2019技术大赛预选赛 writeup

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2019技术大赛预选赛 writeup

第一题简单

base64加字符偏移就可以了

第二题 维吉利亚加密

代码语言：javascript代码运行次数：0运行复制

Vvr Ifnvaus Bdwokv Gbtrzsa Vkqgofntja rrlznxk eflvkozjcdue rs “mzg oez ff pjkhvtx ok kqzioeg vgfsf.” Zyil au vvykokaeoyrp avuwfnzv, bnl fcry eom ucdgaie mzg qhxiegl dfrguta gh huk wixdf ce oks ijggrtk-dtq uqvketbxkq sulnwsvwbtj. Taw fssoeimaqb sutulwu gbrvlr gp huk towwu hugk htng prke ulwf tbx teglwfvkj th wpoorv sxutsg ifmfmpwpgkihf. Dig iiyilquegghr fqknjryl wpqbsgalkgg zath fgts gnrn mzkg: vz uetdu kvzy mxujoaojml xqf rtjukapu vtkezjkhl, zvcafkehkj fhj glpnrnzapu fktrxl msly, grhlqqbrj fhj cignvnmaeogoeg nkgff, kcevltcaot anuvwbtj agv gzrikihfu, rvmzttd eofn, rnw eqfr. Cztagwh nzkefhvwam ko ijqjvjv a vgodykke vzcfnikekabogofn, pw ychru stq vvnz dowwtb pxppmgifnvyy bfxcybvs mzg ggauy hx oognvmtlkqnr kevzpwdavs ygt grilrbfi rvmzttd kbsuimtlkca, ypsmwog, ntu dbkvfvhltxv eczvlttlkcay rgtapgg guvxjuoeorl tlvopqj. fesi{3osir4d633096u273734wct73r6985l4wc2us213}

直接暴力破解

Mac下可以用 这个工具

Clear text using key "congrats":

flag在最后flag{3afca4d633096b273734eaf73e6985f4fc2ba213}

代码语言：javascript代码运行次数：0运行复制

The Concise Oxford English Dictionary defines cryptography as “the art of writing or solving codes.” This is historically accurate, but does not capture the current breadth of the field or its present-day scientific foundations. The definition focuses solely on the codes that have been used for centuries to enable secret communication. But cryptography nowadays encompasses much more than this: it deals with mechanisms for ensuring integrity, techniques for exchanging secret keys, protocols for authenticating users, electronic auctions and elections, digital cash, and more. Without attempting to provide a complete characterization, we would say that modern cryptography involves the study of mathematical techniques for securing digital information, systems, and distributed computations against adversarial attacks. flag{3afca4d633096b273734eaf73e6985f4fc2ba213}

第三题 音频加密

工具打开，切换到频谱图

解压得到

还缺少part2

直接二进制打开图片

第四题 USB 键盘

按照

tshark.exe -r usb1.pcap -T fields -e usb.capdata > usbdata.txt

获取到具体的data

代码语言：javascript代码运行次数：0运行复制

mappings = { 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:"'", 0x36:",",  0x37:"." }
nums = []
keys = """00:00:00:00:00:00:00:00
20:00:00:00:00:00:00:00
20:00:17:00:00:00:00:00
00:00:17:00:00:00:00:00
00:00:00:00:00:00:00:00
00:00:0b:00:00:00:00:00
00:00:00:00:00:00:00:00
00:00:0c:00:00:00:00:00
00:00:0c:16:00:00:00:00
00:00:16:00:00:00:00:00
00:00:00:00:00:00:00:00
20:00:00:00:00:00:00:00
20:00:0c:00:00:00:00:00
20:00:00:00:00:00:00:00
00:00:00:00:00:00:00:00
00:00:16:00:00:00:00:00
00:00:00:00:00:00:00:00
20:00:00:00:00:00:00:00
20:00:10:00:00:00:00:00
00:00:10:00:00:00:00:00
00:00:00:00:00:00:00:00
00:00:1c:00:00:00:00:00
00:00:00:00:00:00:00:00
20:00:00:00:00:00:00:00
20:00:13:00:00:00:00:00
20:00:00:00:00:00:00:00
00:00:00:00:00:00:00:00
00:00:04:00:00:00:00:00
00:00:00:00:00:00:00:00
00:00:16:00:00:00:00:00
00:00:00:00:00:00:00:00
00:00:16:00:00:00:00:00
00:00:00:00:00:00:00:00
00:00:1a:00:00:00:00:00
00:00:1a:12:00:00:00:00
00:00:12:00:00:00:00:00
00:00:00:00:00:00:00:00
00:00:15:00:00:00:00:00
00:00:00:00:00:00:00:00
00:00:07:00:00:00:00:00
00:00:00:00:00:00:00:00
20:00:00:00:00:00:00:00
20:00:1e:00:00:00:00:00
20:00:00:00:00:00:00:00
00:00:00:00:00:00:00:00
00:00:28:00:00:00:00:00
00:00:00:00:00:00:00:00"""
keys = keys.split("\n")
import sys
import os

DataFileName = "usb.dat"

presses = []

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>"}

# handle
result = ""
for press in keys:
    Bytes = press.split(":")
    if Bytes[0] == "00":
        if Bytes[2] != "00" and Bytes[3] == "00":
            result += normalKeys[Bytes[2]]
    elif Bytes[0] == "20":  # shift key is pressed.
        if Bytes[2] != "00" and Bytes[3] == "00":
            result += shiftKeys[Bytes[2]]
    else:
        print("[-] Unknow Key : %s" % (Bytes[0]))
print("[+] Found : %s" % (result))

输出[+] Found : TthisIsMmyPassword!<RET>

加上一些猜测 正确密码是ThisIsMyPassword!

解压获得flag{ccc919529c01014af868bfa8d9c1c00a3fc2348f}

第五题 多层 zip解压

比较绕，先第一层暴力19900000000开始枚举

解压得到level2.zip和readme.txt

readme没有啥用

Please notice that the telephone number in level1 is a fake one, do not try to call it.

看到level2.zip里面也有一个readme.txt

用zip明文attack，上AZPR

接触level3.zip

这回没啥线索了，可能是个伪加密

终于得到一个图片flag.png

但是看内容

？？？

绕了三层还没给结果。。。

用TweakPng打开提示CRC错误

修改一下图片高度

然后

这个真是。。。

本文参与 腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。 原始发表：2024-10-11，如有侵权请联系 cloudcommunity@tencent 删除前往查看加密音频zip二进制工具

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