将其返回给调用应用程序。此句柄用于对 CryptHashData 和 CryptHashSessionKey 的后续调用,以哈希会话密钥和其他数据流。
BOOL CryptCreateHash([in]  HCRYPTPROV hProv,[in]  ALG_ID     Algid,[in]  HCRYPTKEY  hKey,[in]  DWORD      dwFlags,[out] HCRYPTHASH *phHash);


注意这里Algid是标识要使用的哈希算法的参数,通过不同的值的传入选择不同的hash算法,可查下面的链接:ALG_ID (Wincrypt.h) - Win32 apps | Microsoft Docshttps://docs.microsoft.com/en-us/windows/win32/seccrypto/alg-id
[Rev赛题复现]DASCTF Apr X FATE 2022
[Rev赛题复现]DASCTF Apr X FATE 2022

CryptHashData 函数将数据添加到指定的哈希对象。此函数和CryptHashSessionKey可以多次调用,以计算长数据流或不连续数据流的哈希值。
BOOL CryptHashData([in] HCRYPTHASH hHash,[in] const BYTE *pbData,[in] DWORD      dwDataLen,[in] DWORD      dwFlags);


CryptGetHashParam 函数检索控制哈希对象操作的数据。可以使用此函数检索实际的哈希值。
BOOL CryptGetHashParam([in]      HCRYPTHASH hHash,[in]      DWORD      dwParam,[out]     BYTE       *pbData,[in, out] DWORD      *pdwDataLen,[in]      DWORD      dwFlags);


CryptEncrypt 函数对数据进行加密。用于加密数据的算法由 CSP 模块持有的密钥指定,并由 hKey 参数引用。
BOOL CryptEncrypt([in]      HCRYPTKEY  hKey,[in]      HCRYPTHASH hHash,[in]      BOOL       Final,[in]      DWORD      dwFlags,[in, out] BYTE       *pbData,[in, out] DWORD      *pdwDataLen,[in]      DWORD      dwBufLen);

程序大概的逻辑就是这样:
[Rev赛题复现]DASCTF Apr X FATE 2022
然后我们就通过动调去拿数据,这里有两种方式,一种是通过ida patch反调试函数的方式,一种是通过od 的sharp od插件直接绕过。

绕反调试方法一:先用ida绕过反调试


对比各种反调试和去IAT表找导入函数没有找到,后面在strings界面可以发现ZwSetInformationThread反调试的特征ZwSetInformationThread - CTF Wiki (ctf-wiki.org)(https://ctf-wiki.org/reverse/windows/anti-debug/zwsetinformationthread/
 
ZwSetInformationThread通过为线程设置 ThreadHideFromDebugger,可以禁止线程产生调试事件。

绕过: ZwSetInformationThread 函数的第 2 个参数为 ThreadHideFromDebugger,其值为 0x11。调试执行到该函数时,若发现第 2 个参数值为 0x11,跳过或者将 0x11 修改为其他值即可。
[Rev赛题复现]DASCTF Apr X FATE 2022
看来是自己实现调用dll导入的。
[Rev赛题复现]DASCTF Apr X FATE 2022
类似于这种写法:[Rev赛题复现]DASCTF Apr X FATE 2022
我们可以先在调用处下一个断点,跑起来之后再修改patch 0x11 改掉,我这里patch成了0x9。
[Rev赛题复现]DASCTF Apr X FATE 2022
然后继续下断点拿密文。
[Rev赛题复现]DASCTF Apr X FATE 2022
[Rev赛题复现]DASCTF Apr X FATE 2022
[Rev赛题复现]DASCTF Apr X FATE 2022
然后再扔到md5解密网站解密。
[Rev赛题复现]DASCTF Apr X FATE 2022
类似的拿到sha1解密后的key后四位 https://crackstation.net/
[Rev赛题复现]DASCTF Apr X FATE 2022
key:NocTuRne
 
