攻防世界 m0_01讲解

我的另一篇文章有做杂项的基本思路

希望对你有帮助CTF MISC解题思路总结篇_Dadeln。的博客-CSDN博客

打开附件解压出来为一个流量文件，应该是个流量分析题

用wireshark打开看

经典的usb流量分析题

随便打开一个数据包查看   发现是八字节 为键盘流量

如果是四字节则为鼠标流量

在wireshark的tshark工具下将数据提取为文本文件

得到1.txt文件

接下来用用键盘的脚本跑出来

import os
os.system("tshark -r test.pcapng -T fields -e usb.capdata > usbdata.txt")
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>"}


nums = []
keys = open('1.txt')#你导出的数据文档
for line in keys:
    #print(line)
    if len(line)!=17: #首先过滤掉鼠标等其他设备的USB流量
         continue
    nums.append(line[0:2]+line[4:6]) #取一、三字节
    #print(nums)
keys.close()
output = ""
for n in nums:
    if n[2:4] == "00" :
        continue

    if n[2:4] in normalKeys:
        if n[0:2]=="02": #表示按下了shift
            output += shiftKeys [n[2:4]]
        else :
            output += normalKeys [n[2:4]]
    else:
        output += '[unknown]'
print('output :' + output)

得到884080810882108108821042084010421

观察这段数字只有01248，上网查了一下是云影密码​​​​​​​

接下来用解密代码跑出来

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])

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)

所以flag就是

flag{THISISFLAG}

恭喜得到flag