长安“战疫”赛2021-Misc
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54
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WriteUp

1025 字
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5 分钟

题目复现环境：【云演】–在线信息安全教育平台，让攻防更简单！ (yunyansec.com)

西安加油

wireshark打开压缩包里的secret.pcap，导出HTTP对象保存到一个文件夹里

image-20220108183802244

用Sublime打开文件夹一个个查看文件，发现hint.txt、secret.txt里含有编码后的字符串

用base64解码secret.txt，发现压缩包文件头

image-20220108184017758

导出为文件，改后缀为zip，发现里面有许多图片，是拼图类型

image-20220108184112359

用base32解码hint.txt，发现有提示哪张图片是数字，猜想就是拼图的顺序

image-20220108183833616

依次按照hint提示的顺序拼图，拼出来大致如此

71bb5e1957852441085e9615fce126e

cazy{make_XiAN_great_Again}

朴实无华的取证

解压得到xp_sp3.raw镜像，猜想是内存取证，用volatility进行取证

┌──(kali㉿kali)-[~/Desktop]
└─$ volatility -f xp_sp3.raw imageinfo
Volatility Foundation Volatility Framework 2.6
INFO    : volatility.debug    : Determining profile based on KDBG search...
          Suggested Profile(s) : WinXPSP2x86, WinXPSP3x86 (Instantiated with WinXPSP2x86)
                     AS Layer1 : IA32PagedMemoryPae (Kernel AS)
                     AS Layer2 : FileAddressSpace (/home/kali/Desktop/xp_sp3.raw)
                      PAE type : PAE
                           DTB : 0x764000L
                          KDBG : 0x8054e2e0L
          Number of Processors : 2
     Image Type (Service Pack) : 3
                KPCR for CPU 0 : 0xffdff000L
                KPCR for CPU 1 : 0xf8757000L
             KUSER_SHARED_DATA : 0xffdf0000L
           Image date and time : 2021-12-27 02:37:41 UTC+0000
     Image local date and time : 2021-12-27 10:37:41 +0800

对镜像进行基本信息的取证，随后查看是否有关键词为flag的文件

┌──(kali㉿kali)-[~/Desktop]
└─$ volatility -f xp_sp3.raw --profile=WinXPSP2x86 filescan | grep 'flag'  
Volatility Foundation Volatility Framework 2.6
0x00000000017ad6a8      2      0 R--rw- \Device\HarddiskVolume1\Documents and Settings\Administrator\桌面\flag.zip
0x00000000018efcb8      1      0 RW-rw- \Device\HarddiskVolume1\Documents and Settings\Administrator\Recent\flag.lnk
0x0000000001b34f90      1      1 R--r-- \Device\HarddiskVolume1\Documents and Settings\Administrator\桌面\flag.zip
0x0000000001e65028      1      0 R--rw- \Device\HarddiskVolume1\Documents and Settings\Administrator\桌面\flag.png

发现有后缀名为png和zip的flag文件，将其导出

volatility -f xp_sp3.raw --profile=WinXPSP2x86 dumpfiles -D ./ -Q 0x0000000001e65028 -n
volatility -f xp_sp3.raw --profile=WinXPSP2x86 dumpfiles -D ./ -Q 0x0000000001b34f90 -n

将png图片里面的类似flag的字符提取，为FDCB[8LDQ?ZL00?FHUWDLQOB?VXFFHHG?LQ?ILJKWLQJ?WKH?HSLGHPLF]

压缩包有密码，在对镜像的记事本取证时发现密码

┌──(kali㉿kali)-[~/Desktop]
└─$ volatility -f xp_sp3.raw --profile=WinXPSP2x86 notepad              
Volatility Foundation Volatility Framework 2.6
Process: 2976
Text:
?
​
Text:
?
​
Text:
​
​
Text:
?
​
Text:
??????????????
20211209(encrypt)
???????????????????????
????!?????
????!???

