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')
fp.seek(16)
co = marshal.load(fp)
dis.dis(co)

 

使用脚本跑出字节码

发现花指令

去除花指令并修改字节码长度,使用uncompy6跑出源码

_map = [
[
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 5, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1], [1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1], [1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1], [1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1], [1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1], [1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1], [1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1], [1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1], [1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1], [1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1], [1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1], [1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1], [1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1], [1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1], [1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1], [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1], [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1], [1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1], [1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1], [1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1], [1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1], [1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1], [1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1], [1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1], [1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1], [1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1], [1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1], [1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1], [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 7, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]] def maze():
x = 1
y = 1
step = input()
for i in range(len(step)):
if step[i] == 'w':
x -= 1
else:
if step[i] == 's':
x += 1
else:
if step[i] == 'a':
y -= 1
else:
if step[i] == 'd':
y += 1
else:
return False
if _map[x][y] == 1:
return False
if x == 29 and y == 29:
return True def main():
print('Welcome To VNCTF2022!!!')
print('Hello Mr. X, this time your mission is to get out of this maze this time.(FIND THAT 7!)')
print('you are still doing the mission alone, this tape will self-destruct in five seconds.')
if maze():
print('Congratulation! flag: VNCTF{md5(your input)}')
else:
print("Sorry, we won't acknowledge the existence of your squad.") if __name__ == '__main__':
main()

DFS算法

地图非常大,使用DFS算法

map1 = [[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1], [1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1], [1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1,
1, 1, 1, 1, 1, 1, 0, 1], [1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0,
0, 1, 0, 1, 0, 0, 0, 1, 0, 1], [1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1,
1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1], [1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1], [1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1], [1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0,
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1], [1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1,
0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1], [1, 0, 0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1], [1, 1, 1, 1, 1, 1, 1,
0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1], [1, 0, 0, 0, 0,
0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1], [1, 0, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1], [1,
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0,
1], [1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1,
1, 1, 1], [1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 1], [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1,
1, 0, 1, 1, 1, 0, 1], [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0,
0, 0, 1, 0, 1, 0, 0, 0, 1], [1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0,
1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1], [1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0,
1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1], [1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1,
1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1], [1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0,
0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1], [1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1], [1, 0, 1, 0, 0, 0, 1, 0, 1, 0,
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1], [1, 0, 1, 1, 1, 0, 1, 0,
1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1], [1, 0, 0, 0, 1, 0,
1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1], [1, 1, 1, 1,
1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1], [1, 0,
0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1],
[1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1,
0, 1], [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0,
0, 0, 0, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1]]
flag = ""
map2 = [[0 for i in range(len(map1))] for i in range(len(map1)) ]
def DFS(x,y):
global flag
if x == len(map1) - 2 and y == len(map1) - 2: #判断边界
print(flag)
if map1[x+1][y] == 0 and map2[x+1][y] == 0:
map2[x][y] = 1
flag += 's'
DFS(x+1,y)
flag = flag[:-1]
map2[x][y] = 0
if map1[x-1][y] == 0 and map2[x-1][y] == 0:
map2[x][y] = 1
flag += 'w'
DFS(x-1,y)
flag = flag[:-1]
map2[x][y] = 0
if map1[x][y+1] == 0 and map2[x][y+1] == 0:
map2[x][y] = 1
flag += 'd'
DFS(x,y+1)
flag = flag[:-1]
map2[x][y] = 0
if map1[x][y-1] == 0 and map2[x][y-1] == 0:
map2[x][y] = 1
flag += 'a'
DFS(x,y-1)
flag = flag[:-1]
map2[x][y] = 0
y=1
x=1
DFS(x,y)

得出

ssssddssaassddddwwwwddwwddddddwwddddddssddwwddddddddssssaawwaassaassaassddssaassaawwwwwwaaaaaaaassaassddddwwddssddssssaassddssssaaaaaawwddwwaawwwwaassssssssssssddddssddssddddddddwwaaaaaawwwwddssddwwwwwwwwddssddssssssssddddss、

getflag

VNCTF{801f190737434100e7d2790bd5b0732e}

cm1

源码分析

APK文件,直接拖入GDA中查看

拖入GDA中查看main函数,先看导入库

发现Main$1、IHeihei及FileUtils等

安装之后查看,基本就是一个check界面

这里提示打开了asset目录下某个dex文件

这里使用FileUtils

发现调用了里面的Haha

跟入查看发现有异或,1024为一组

脚本处理

将cm1解压后进入到asset目录找到ooo

key = 'vn2022'
f = open('ooo','rb')
f = f.read()
with open('out','wb') as h:
for i in range(len(f)):
h.write((f[i]^ord(key[i % 1024 % len(key)])).to_bytes(1,byteorder='little',
signed=False)) print('ok')

写python脚本模拟

发现是个dex文件

使用dex2jar转换为jar文件进行分析

找到haha是一个xxtea加密,写xxtea脚本即可getflag

【VNCTF2022】Reverse wp的更多相关文章

  1. 【leetcode80】Reverse Vowels of a String(元音字母倒叙)

    题目描述: 写一个函数,实现输入一个字符串,然后把其中的元音字母倒叙 注意 元音字母包含大小写,元音字母有五个a,e,i,o,u 原文描述: Write a function that takes a ...

