CATCTF-RCatFly Writeup Analysis 查看文件类型 ~/CTF_Challenges/Reverse/CatFly$ file CatFly CatFly: ELF 64-bit LSB pie executable, x86-64, version 1 (SYSV), dynamicall y linked, interpreter /lib64/ld-linux-x86-64.so.2, BuildID[sh 2025-08-26 Reverse #CTF #Reverse
XCTF-easyRE1 Writeup Analysis 拿到附件用IDA打开F5出伪代码 int __cdecl main(int argc, const char **argv, const char **envp) { char s[256]; // [esp+1Ch] [ebp-10Ch] BYREF <!--more--> unsigned int v5; // [esp+11Ch] [ebp 2025-08-26 Reverse #CTF #Reverse
Reversing-x64Elf-100 Writeup Analysis 拿到附件用file确定一下文件类型得知是64位ELF文件,用IDA分析: mian函数伪代码如下: __int64 __fastcall main(int a1, char **a2, char **a3) { char s[264]; <!--more--> // [rsp+0h] [rbp-110h] BYREF unsigned __int 2025-08-26 #CTF #Reverse
CatCTF-Cat_Jump-Miao Writeup Cat_Jump 直接用vscode打开搜索catctf{ Miao~ 附件内容是一张小猫的图片 分 析图片中隐含的内容,得到一个wav文件 在Audacity中打开,查看频谱图,得到密码CatCTF 2025-08-12 misc #CTF #misc
XSCTF联合招新赛-真是阳间题 Writeup 附件内容: 艾尔登法环正式发售在即,迫不及待的mjj身上好像有蚂蚁在爬,写下了一句谜语: 320735797564158713612246651442515296165461341072833714227175027312499510574705399164081706635234365 7398947248938255086358418100814441196784643527787764297 2025-08-12 misc #CTF #misc
Tomcat RCE漏洞CVE-2025-24813复现 前置知识 JAVA序列化与反序列化简述 JAVA序列化是JAVA语言内置的一种对象持久化的机制,它可以将JAVA对象转化为字节流,以便于在网络中进行传输,或者保存到本地进行持久化(类似于PHP序列化)。反序列化则是序列化的逆过程,将字节流转化回JAVA对象。 序列化过程中,JAVA会将对 象的类型、属性以及属性的值等信息一起保存下来,反序列化时,JAVA会根据这些信息重新构建出原来的对象。这就 2025-08-05 vuln analysis #vuln analysis
SWPUCTF2021新生赛Web Writeups [SWPUCTF 2021 新生赛]gift_F12 Ctrl+U查看前端源码再Ctrl+F搜搜Flag即可 [SWPUCTF 2021 新生赛]jicao <?php highlight_file('index.php'); include("flag.php") <!--more--> ; $id=$_POST['id 2025-07-21 Web #CTF #Web
XSCTF联合招新赛-babyFibo Writeup 分析 附件内容: import os import libnum from secret import flag def fibo(n): assert n >= 0 if n < 2: return n return fib <!--more--> o(n-1) + fibo(n-2) s = fibo(1000) m = l 2025-07-21 Crypto #Crypto #CTF
XSCTF联合招新赛-初识RSA Writeup 分析 附件内容: from Crypto.Util.number import bytes_to_long,inverse,getPrime from flag import flag m = bytes_to_long(flag) p = getPrime(1024) q = getPrime(1024) n = p*q print(n) e = 65537 c = pow(m,e,n) 2025-07-21 Crypto #Crypto #CTF
XSCTF联合招新赛-二元一次方程组 Writeup 分析: import libnum from Crypto.Util import number from secret import flag size = 256 e = 65537 p = number.getPrime(size) q = number.getPrime(size) avg = (p+q)/2 n = p*q m = libnum.s2n(flag) c = pow 2025-07-21 Crypto #Crypto #CTF
XSCTF联合招新赛-baigeiRSA Writeup 拿到附件后查看内容 import libnum from Crypto.Util import number from secret import flag size = 128 e = 65537 p = number.getPrime(size) q = number. <!--more--> getPrime(size) n = p*q m = libnum.s2n(fla 2025-07-21 Crypto #Crypto #CTF