md5(key):C0804C74E05B4C7440AC4D7480954C74


绕反调试方法二:OD odsharp插件直接绕

[Rev赛题复现]DASCTF Apr X FATE 2022
之后就是模拟调用win32 的aes解密api来解密的过程了。
#include <Windows.h>#include <stdio.h>#include <wincrypt.h>
int main(){ BYTE pbData[] = {0x5c,0x53,0xa4,0xa4,0x1d,0x52,0x43,0x7a,0x9f,0xa1,0xe9,0xc2,0x6c,0xa5,0x90,0x90,0x0}; //key_buf BYTE flag_encrypt[] = {0x5B, 0x9C, 0xEE, 0xB2, 0x3B, 0xB7, 0xD7, 0x34, 0xF3, 0x1B, 0x75, 0x14, 0xC6, 0xB2, 0x1F, 0xE8, 0xDE, 0x33, 0x44, 0x74, 0x75, 0x1B, 0x47, 0x6A, 0xD4, 0x37, 0x51, 0x88, 0xFC, 0x67, 0xE6, 0x60, 0xDA, 0x0D, 0x58, 0x07, 0x81, 0x43, 0x53, 0xEA, 0x7B, 0x52, 0x85, 0x6C, 0x86, 0x65, 0xAF, 0xB4,0x0}; DWORD dwDataLen = 0x10; DWORD ddwDataLen; DWORD* pdwDataLen = &ddwDataLen; *pdwDataLen = 0x20;

BOOL v6; // [esp+4h] [ebp-18h] HCRYPTKEY phKey; // [esp+Ch] [ebp-10h] BYREF HCRYPTPROV phProv; // [esp+10h] [ebp-Ch] BYREF HCRYPTHASH phHash; // [esp+14h] [ebp-8h] BYREF
phProv = 0; phHash = 0; phKey = 0; v6 = CryptAcquireContextA(&phProv, 0, 0, 0x18u, 0xF0000000); if (v6) { v6 = CryptCreateHash(phProv, 0x8003u, 0, 0, &phHash); if (v6) { v6 = CryptHashData(phHash, pbData, dwDataLen, 0); if (v6) { v6 = CryptDeriveKey(phProv, 0x660Eu, phHash, 1u, &phKey);// key的md5值再生成aes密钥 if (v6) v6 = CryptDecrypt(phKey, 0, 1, 0, flag_encrypt, pdwDataLen); printf("%s", flag_encrypt); } } } if (phKey) CryptDestroyKey(phKey); if (phHash) CryptDestroyHash(phHash); if (phProv) CryptReleaseContext(phProv, 0); return v6;}


拿到flag!
[Rev赛题复现]DASCTF Apr X FATE 2022


补充方法:Hook Windows API 求解


另外经mas0n师傅补充,附上frida来hook求解的方法。
  var baseAddr = Process.findModuleByName('Crackme_1.exe');
// input 32 length flag, e.g. 11111111111111111111111111111111// key: NocTuRne// frida attach -p 48964 -l agenthook_win.js

// memcmpvar hookAddr = ptr(0x0109D4BC);Interceptor.attach(hookAddr, { onEnter: function(args) { let Buf1 = args[0]; let Buf2 = args[1]; let Size = args[2]; console.log("-----n[Size]n", Size); let size = Size.toInt32(); console.log("-----n[Buf1]n", Buf1.readByteArray(size)); console.log("-----n[Buf2]n", Buf2.readByteArray(size)); console.log("---------------------------"); }, onLeave: function(arg) { return arg; }})
var libAddr = Process.findModuleByName('ADVAPI32.dll');var fn_CryptEncrypt = libAddr.getExportByName("CryptEncrypt");var fn_CryptDecrypt = libAddr.getExportByName("CryptDecrypt");
var flag = null;Interceptor.replace(fn_CryptEncrypt, fn_CryptDecrypt);Interceptor.attach(fn_CryptDecrypt, { onEnter: function(args) { args[4].writeByteArray([0x5b,0x9c,0xee,0xb2,0x3b,0xb7,0xd7,0x34,0xf3,0x1b,0x75,0x14,0xc6,0xb2,0x1f,0xe8,0xde,0x33,0x44,0x74,0x75,0x1b,0x47,0x6a,0xd4,0x37,0x51,0x88,0xfc,0x67,0xe6,0x60,0xda,0x0d,0x58,0x07,0x81,0x43,0x53,0xea,0x7b,0x52,0x85,0x6c,0x86,0x65,0xaf,0xb4]); args[5].writeInt(0x40); flag = args[4]; console.log("hook fn_CryptDecrypt"); return args; }, onLeave: function(arg) { console.log(flag.readCString()); return arg; }})


