writeup of difficult_programming_language
这是我第一次参与CTF的出题，出的不是很好，看到有些师傅说过于脑洞和套路化了，在这里先向各位师傅道个歉orz
这题我给了一个usb流量包，用tshark命令可以将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
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