2021年津门杯国际网络安全创新大赛 - Misc

• m0usb
• m1bmp
• tunnel

m0usb

• 查看流量包，发现是 USB 的流量，USB 协议的数据部分在 Leftover Capture Data 域中，先用 tshark 将流量提取出来

tshark -r usb.pcapng -T fields -e usb.capdata > usbdata.txt
如果提取出来的数据有空行可以将命令改为如下形式
tshark -r usb.pcapng -T fields -e usb.capdata | sed '/^\s*$/d' > usbdata.txt

• 击键信息对照表可以参考：https://max.book118.com/html/2017/0407/99227972.shtm

• 提取出来的数据可能会带冒号，也可能不带（有可能和wireshark的版本相关），但是一般的脚本都会按照有冒号的数据来识别，这里参考一个添加冒号的脚本

f=open('usbdata.txt','r')
fi=open('out.txt','w')
while 1:
    a=f.readline().strip()
    if a:
        if len(a)==16: # 鼠标流量的话len改为8
            out=''
            for i in range(0,len(a),2):
                if i+2 != len(a):
                    out+=a[i]+a[i+1]+":"
                else:
                    out+=a[i]+a[i+1]
            fi.write(out)
            fi.write('\n')
    else:
        break

fi.close()

• 鼠标流量转换数据

normalKeys = {
    "04":"a", "05":"b", "06":"c", "07":"d", "08":"e",
    "09":"f", "0a":"g", "0b":"h", "0c":"i", "0d":"j",
     "0e":"k", "0f":"l", "10":"m", "11":"n", "12":"o",
      "13":"p", "14":"q", "15":"r", "16":"s", "17":"t",
       "18":"u", "19":"v", "1a":"w", "1b":"x", "1c":"y",
        "1d":"z","1e":"1", "1f":"2", "20":"3", "21":"4",
         "22":"5", "23":"6","24":"7","25":"8","26":"9",
         "27":"0","28":"<RET>","29":"<ESC>","2a":"<DEL>", "2b":"\t",
         "2c":"<SPACE>","2d":"-","2e":"=","2f":"[","30":"]","31":"\\",
         "32":"<NON>","33":";","34":"'","35":"<GA>","36":",","37":".",
         "38":"/","39":"<CAP>","3a":"<F1>","3b":"<F2>", "3c":"<F3>","3d":"<F4>",
         "3e":"<F5>","3f":"<F6>","40":"<F7>","41":"<F8>","42":"<F9>","43":"<F10>",
         "44":"<F11>","45":"<F12>"}
shiftKeys = {
    "04":"A", "05":"B", "06":"C", "07":"D", "08":"E",
     "09":"F", "0a":"G", "0b":"H", "0c":"I", "0d":"J",
      "0e":"K", "0f":"L", "10":"M", "11":"N", "12":"O",
       "13":"P", "14":"Q", "15":"R", "16":"S", "17":"T",
        "18":"U", "19":"V", "1a":"W", "1b":"X", "1c":"Y",
         "1d":"Z","1e":"!", "1f":"@", "20":"#", "21":"$",
          "22":"%", "23":"^","24":"&","25":"*","26":"(","27":")",
          "28":"<RET>","29":"<ESC>","2a":"<DEL>", "2b":"\t","2c":"<SPACE>",
          "2d":"_","2e":"+","2f":"{","30":"}","31":"|","32":"<NON>","33":"\"",
          "34":":","35":"<GA>","36":"<","37":">","38":"?","39":"<CAP>","3a":"<F1>",
          "3b":"<F2>", "3c":"<F3>","3d":"<F4>","3e":"<F5>","3f":"<F6>","40":"<F7>",
          "41":"<F8>","42":"<F9>","43":"<F10>","44":"<F11>","45":"<F12>"}
output = []
keys = open('out.txt')
for line in keys:
    try:
        if line[0]!='0' or (line[1]!='0' and line[1]!='2') or line[3]!='0' or line[4]!='0' or line[9]!='0' or line[10]!='0' or line[12]!='0' or line[13]!='0' or line[15]!='0' or line[16]!='0' or line[18]!='0' or line[19]!='0' or line[21]!='0' or line[22]!='0' or line[6:8]=="00":
             continue
        if line[6:8] in normalKeys.keys():
            output += [[normalKeys[line[6:8]]],[shiftKeys[line[6:8]]]][line[1]=='2']
        else:
            output += ['[unknown]']
    except:
        pass

keys.close()

flag=0
print("".join(output))
for i in range(len(output)):
    try:
        a=output.index('<DEL>')
        del output[a]
        del output[a-1]
    except:
        pass

for i in range(len(output)):
    try:
        if output[i]=="<CAP>":
            flag+=1
            output.pop(i)
            if flag==2:
                flag=0
        if flag!=0:
            output[i]=output[i].upper()
    except:
        pass

print ('output: ' + "".join(output))