奇怪的交易


拖进ida里发现经过upx加壳,直接upx -d脱壳。
 
接着再使用pyinstxtractor解包得到一堆文件(比较坑的地方是本地python环境必须与源程序的python环境相同才能解包PYZ-00.pyz),下面是彻底解包后的几个关键的文件。
[Rev赛题复现]DASCTF Apr X FATE 2022
奇怪的交易.pyc文件内容如下:

python反编译 - 在线工具 (tool.lu)(https://tool.lu/pyc/
[Rev赛题复现]DASCTF Apr X FATE 2022
样其实逻辑很明显还是有问题的,题目特意用的Python3.10版本,导致反编译结果会不正确。
 
通过pycdump可以dump出opcode,对比进行变量名和代码逻辑的修复。
 
可以参考下面的文章:
 
Python字节码文档 Python字节码详解(介绍了Python的特有类型以及遍历等操作)(https://blog.csdn.net/weixin_46263782/article/details/120930191

[Rev赛题复现]DASCTF Apr X FATE 2022

修复后的 奇怪的交易.py
from cup import *from libnum import *
if __name__ == '__main__': flag = input('请输入flag') pub_key = [ 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m = libnum.s2n(flag) c = str(pow(m, pub_key[1], pub_key[0])) # 极长的一串东西 store = [] cipher = [3532577106, 1472742623, 3642468664, 4193500461, 2398676029, 617653972, 1474514999, 1471783658, 1012864704, 3615627536, 993855884, 438456717, 3358938551, 3906991208, 198959101, 3317190635, 3656923078, 613157871, 2398768861, 97286225, 2336972940, 1471645170, 3233163154, 583597118, 2863776301, 3183067750, 1384330715, 2929694742, 3522431804, 2181488067, 3303062236, 3825712422, 145643141, 2148976293, 2940910035, 506798154, 994590281, 2231904779, 3389770074, 2814269052, 1105937096, 1789727804, 3757028753, 2469686072, 1162286478, 680814033, 2934024098, 2162521262, 4048876895, 2121620700, 4240287315, 2391811140, 3396611602, 3091349617, 3031523010, 2486958601, 3164065171, 1285603712, 798920280, 2337813135, 4186055520, 3523024366, 1077514121, 1436444106, 2731983230, 1507202797, 500756149, 198754565, 2382448647, 880454148, 1970517398, 3217485349, 1161840191, 560498076, 1782600856, 2643721918, 1285196205, 788797746, 1195724574, 4061612551, 103427523, 2502688387, 4147162188, 617564657, 978211984, 1781482121, 2205798970, 3939973102, 3826603515, 659557668, 2582884932, 1561884856, 2217488804, 1189296962, 169145316, 2781742156, 1323893433, 824667876, 408202876, 3759637634, 4094868412, 1508996065, 162419237, 3732146944, 3083560189, 3955940127, 2393776934, 2470191468, 3620861513, 481927014, 2756226070, 3154651143, 1261069441, 2063238535, 2222237213, 101459755, 3159774417, 1721190841, 1078395785, 176506553, 3552913423, 1566142515, 1938949000, 1499289517, 3315102456, 829714860, 3843359394, 952932374, 1283577465, 2045007203, 3957761944, 3767891405, 2917089623, 3296133521, 482297421, 1734231412, 3670478932, 2575334979, 2827842737, 3413631016, 1533519803, 4008428470, 3890643173, 272960248, 317508587, 3299937500, 2440520601, 27470488, 1666674386, 1737927609, 750987808, 2385923471, 2694339191, 562925334, 2206035395]
i = 0 # rsa 生成的密文遍历加密 while i < len(c): # i<155 index = 0 for ii in c[i:i + 4]: index = (index << 8) + ord(ii) store.append(index)
i += 4 if not i < len(c): key = [54, 54, 54, 54] store_len = len(store) res = encrypt(store_len, store, key) if store == cipher: print('You are right!') input('') quit() else: print('Why not drink a cup of tea and have a rest?')
continue