  2. 【leetcode】Reverse Nodes in k-Group

    Reverse Nodes in k-Group Given a linked list, reverse the nodes of a linked list k at a time and ret ...

  3. 【leetcode】Reverse Linked List II

    Reverse Linked List II Reverse a linked list from position m to n. Do it in-place and in one-pass. F ...

  4. 【leetcode】Reverse Nodes in k-Group (hard)☆

    Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If ...

  5. 【leetcode】Reverse Linked List II (middle)

    Reverse a linked list from position m to n. Do it in-place and in one-pass. For example:Given 1-> ...

  6. 【leetcode】Reverse Linked List(easy)

    Reverse a singly linked list. 思路:没啥好说的.秒... ListNode* reverseList(ListNode* head) { ListNode * rList ...

  7. 【leetcode】Reverse Bits(middle)

    Reverse bits of a given 32 bits unsigned integer. For example, given input 43261596 (represented in ...

  8. 【leetcode】Reverse Words in a String(hard)☆

    Given an input string, reverse the string word by word. For example,Given s = "the sky is blue& ...

  9. 【leetcode】Reverse Integer(middle)☆

    Reverse digits of an integer. Example1: x = 123, return 321Example2: x = -123, return -321 总结:处理整数溢出 ...

随机推荐

  1. suse 12 升级 OpenSSH-7.2p2 到 OpenSSH-8.4p1

    文章目录 1.查看当前当前环境信息 1.1.查看openssh当前版本 1.2.查看当前linux发行版 2.部署telnet-server 2.1.下载telnet-server 2.2.配置tel ...

  2. JUC并发工具类之 CountDownLatch等待多线程完成

    上篇JUC同步工具之Semaphore - 池塘里洗澡的鸭子 - 博客园 (cnblogs.com)示例中,资源释放一个线程就可以退出然后另一个线程可以使用了,那如果需要所有规定数量的资源同时释放了才 ...

  3. Spring Boot部署之jar包运行

    上篇阐述了Spring Boot war部署项目,本篇阐述另一种运行方式:jar包运行. 一.打jar包 1.修改pom.xml配置 2.执行package(对于module执行package之前需要 ...

  4. 聊聊MySQL的加锁规则《死磕MySQL系列 十五》

    大家好,我是咔咔 不期速成,日拱一卒 本期来聊聊MySQL的加锁规则,知道这些规则后可以判断SQL语句的加锁范围,同时也可以写出更好的SQL语句,防止幻读问题的产生,在能力范围内最大程度的提升MySQ ...

  5. service与systemctl命令比较

    本文将比较 linux 的 service 和 systemctl 命令,先分别简单介绍这两个命令的基础用法,然后进行比较. 从 CentOS 7.x 开始,CentOS 开始使用 systemd 服 ...

  6. 扫盲贴:2021 CSS 最冷门特性都是些啥?

    最近几年 CSS 界的大事之一是每年年底的 <State Of CSS>,也就是 CSS 现状调查,去年年底发布了<State Of CSS 2021>.其中关于特性这一章,会 ...

  7. (反射+内省机制的运用)简单模拟spring IoC容器的操作

    简单模拟spring IoC容器的操作[管理对象的创建.管理对象的依赖关系,例如属性设置] 实体类Hello package com.shan.hello; public class Hello { ...

  8. pytest(5)-断言

    前言 断言是完整的测试用例中不可或缺的因素,用例只有加入断言,将实际结果与预期结果进行比对,才能判断它的通过与否. unittest 框架提供了其特有的断言方式,如:assertEqual.asser ...

  9. AQS源码二探-JUC系列

    本文已在公众号上发布,感谢关注,期待和你交流. AQS源码二探-JUC系列 共享模式 doAcquireShared 这个方法是共享模式下获取资源失败,执行入队和等待操作,等待的线程在被唤醒后也在这个 ...

  10. [Python]从哪里开始学习写代码(未完待续)

    预警:这只是我在学习中的一点感受,可能并不完全准确,也不包括面向对象编程的思想(我还不太懂),也有水文的嫌疑,大佬请温和批评指正或者绕道. 计算机语言 语言,是用来交流的.计算机是不能直接听懂人的语言 ...