2018 CISCN reverse

这题比赛的时候没做出来,主要是心态崩了看不下去。。赛后看了下网上的wp发现不难,是自己想复杂了。这里将我的思路和exp放出来,希望大家一起交流学习。

main函数

它首先是check了输入的前六个字符是否与“CISCN{”匹配,接着使用strtok函数将字符串以“_”分割为三部分,然后分别对这三部分check。

sub_4012DE函数

关键部分如下

将第一部分的字符串经过以上变换后与一串MD5值5BH8170528842F510K70EGH31F44M24B比较。

那么我们可以直接逆出原本的md5,这个函数的脚本如下。

def change1(str0):
#str0即要逆的md5
str00 = ''
for i in range(len(str0)):
temp = ord(str0[i])-i%10
if temp <= ord('A') + 5 and temp >= ord('A'):
str00 += chr(temp)
else:
str00 += str0[i] return str00

得到5AF8170528842C510D70EFF31A44E24A ,在线解密得到tima

sub_401411函数

这个函数相较上个只是多了个亦或的过程,同样可逆,脚本如下。

def change2(str0):
#str0是已经经过change1处理的md5
byte_603860 = [0x92,0x84,0x3d,0xa7,0x14,0xf2,0xfb,0x4b,0xee,0x8a,0xc2,0xc3,0x76,0x68,0x13,0x1e]
str2 = '['
for i in range(32):
if i%2 == 0:
str2 += '0x' + str0[i]
elif i != len(str1) - 1 :
str2 += str0[i] + ','
else:
str2 += str0[i] + ']'
#print str2
str2 = eval(str2)
str2_2 = ''
for i in range(len(str2)):
str2_2 += str( hex(str2[i] ^ byte_603860[i])[2:] )
return str2_2

得到c87c2aa23c76d71ae3fa2d306c2cf154 ,在线解密得到yefb

sub_401562函数

这个函数除了有sub_401411的全部加密过程,还会生成一个flag文件,但由于其中未知数太多,所以不采用逆向全部过程,生成flag文件的代码如下。

可以看到,我们只需爆破出v15,v16的值即可得到正确的flag文件,爆破脚本如下,这里我只取了前500个byte,能识别出文件格式即可,其实更少也行。

(这里使用了python的库filetype,pip安装即可)