解压后打开文本发现里面有段加密的脚本

//幼儿园水平的加密（部分）
void Encrypt(string& str)
{
    for(int i = 0; i < str.length(); i++)
    {
        if(str[i] >='a'&& str[i]<='w')
            str[i]+=3;
        else if(str[i]=='x')
            str[i]='a';
        else if(str[i]=='y')
            str[i]='b';
        else if(str[i]=='z')
            str[i]='c'; 
        else if(str[i]=='_')
            str[i]='|';
        str[i] -= 32;
    }
}

综合以上对其转换成python解密flag

def Encrypt(str):
    for i in range(len(str)):
        str[i] = chr(ord(str[i]) + 32)
        if ((ord(str[i]) >= ord('d')) and (ord(str[i]) <= ord('z'))):
            str[i] = chr(ord(str[i]) - 3)
        elif (str[i] == 'a'):
            str[i] = 'x'
        elif (str[i] == 'b'):
            str[i] = 'y'
        elif (str[i] == 'c'):
            str[i] = 'z'
        elif (str[i] == '|'):
            str[i] = '_'
​
    print(''.join(str))
​
​
Encrypt(list('FDCB[8LDQ?ZL00?FHUWDLQOB?VXFFHHG?LQ?ILJKWLQJ?WKH?HSLGHPLF]'))
​

cazy{Xian_will_certainly_succeed_in_fighting_the_epidemic}

无字天书

解压压缩包得到流量捕获包，导出http对象到一个文件夹

用sublime text一个个查看导出的文件，在一个文件发现504B开头，明显zip压缩包，将其复制成16进制导出成压缩包

image-20220110211204919

解压后有两个，flag.txt和key.ws，打开全是空白，全选复制有东西但是看不见

以为是零宽字符隐写，但复制到在线网站解不出，在kali里打开并没有200C之类的

用16进制查看，发现全是20 09 0A之类

image-20220110211814446

在比赛时做到这就不会了

复现

ws文件在比赛时怎么想都不知是什么

看wp才知道是whitespace，储备量不够

Whitelips the Whitespace IDE (vii5ard.github.io)

将key.ws内容全选复制上去，run一下得到XiAnWillBeSafe

image-20220110212059339

然后flag.txt是snow，用snow工具

D:\CTF\ctftool\图片隐写\snow>SNOW.EXE -p XiAnWillBeSafe -C flag.txt
cazy{C4n_y0u_underSt4nd_th3_b0oK_With0ut_Str1ng}