发现从cup包导入了一个encrypt函数。
 
以下是对经key加密后的cup.pyc.encrypted的解密脚本
[原创]Python逆向——Pyinstaller逆向-软件逆向-看雪论坛-安全社区|安全招聘|bbs.pediy.com
https://bbs.pediy.com/thread-271253.htm
      #!/usr/bin/env python3import tinyaesimport zlib
CRYPT_BLOCK_SIZE = 16
# 从crypt_key.pyc获取key,也可自行反编译获取key = bytes('0000000000000tea', 'utf-8')
inf = open('cup.pyc.encrypted', 'rb') # 打开加密文件outf = open('output.pyc', 'wb') # 输出文件
# 按加密块大小进行读取iv = inf.read(CRYPT_BLOCK_SIZE)
cipher = tinyaes.AES(key, iv)
# 解密plaintext = zlib.decompress(cipher.CTR_xcrypt_buffer(inf.read()))
# 补pyc头(最后自己补也行)outf.write(b'x6fx0dx0dx0a')
# 写入解密数据outf.write(plaintext)
inf.close()outf.close()

解密得到发现是一个python实现的xxtea加密,最基础的版本,甚至连key都没变。
 
python实现xxtea加解密参考链接
https://www.icode9.com/content-1-1126418.html

#!/usr/bin/env python# visit https://tool.lu/pyc/ for more informationimport libnumfrom ctypes import *
def MX(z, y, total, key, p, e): temp1 = (z.value >> 5 ^ y.value << 2) + (y.value >> 3 ^ z.value << 4) temp2 = (total.value ^ y.value) + (key

^ z.value) return c_uint32(temp1 ^ temp2)

def encrypt(ᘗ, ᘖ, ᘘ): ᘜ = 0x9E3779B9L ᘛ = 6 + 52 // ᘗ total = c_uint32(0) ᘔ = c_uint32(ᘖ[ᘗ - 1]) ᘕ = c_uint32(0) if ᘛ > 0: total.value += ᘜ ᘕ.value = total.value >> 2 & 3 ᘚ = c_uint32(ᘖ[0]) ᘖ[ᘗ - 1] = c_uint32(ᘖ[ᘗ - 1] + MX(ᘔ, ᘚ, total, ᘘ, ᘗ - 1, ᘕ).value).value ᘔ.value = ᘖ[ᘗ - 1] ᘛ -= 1 if not ᘛ > 0: return ᘖ


先解密xxtea得到结果rsa加密的密文c。
from ctypes import *

def MX(z, y, total, key, p, e): temp1 = (z.value>>5 ^ y.value<<2) + (y.value>>3 ^ z.value<<4) temp2 = (total.value ^ y.value) + (key[(p&3) ^ e.value] ^ z.value)
return c_uint32(temp1 ^ temp2)

def encrypt(n, v, key): delta = 0x9e3779b9 rounds = 6 + 52//n
total = c_uint32(0) z = c_uint32(v[n-1]) e = c_uint32(0)
while rounds > 0: total.value += delta e.value = (total.value >> 2) & 3 for p in range(n-1): y = c_uint32(v