import filetype

data = [0xc7,0xb7,0xc7,0x8f,0x38,0x7f,0x72,0x29,0x71,0x29,0x38,0x6e,0x39,0x6e,0x39,0x43,0x39,0x43,0x38,0x6f,0xc7,0xb4,0x38,0x2c,0x38,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x90,0xe3,0x6f,0x7b,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0xc7,0xaf,0x38,0x7e,0x30,0x6f,0x2e,0x6f,0xb8,0x6c,0x39,0x4e,0x38,0x6d,0x29,0x6e,0x3b,0x7e,0x39,0x90,0xfc,0x6f,0x27,0x6f,0x38,0x6e,0x3d,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6f,0x38,0x6f,0x38,0x6f,0x38,0x6f,0x38,0x6e,0x3a,0x6c,0x3c,0x6a,0x3e,0x68,0x30,0x66,0x32,0x64,0xc7,0xab,0x38,0xda,0x28,0x6f,0x3a,0x6e,0x3b,0x6c,0x3a,0x6b,0x3b,0x6a,0x3d,0x6b,0x3c,0x6f,0x38,0x6e,0x45,0x6e,0x3a,0x6c,0x38,0x6b,0x29,0x6a,0x2a,0x4e,0x9,0x2e,0x3e,0x7c,0x69,0xe,0x3f,0x4d,0x49,0x7b,0xa,0xee,0xa9,0xce,0x30,0x4c,0x7a,0xde,0xf9,0x7a,0x6a,0xbe,0xc8,0x4b,0xb,0xd,0x4a,0xed,0x31,0x65,0x2e,0x78,0x20,0x76,0x22,0x4a,0x1e,0x48,0x10,0x46,0x12,0x5b,0xd,0x59,0xf,0x57,0x1,0x55,0x7b,0x2b,0x7d,0x29,0x7f,0x27,0x71,0x25,0x6b,0x3b,0x6d,0x39,0x6f,0x37,0x61,0x35,0x5b,0xb,0x5d,0x9,0x5f,0x7,0x51,0x5,0x4b,0x1b,0x4d,0x19,0x4f,0x17,0x41,0x15,0xbb,0xeb,0xbd,0xe9,0xbf,0xe7,0xb1,0xe5,0xaa,0xfc,0xac,0xfa,0xae,0xf8,0xa0,0xf6,0xa2,0xcd,0x9b,0xcb,0x9d,0xc9,0x9f,0xc7,0x91,0xc5,0x8a,0xdc,0x8c,0xda,0x8e,0xd8,0x80,0xd6,0x82,0xad,0xfb,0xab,0xfd,0xa9,0xff,0xa7,0xf1,0xa5,0xea,0xbc,0xec,0xba,0xee,0xb8,0xe0,0xb6,0xe2,0x8e,0xda,0x8c,0xdc,0x8a,0xde,0x88,0xd0,0x86,0xd2,0x9e,0xca,0x9c,0xcc,0x9a,0xce,0x98,0xc0,0x96,0xc2,0x90,0xfc,0x6f,0x27,0x6e,0x38,0x6c,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6e,0x39,0x6f,0x38,0x6f,0x38,0x6f,0x38,0x6e,0x3a,0x6c,0x3c,0x6a,0x3e,0x68,0x30,0x66,0x32,0x64,0xc7,0xab,0x38,0xda,0x29,0x6f,0x3a,0x6e,0x3a,0x6b,0x3c,0x6c,0x3c,0x68,0x3d,0x6b,0x3c,0x6f,0x39,0x6d,0x4f,0x6f,0x39,0x6d,0x3b,0x7e,0x3c,0x6a,0x19,0x5e,0x3e,0x7d,0x79,0x3e,0x3f,0xe,0x49,0x7c,0x1a,0x5d,0xb9,0x67,0x2c,0x2d,0xa9,0xce,0x89,0xae,0x31,0x4c,0xb,0x3d,0xc8,0x7a,0x5a,0x1d,0xe9,0x65,0x2e,0x4b,0xc,0x8e,0x1d,0x9e,0x2f,0x77,0x21,0x75,0x1e,0x48,0x10,0x46,0x12,0x5a]

for i in range(256):
for j in range(256): result = ''
for k in range(len(data)):
if k&1 :
result += chr( data[k]^i )
else:
result += chr( data[k]^j )
a = open('re_guess','w')
a.write(result)
a.close()
kind = filetype.guess('re_guess')
if kind is None:
continue
else:
print i,j,kind.extension

结果如下

23 216 Z
42 216 Z
76 56 mp3
111 56 jpg
150 226 ps
237 138 exe
250 133 bmp

jpg很可疑,于是生成完整文件看看。

x = open('data.txt','r').read().replace('\n','')
data = eval('[' + x + ']') i = 111
j = 56 a = open('flag.jpg','w')
temp = ''
for k in range(len(data)):
if k&1:
temp += chr( data[k] ^ i )
else:
temp += chr( data[k] ^ j )
a.write(temp)
a.close()

idc提取data.txt的脚本如下(shift+F2打开Execute script)

auto addr1 = 0x006020E0;
auto i,x; for(i=0; i < 6016 ; i ++ )
{
Message("0x%x,",Byte(i+addr1));
}

得到第三部分的flag

验证

将以上得到的三部分以下划线拼接得到

CISCN{tima_yefb_MayDetyU$hhtIm2}

运行结果如下图

作者: LB919

出处:http://www.cnblogs.com/L1B0/

如有转载,荣幸之至!请随手标明出处;

2018 CISCN reverse wp的更多相关文章

  1. 2019 湖湘杯 Reverse WP

    0x01 arguement 下载链接:https://www.lanzous.com/i7atyhc 1.准备 获取到信息: 32位的文件 upx加密文件 在控制台打开文件 使用"upx ...

  2. Google Capture The Flag 2018 (Quals) - Reverse - Beginner's Quest - Gatekeeper

    参考链接:https://ctftime.org/task/6264 题目 It's a media PC! All fully purchased through the online subscr ...

  3. 【VNCTF2022】Reverse wp

    babymaze 反编译源码 pyc文件,uncompy6撸不出来,看字节码 import marshal, dis fp = open(r"BabyMaze.pyc", 'rb' ...