binary

解压得到名为234的文件，没有后缀名

用010editor查看，发现文件头为CA FE BA BE，class文件的魔数

image-20220110213609858

用jadx反编译

image-20220110214142689

将源码复制出来，转python

s = [77, 68, 65, 119, 77, 68, 65, 119, 77, 68, 69, 119, 77, 84, 69, 120, 77, 68, 65, 119, 77, 68, 65, 119, 77, 68, 65, 120, 77, 84, 69, 120, 77, 84, 69, 119, 77, 84, 69, 120, 77, 68, 65, 119, 77, 68, 65, 119, 77, 70, 120, 117, 77, 68, 69, 120, 77, 84, 69, 120, 77, 68, 69, 119, 77, 84, 69, 119, 77, 84, 65, 120, 77, 68, 69, 120, 77, 84, 69, 120, 77, 68, 65, 119, 77, 84, 69, 120, 77, 68, 69, 120, 77, 68, 69, 120, 77, 84, 69, 120, 77, 70, 120, 117, 77, 68, 69, 119, 77, 68, 65, 120, 77, 68, 69, 119, 77, 68, 65, 119, 77, 84, 69, 120, 77, 84, 65, 119, 77, 68, 69, 120, 77, 84, 65, 120, 77, 68, 69, 120, 77, 68, 69, 120, 77, 68, 69, 119, 77, 68, 65, 120, 77, 70, 120, 117, 77, 68, 69, 119, 77, 68, 65, 120, 77, 68, 69, 120, 77, 68, 65, 119, 77, 68, 65, 120, 77, 84, 65, 119, 77, 68, 69, 120, 77, 84, 65, 119, 77, 68, 65, 119, 77, 84, 65, 120, 77, 68, 69, 119, 77, 68, 65, 120, 77, 70, 120, 117, 77, 68, 69, 119, 77, 68, 65, 120, 77, 68, 69, 120, 77, 84, 65, 120, 77, 84, 65, 120, 77, 84, 65, 119, 77, 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84, 65, 120, 77, 84, 69, 119, 77, 84, 65, 120, 77, 84, 69, 119, 77, 84, 69, 120, 77, 84, 69, 119, 77, 86, 120, 117, 77, 68, 65, 120, 77, 68, 69, 119, 77, 68, 69, 119, 77, 68, 69, 120, 77, 84, 69, 120, 77, 84, 69, 119, 77, 84, 69, 120, 77, 68, 65, 119, 77, 68, 69, 120, 77, 68, 65, 120, 77, 68, 69, 119, 77, 68, 65, 120, 77, 70, 120, 117, 77, 68, 65, 120, 77, 68, 65, 119, 77, 84, 69, 119, 77, 84, 69, 120, 77, 68, 69, 120, 77, 68, 69, 120, 77, 68, 65, 120, 77, 84, 65, 119, 77, 84, 69, 119, 77, 68, 69, 120, 77, 68, 65, 120, 77, 84, 69, 119, 77, 86, 120, 117, 77, 84, 69, 120, 77, 68, 69, 119, 77, 68, 69, 120, 77, 68, 65, 119, 77, 84, 69, 120, 77, 84, 69, 120, 77, 84, 65, 120, 77, 84, 65, 120, 77, 68, 65, 120, 77, 84, 65, 119, 77, 68, 65, 119, 77, 68, 65, 120, 77, 70, 120, 117, 77, 68, 65, 119, 77, 68, 69, 120, 77, 84, 65, 120, 77, 68, 69, 119, 77, 68, 65, 120, 77, 84, 69, 119, 77, 68, 65, 119, 77, 68, 69, 119, 77, 84, 69, 119, 77, 84, 69, 120, 77, 84, 69, 120, 77, 68, 69, 120, 77, 86, 120, 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for i in s:
    print(chr(i),end='')

得到一串字符串，看到最后有两个=，转base64

得到一串01，是二维码

0000000101110000000011111101110000000\n0111110101101010111110001110110111110\n0100010100001111000111010110110100010\n0100010110000011000111000001010100010\n0100010111011011001101101011110100010\n0111110101110100000001001000010111110\n0000000101010101010101010101010000000\n1111111100100000000100110011111111111\n1100010101010000101111110100000011000\n0101101000110010010000100110101011101\n1011000001001111001100011010000010010\n1110111111110010101101000110101011100\n1010110001110000000110100000000000010\n0110101001000100011011101011101111101\n0010100100111111101110000110010100010\n0010001101110110110011001100110011101\n1110100110001111111011010011000000010\n0000111010100011100000101101111110111\n1101100110101101001100010100110000100\n0101001001111001000001001110010010111\n0101010011000111000110010000010101000\n1001101111101110110010011111101011101\n1101100010111000000101110110001011010\n0011001000111101100011110100100111101\n0101000001110101110110101111110100010\n0101011011001001000000110100010011111\n0110100010001110010110011011111001100\n0111001111100000010110110111001111100\n0100110010110010100010111011000000000\n1111111101011001110011100101011101011\n0000000111000111011010110001010100100\n0111110111001101010110101100011101111\n0100010100110000110011010000000000010\n0100010101111101100011111111110100111\n0100010101101111111100000010101010110\n0111110111111000101101001111000110110\n0000000111111011110110000000100011000