) v

= c_uint32(v

+ MX(z,y,total,key,p,e).value).value z.value = v

y = c_uint32(v[0]) v[n-1] = c_uint32(v[n-1] + MX(z,y,total,key,n-1,e).value).value z.value = v[n-1] rounds -= 1
return v

def decrypt(n, v, key): delta = 0x9E3779B9 rounds = 6 + 52//n
total = c_uint32(rounds * delta) y = c_uint32(v[0]) e = c_uint32(0)
while rounds > 0: e.value = (total.value >> 2) & 3 for p in range(n-1, 0, -1): z = c_uint32(v[p-1]) v

= c_uint32((v

- MX(z,y,total,key,p,e).value)).value y.value = v

z = c_uint32(v[n-1]) v[0] = c_uint32(v[0] - MX(z,y,total,key,0,e).value).value y.value = v[0] total.value -= delta rounds -= 1
return v

# test if __name__ == "__main__": # 该算法中每次可加密不只64bit的数据,并且加密的轮数由加密数据长度决定
k = [54, 54, 54, 54] n = 155
res=[3532577106, 1472742623, 3642468664, 4193500461, 2398676029, 617653972, 1474514999, 1471783658, 1012864704, 3615627536, 993855884, 438456717, 3358938551, 3906991208, 198959101, 3317190635, 3656923078, 613157871, 2398768861, 97286225, 2336972940, 1471645170, 3233163154, 583597118, 2863776301, 3183067750, 1384330715, 2929694742, 3522431804, 2181488067, 3303062236, 3825712422, 145643141, 2148976293, 2940910035, 506798154, 994590281, 2231904779, 3389770074, 2814269052, 1105937096, 1789727804, 3757028753, 2469686072, 1162286478, 680814033, 2934024098, 2162521262, 4048876895, 2121620700, 4240287315, 2391811140, 3396611602, 3091349617, 3031523010, 2486958601, 3164065171, 1285603712, 798920280, 2337813135, 4186055520, 3523024366, 1077514121, 1436444106, 2731983230, 1507202797, 500756149, 198754565, 2382448647, 880454148, 1970517398, 3217485349, 1161840191, 560498076, 1782600856, 2643721918, 1285196205, 788797746, 1195724574, 4061612551, 103427523, 2502688387, 4147162188, 617564657, 978211984, 1781482121, 2205798970, 3939973102, 3826603515, 659557668, 2582884932, 1561884856, 2217488804, 1189296962, 169145316, 2781742156, 1323893433, 824667876, 408202876, 3759637634, 4094868412, 1508996065, 162419237, 3732146944, 3083560189, 3955940127, 2393776934, 2470191468, 3620861513, 481927014, 2756226070, 3154651143, 1261069441, 2063238535, 2222237213, 101459755, 3159774417, 1721190841, 1078395785, 176506553, 3552913423, 1566142515, 1938949000, 1499289517, 3315102456, 829714860, 3843359394, 952932374, 1283577465, 2045007203, 3957761944, 3767891405, 2917089623, 3296133521, 482297421, 1734231412, 3670478932, 2575334979, 2827842737, 3413631016, 1533519803, 4008428470, 3890643173, 272960248, 317508587, 3299937500, 2440520601, 27470488, 1666674386, 1737927609, 750987808, 2385923471, 2694339191, 562925334, 2206035395]
res = decrypt(n, res, k)
# print(res) for i in res: print(chr(i>>24),end="") print(chr((i&0x00ff0000)>>16),end="") print(chr((i&0x0000ff00)>>8),end="") print(chr(i&0x000000ff),end="")
#c= 10610336534759505889607399322387179316771488492347274741918862678692508953185876570981227584004676580623553664818853686933004290078153620168054665086468417541382824708104480882577200529822968531743002301934310349005341104696887943182074473298650903541494918266823037984054778903666406545980557074219162536057146090758158128189406073809226361445046225524917089434897957301396534515964547462425719205819342172669899546965221084098690893672595962129879041507903210851706793788311452973769358455761907303633956322972510500253009083922781934406731633755418753858930476576720874219359466503538931371444470303193503733920039