  4. 航遇项目react踩坑

    1.iconfont应用: a.正常用法如下 <span className='iconfont' > iconfont的代码,例如: </span> b.react不能动态 ...

  5. Django的View(视图)和路由系统

    一.Django的View(视图) 1.介绍 一个视图函数(类),简称视图,是一个简单的Python 函数(类),它接受Web请求并且返回Web响应. 响应可以是一张网页的HTML内容,一个重定向,一 ...

  6. Django urls 路由

    写url和视图的的对应关系 from django.conf.urls import url from django.contrib import admin from app名 import vie ...

  7. URL的命名和反向解析

    1. 分组 url(r'^del_publisher/(\d+)', views.del_publisher), 匹配到参数,按照位置参数的方式传递给视图函数 视图函数需要定义形参接收变量 2. 命名 ...

  8. CTF各种资源:题目、工具、资料

    目录 题目汇总 Reverse 签到题 Web Web中等难度 Crypto 基础网站 各类工具 综合 Web Payloads 逆向 Pwn 取证 题目汇总 这里收集了我做过的CTF题目 Rever ...

  9. python中列表常用的几个操作函数

    # coding=utf-8#在列表末尾添加新的对像#实例展现函数append()的用法aList=[456,'abc','zara','ijk',2018]aList.append(123)prin ...

随机推荐

  1. Oracle生成查询包含指定字段名对应的所有数据表记录语句

    应用场合:已知字段名字,查询数据库中所有数据表中包含该字段名的所有数据表 操作办法:指定字段名,数据库表用户,执行下面查询语句即可 --Oracle生成查询包含指定字段名对应的所有数据表记录语句 de ...

  2. 利用HTP工具包开发报表

    利用这种方式的优点是不需要跑请求就可以打印报表 工具包中常用程序说明 htp.print 语法   htp.print (cbuf | dbuf | nbuf); 作用   generates a l ...

  3. C++对象模型(二):The Semantics of Copy Constructors(拷贝构造函数之编译背后的行为)

    本文是 Inside The C++ Object Model's Chapter 2  的部分读书笔记. 有三种情况,需要拷贝构造函数: 1)object直接为另外一个object的初始值 2)ob ...

  4. javascript语法之String对象

    学习String类就是学习它的一些方法,主要用到方法全部罗列出来.如下: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transition ...

  5. 【Unity Shaders】Diffuse Shading——在Surface Shader中使用properties

    本系列主要参考<Unity Shaders and Effects Cookbook>一书(感谢原书作者),同时会加上一点个人理解或拓展. 这里是本书所有的插图.这里是本书所需的代码和资源 ...

  6. 【Android 应用开发】Android屏幕适配解析 - 详解像素,设备独立像素,归一化密度,精确密度及各种资源对应的尺寸密度分辨率适配问题

    . 作者 :万境绝尘 转载请注明出处 : http://blog.csdn.net/shulianghan/article/details/19698511 . 最近遇到了一系列的屏幕适配问题, 以及 ...

  7. ViewPager适配器学习记录( pageAdapter和FragmentPagerAdapter/FragmentStatePagerAdapter))

    1.概述 ViewPager,顾名思义实现控件的滚动功能,是Support-v4的包中类,使用前要先导包.使用的时候跟listView有点相似,需要设置对应的适配器,通常有俩大类 [pageAdapt ...

  8. 漫谈程序员(十八)windows中的命令subst

    漫谈程序员(十八)windows中的命令subst 用法格式 一.subst [盘符] [路径]  将指定的路径替代盘符,该路径将作为驱动器使用 二.subst /d 解除替代 三.不加任何参数键入  ...

  9. 【一天一道LeetCode】#39. Combination Sum

    一天一道LeetCode系列 (一)题目 Given a set of candidate numbers (C) and a target number (T), find all unique c ...

  10. 粒子系统属性Life,发射速率和总数的关系

    提示,粒子系统的life,发射速率以及总粒子数是相互影响的. 如果你要发射器射出粒子流然后停顿一会,你将简单的必须确保lifetime和发射速率相匹配以至于在发射出的粒子达到粒子总数之前一些粒子是活跃 ...