CTFHub--Misc部分wp

一叶障目  

just a rar

鸡你太美

你猜我是个啥

藏藏藏

认真你就输了 

签个到 

difficult programming language

 

一叶障目  

放到winhex中调一下高度

 

 

答案： xaflag{66666}

 

just a rar

下载下来题目，解压压缩包，4位数密码爆破

密码：2016

 

 

 flag{Wadf_123}

 

 鸡你太美

第2个打篮球的gif,副本打不开，通过winhex打开后，发现缺少gif开头

添加4个0字节符，输入gif开头格式：47 49 46 38

保存打开

 

 

 flag如图

 

你猜我是个啥

正常打开出现错误，尝试用winhex打开

 

 

 

 

发现文件底部就有flag

 

 

 提交即可

 

藏藏藏

 

解压改后缀就ok了,这一类题多是这种套路

打开word,扫码即可

 

 

认真你就输了

 

 

 

 

 解压发现是个.xls文件，打开后，提示flag就在该表格下，找了一小会儿，也没发现什么

我尝试下了用txt打开，看到里面好多独立的文件，改后缀zip，解压

 

 

 直接搜索flag,秒出答案

flag{M9eVfi2Pcs#}

 

签个到

 

 

 一直解压完事了，到这里，改后缀名为.png    二维码扫描即可

xaflag{i_am_a_tupian}

 

difficult programming language

题给了一个usb流量包，用tshark命令，将Leftover Capture Date域中的usb协议数据提取出来

tshark -r difficult_programming_language.pcap -T fields -e usb.capdata > usbdata.txt

　

 

 

 说实话，这个题有点难，复现失败了，不知道是怎么通过usb协议文档可以查找到值和键位的关系

下面贴上MiGo出题人师傅的wp    https://www.dazhuanlan.com/2020/01/30/5e326194b93d1

借着把后面的wp补充完整一下

    提取出数据后我们通过usb协议文档可以查找到值和键位的关系，编写脚本后得到结果   

D’`;M?!mZ4j8hgSvt2bN);^]+7jiE3Ve0A@Q=|;)sxwYXtsl2pongOe+LKa’e^]a`_X|V[Tx;:VONSRQJn1MFKJCBfFE>&<`@9!=<5Y9y7654-,P0/o-,%I)ih&%$#z@xw|{ts9wvXWm3~

 

联系题目 difficult programming language ,可以搜索到这是一段malbolge语言的代码，找一个在线编译网站跑一下即可得到flag.

hctf{m4lb0lGe}

附上MiGo师傅的脚本

#!/usr/bin/env python

import sys
import os

normalkeys = { 0x04:"a",  0x05:"b",  0x06:"c", 0x07:"d", 0x08:"e", 0x09:"f", 0x0A:"g",  0x0B:"h", 0x0C:"i",  0x0D:"j", 0x0E:"k", 0x0F:"l", 0x10:"m", 0x11:"n",0x12:"o",  0x13:"p", 0x14:"q", 0x15:"r", 0x16:"s", 0x17:"t", 0x18:"u",0x19:"v", 0x1A:"w", 0x1B:"x", 0x1C:"y", 0x1D:"z",0x1E:"1", 0x1F:"2", 0x20:"3", 0x21:"4", 0x22:"5",  0x23:"6", 0x24:"7", 0x25:"8", 0x26:"9", 0x27:"0", 0x28:"n", 0x2a:"[DEL]",  0X2B:"    ", 0x2C:" ",  0x2D:"-", 0x2E:"=", 0x2F:"[",  0x30:"]",  0x31:"\", 0x32:"`", 0x33:";",  0x34:"'",0x35:"`", 0x36:",",  0x37:"." , 0x38:"/"}

shiftkeys = { 0x04:"A",  0x05:"B",  0x06:"C", 0x07:"D", 0x08:"E", 0x09:"F", 0x0A:"G",  0x0B:"H", 0x0C:"I",  0x0D:"J", 0x0E:"K", 0x0F:"L", 0x10:"M", 0x11:"N",0x12:"O",  0x13:"P", 0x14:"Q", 0x15:"R", 0x16:"S", 0x17:"T", 0x18:"U",0x19:"V", 0x1A:"W", 0x1B:"X", 0x1C:"Y", 0x1D:"Z", 0x1E:"!", 0x1F:"@", 0x20:"#", 0x21:"$", 0x22:"%",  0x23:"^", 0x24:"&", 0x25:"*", 0x26:"(", 0x27:")", 0x28:"n", 0x2a:"[DEL]",  0X2B:"    ", 0x2C:" ",  0x2D:"_", 0x2E:"+", 0x2F:"{",  0x30:"}",  0x31:"|", 0x32:"~", 0x33:":",  0x34:""",0x35:"~", 0x36:"<",  0x37:">", 0x38:"?" }

nums = []
shift_press = []
keys = open('usbdata.txt')
for line in keys:
    shift_press.append(line[1])
    nums.append(int(line[6:8],16))
keys.close()

output = ""
m = 0
for n in nums:
    if n == 0 :
        m += 1
        continue
    if n in shiftkeys:
        if shift_press[m] == '2' : #shift is pressed
            output += shiftkeys[n]
            m += 1 
        elif shift_press[m] == '0' :
            output += normalkeys[n]
            m += 1
print 'output :n' + output

 

菜鸟的路，还很长，干就完了，