每一个\n之间是37个01字符，明显每一个\n就换行，共36个\n，即37行

所以可以形成正方形，为二维码，用脚本转

s = '0000000101110000000011111101110000000\n0111110101101010111110001110110111110\n0100010100001111000111010110110100010\n0100010110000011000111000001010100010\n0100010111011011001101101011110100010\n0111110101110100000001001000010111110\n0000000101010101010101010101010000000\n1111111100100000000100110011111111111\n1100010101010000101111110100000011000\n0101101000110010010000100110101011101\n1011000001001111001100011010000010010\n1110111111110010101101000110101011100\n1010110001110000000110100000000000010\n0110101001000100011011101011101111101\n0010100100111111101110000110010100010\n0010001101110110110011001100110011101\n1110100110001111111011010011000000010\n0000111010100011100000101101111110111\n1101100110101101001100010100110000100\n0101001001111001000001001110010010111\n0101010011000111000110010000010101000\n1001101111101110110010011111101011101\n1101100010111000000101110110001011010\n0011001000111101100011110100100111101\n0101000001110101110110101111110100010\n0101011011001001000000110100010011111\n0110100010001110010110011011111001100\n0111001111100000010110110111001111100\n0100110010110010100010111011000000000\n1111111101011001110011100101011101011\n0000000111000111011010110001010100100\n0111110111001101010110101100011101111\n0100010100110000110011010000000000010\n0100010101111101100011111111110100111\n0100010101101111111100000010101010110\n0111110111111000101101001111000110110\n0000000111111011110110000000100011000'
s = s.split('\n')
​
from PIL import Image
pic = Image.new('RGB',(37,37),(255,255,255))
for i in range(37):
    for j in range(37):
        if(s[i][j] == '0'):
            pic.putpixel((j,i),(0,0,0))
pic.show()
pic.save('fllllag.png')

扫描二维码得到flag

ez_Encrypt

对流量包分析追踪流时，发现第10个流有写入一串字符到web123

image-20220111225229433

复制这串字符base64解码发现zip文件头，导出为压缩包，发现里面有ThinkPHP源码

用D盾扫描发现其有base64_decode加密的PHP文件，在app/controller/Index.php

image-20220111225731346

查看发现一大串base64加密字符以及有混淆字符，用在线网站解

php解密加密|php混淆破解|phpjm破解|phpdp神盾破解|php威盾破解|php微盾破解|tianyiw破解|php源码破解|php在线破解|php反编译|zend6解密|Zend Guard 6 破解 (zhaoyuanma.com)

image-20220111230302264

解密完下载的php有flag

image-20220111230401618

Ez_Steg

压缩包注释提示六个数字，用工具跑一下就出来，220101

image-20220111214653894

pyc文件提示steg，是Pyc字节码隐写，用工具

AngelKitty/stegosaurus: A steganography tool for embedding payloads within Python bytecode. (github.com)

这个工具不知为什么python3.9跑不出来，用服务器的python3.6可以跑的出来

[root@VM-4-17-centos stegosaurus]# python3 -m stegosaurus steg.pyc -x
Extracted payload: TheKey:St3g1sV3ryFuNny

另一个emoji.txt全是emoji，那么就只能是emoji-aes了

emoji-aes (aghorler.github.io)

flag：cazy{Em0j1s_AES_4nd_PyC_St3g_D0_yoU_l1ke}

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|´・ω・)ノ
ヾ(≧∇≦*)ゝ
(☆ω☆)
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(/ω＼)
∠( ᐛ 」∠)＿
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(ノ°ο°)ノ
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ヾ(´･ ･｀｡)ノ"
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(ó﹏ò｡)
Σ(っ °Д °;)っ
( ,,´･ω･)ﾉ"(´っω･｀｡)
╮(╯▽╰)╭
o(*////▽////*)q
＞﹏＜
( ๑´•ω•) "(ㆆᴗㆆ)
😂
😀
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😊
🙂
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Source: github.com/k4yt3x/flowerhd
	

	

		
颜文字
Emoji
小恐龙
花!
	


									

			

			
			
		

	






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