接下来是一个低解密指数 rsa 就可以得到flag了。
import gmpy2from Crypto.PublicKey import RSAimport ContinuedFractions, Arithmeticfrom Crypto.Util.number import long_to_bytes

def wiener_hack(e, n): # firstly git clone https://github.com/pablocelayes/rsa-wiener-attack.git ! frac = ContinuedFractions.rational_to_contfrac(e, n) convergents = ContinuedFractions.convergents_from_contfrac(frac) for (k, d) in convergents: if k != 0 and (e * d - 1) % k == 0: phi = (e * d - 1) // k s = n - phi + 1 discr = s * s - 4 * n if (discr >= 0): t = Arithmetic.is_perfect_square(discr) if t != -1 and (s + t) % 2 == 0: return d return False

def main(): pub_key = [ 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n,1->e
n = pub_key[0] e = pub_key[1] c = 10610336534759505889607399322387179316771488492347274741918862678692508953185876570981227584004676580623553664818853686933004290078153620168054665086468417541382824708104480882577200529822968531743002301934310349005341104696887943182074473298650903541494918266823037984054778903666406545980557074219162536057146090758158128189406073809226361445046225524917089434897957301396534515964547462425719205819342172669899546965221084098690893672595962129879041507903210851706793788311452973769358455761907303633956322972510500253009083922781934406731633755418753858930476576720874219359466503538931371444470303193503733920039 d = wiener_hack(e, n) m = pow(c, d, n) print(long_to_bytes(m)) # flag{You_Need_Some_Tea}

if __name__ == "__main__": main()


FakePica


先用BlackDex脱个壳,然后把解密后的dex文件pull到自己的电脑上后拖进jadx。
 
看到主要是一个登录逻辑:
[Rev赛题复现]DASCTF Apr X FATE 2022
 
题目逻辑非常简单,可以直接Aes解密,但由于主类里有了解密的方法,肯定是要选择更加有意思的方法来玩玩,我这里的做法是直接采用frida来hook。
 
先hook绕过认证:
console.log("Script loaded successfully");Java.perform(function x(){    console.log("inside java perform function");    //定位类    var my_class = Java.use("com.pica.picapica.MainActivity");    console.log("Java.use Successfully");    my_class.check.implementation = function(x,y){        return true;    }})

返回页面如下:
[Rev赛题复现]DASCTF Apr X FATE 2022
 
看到有两个解密相关的方法,果断继续hook。
console.log("Script loaded successfully");Java.perform(function x(){    console.log("inside java perform function");    //定位类    var my_class = Java.use("com.pica.picapica.MainActivity");    console.log("Java.use Successfully");    my_class.check.implementation = function(x,y){        var email=this.decryptByHexString(this.bytesConvertHexString(this.content0.value),this.key.value);        var password=this.decryptByHexString(this.bytesConvertHexString(this.content1.value),this.key.value);        console.log("flag{"+email+password+"}");         return true;    }})

直接拿到flag。
[Rev赛题复现]DASCTF Apr X FATE 2022
奇怪的交易这题附件过大,这里就直接贴我的链接了:

奇怪的交易
http://49.235.65.44:8888/down/QOUeSq0RV5tS



[Rev赛题复现]DASCTF Apr X FATE 2022


看雪ID:t0hka1

https://bbs.pediy.com/user-home-860779.htm

*本文由看雪论坛 t0hka1 原创,转载请注明来自看雪社区

[Rev赛题复现]DASCTF Apr X FATE 2022



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