• 得到一串字符串：884080810882108108821042084010421，发现只有 01248 几个数字，推测是云影密码，解密得到 Flag：flag{THISFLAG}

#!/usr/bin/python
# -*- coding=utf8 -*-

def de_code(c):
    dic = [chr(i) for i in range(ord("A"), ord("Z") + 1)]
    flag = []
    c2 = [i for i in c.split("0")]
    for i in c2:
        c3 = 0
        for j in i:
            c3 += int(j)
        flag.append(dic[c3 - 1])
    return flag

def encode(plaintext):
    dic = [chr(i) for i in range(ord("A"), ord("Z") + 1)]
    m = [i for i in plaintext]
    tmp = [];flag = []
    for i in range(len(m)):
        for j in range(len(dic)):
            if m[i] == dic[j]:
                tmp.append(j + 1)
    for i in tmp:
        res = ""
        if i >= 8:
            res += int(i/8)*"8"
        if i%8 >=4:
            res += int(i%8/4)*"4"
        if i%4 >=2:
            res += int(i%4/2)*"2"
        if i%2 >= 1:
            res += int(i%2/1)*"1"
        flag.append(res + "0")
    print ("".join(flag)[:-1])

if __name__ == '__main__':
    c = input("输入要解密的数字串:")
    print (de_code(c))
    m_code = input("请输入要加密的数字串:")
    encode (m_code)

m1bmp

• 用 binwalk 先查看一下图片，但是没有发现什么东西
• 考虑到是 BMP 图片，猜测存在隐写，用 zsteg 查看一下图片，发现 base64 字符串，解码得到 Flag：flag{l5DGqF1pPzOb2LU919LMaBYS5B1G01FD}

tunnel

• 打开数据包，发现全是有 base64 的 dns 域名请求，使用 tshark 提取 A 记录的域名

tshark -r tunnel.pcap -Y "dns" -T fields -e dns.qry.name > out.txt

• 将 base64 字符串数据提取出来，CyberChef 解密一下，发现是一个 zip 文件

f=open('out.txt','r')
fp=open('zip.txt','w')
tmp=f.readline().strip('\n')
s=''
while tmp[-8:]!='.evil.im':
    tmp=f.readline().strip('\n')
s=tmp[:-8]
while len(s)%4!=0:
    s+='='
fp.write(s)
fp.write('\n')
while 1:
    a=f.readline().strip('\n')
    if a:
        if a[-8:]=='.evil.im':
            if a!=tmp:
                s=a[:-8]
                while len(s)%4!=0:
                    s+='='
                fp.write(s)
                fp.write('\n')
                tmp=a
    else:
        break

• 发现压缩包是加密的，由于前面提取出来了很多 base64 字符串，猜测存在 base64 隐写，解密得到压缩包的密码：password: B@%MG"6FjbS8^c#r，打开压缩包在图片上发现 Flag：flag{D01nt_5py_0nmE}

#!/usr/bin/env python
import re

path = './zip.txt'
b64char = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
with open(path, 'r')as f:
	cipher = [i.strip() for i in f.readlines()]
plaintext = ''
for i in cipher:
	if i[-2] == '=':  # There are 4-bit hidden info while end with two '='
		bin_message = bin(b64char.index(i[-3]))[2:].zfill(4)
		plaintext += bin_message[-4:]
	elif i[-1] == '=':  # There are 2-bit hidden info while end with one '='
		bin_message = bin(b64char.index(i[-2]))[2:].zfill(2)
		plaintext += bin_message[-2:]
plaintext = re.findall('.{8}', plaintext)  # 8bits/group
plaintext = ''.join([chr(int(i,2)) for i in plaintext])
print(plaintext)

标签：津门,plaintext,Misc,len,flag,2021,output,line,txt
来源： https://blog.csdn.net/LYJ20010728/article/details/120295878