BUU MISC刷题记录

附件上传gitee 流程

创建 git 仓库:

mkdir misc
cd misc
git init
touch README.md
git add README.md
git commit -m "first commit"
git remote add origin https://gitee.com/l1yee/misc.git
git push -u origin master

已有仓库?

cd existing_git_repo
git remote add origin https://gitee.com/l1yee/misc.git
git push -u origin master

报错解决:

git 解决push报错：! [rejected] master -> master (fetch first) error: failed to push some refs to

将线上、线下代码进行合并：
git pull --rebase origin master

[MRCTF2020]寻找xxx

题目给出一段录音，初步判断为一个电话号码，后经过dtmf2num工具，得到一个号码，将其发送给公众号得到一张图片，图片内容里就有flag

根据频谱图来

频谱图如下判断

一个拨号音是由一个高频和一个低频同时发音组成
由频谱可得到一个规律：
同一行的低频是一样的，例如789，同一列的高频是一样的，例如2580

分析12个拨号音的频谱图

可以看到一个明显的地方：
第9、10、11位音频的高频是一样的，因此他们应该在同一列，而低频依次降低
根据拨号音频谱结合听到的声音，第9、10、11位拨号应该是085或者852

另一个点：
第2、3、9位的音频低频一样（其中，第3位与第9位是同一拨号音），说明它们在同一行，且第3位的高频比第2的高频低
若第3位的拨号是0，那么第2位的拨号应该是 * 号键，这与hint:纯数字相悖
因此，9、10、11位的拨号是852，第3位的拨号是7

同理，根据高频和低频比较，可以得出其他的位的拨号
最终还原出1~12位的拨号依此为 187485618521

利用DTMF识别软件识别

利用DTMF decoder安卓app(https://github.com/pjasiun/dtmf-decoder/tree/master/bin)

对该声音进行识别获得号码187485618521

但这对音频和环境有一定要求，否则识别会不准确

可参:http://www.ericdshen.com/index.php/archives/5/

[INSHack2017]hiding-in-plain-sight

foremost分离

直接打开jpg看到flag

[INSHack2018]42.tar.xz

一堆压缩包

上shell脚本

while [ "`find . -type f -name '*.tar.xz' | wc -l`" -gt 0 ]; do find -type f -name "*.tar.xz" -exec tar xf '{}' \; -exec rm -- '{}' \;; done;

就能得到flag

[HDCTF2019]信号分析

用Audacity 打开音频分析

观察它的波形，发现了这么一段很有规律的东西，每个周期的东西都是一样的

根据提示文章，已知两长为 1，两短为 0，先短后长为 F，无效码为先长后短

题目不包含同步位，引用一下赵师傅的图：

flag{FFFFFFFF0001}

[XMAN2018排位赛]AutoKey

打开流量包 发现是个usb流量包

usb流量，直接用UsbKeyboardDataHacker一把梭 python UsbKeyboardDataHacker.py attachmeng.pcapng

拿到一串字符

<CAP>a<CAP>utokey('****').decipheer('<CAP>mplrvffczeyoujfjkybxgzvdgqaurkxzolkolvtufblrnjesqitwahxnsijxpnmplshcjbtyhzealogviaaissplfhlfswfehjncrwhtinsmambvexo<DEL>pze<DEL>iz')

其中是转换大小写的键，~~就是删除键，按照这样恢复一下，就还原了出题人一开始敲的字符串

Auto('****').decipheer('MPLRVFFCZEYOUJFJKYBXGZVDGQAURKXZOLKOLVTUFBLRNJESQITWAHXNSIJXPNMPLSHCJBTYHZEALOGVIAAISSPLFHLFSWFEHJNCRWHTINSMAMBVEXPZIZ')

然后就是用autokey爆破

官方脚本 py文件下载地址 http://www.practicalcryptography.com/cryptanalysis/stochastic-searching/cryptanalysis-autokey-cipher/

配置文件下载地址 http://www.practicalcryptography.com/cryptanalysis/text-characterisation/quadgrams/#a-python-implementation

两个网站，一个网站上最下面可以下载break_autokey.py，然后在另一个网站最下面下载english_quadgrams.txt.zip和english_trigrams.txt.zip和ngram_score.py，把两个zip解压并且txt名字分别改为quadgrams.txt和trigrams.txt，然后两个py文件和两个txt放在一起，如果没有装pycipher的装一下（python2环境） pip install pycipher 然后把autokey的密文放在break_autokey.py里面

跑一下py（python2的环境） python2 break_autokey.py 然后就观察，可以看到flag在klen8

这段

HELLOBOYSANDGIRLSYOUARESOSMARTTHATYOUCANFINDTHEFLAGTHATIHIDEINTHEKEYBOARDPACKAGEFLAGISJHAWLZKEWXHNCDHSLWBAQJTUQZDXZQPF

理解一下就知道flag了 flag{JHAWLZKEWXHNCDHSLWBAQJTUQZDXZQPF}

参 http://cbase.top/index.php/2021/04/19/xman2018%E6%8E%92%E4%BD%8D%E8%B5%9Bmisc-autokey/

[QCTF2018]X-man-Keyword

一张图片，上 Stegsolve ，发现图片开头 lsb 里有些东西。

java -jar Stegsolve.jar

用 lsb 解密脚本

ZG4ubmV0L3dlaXhpbl80NTc4NTI4OA==,size_16,color_FFFFFF,t_70)

可以得到：

PVSF{vVckHejqBOVX9C1c13GFfkHJrjIQeMwf}

根据题目提示，尝试了一下，发现是 Nihilist 密码。

写个脚本解密：

import string

enc='PVSF{vVckHejqBOVX9C1c13GFfkHJrjIQeMwf}'
grid='LOVEKFC'+'ABDGHIJMNPQRSTUWXY'
flag=''

for i in enc:
    if i in string.ascii_lowercase:
        index=grid.lower().index(i)
        flag+=string.ascii_lowercase[index]
        continue
    if i in string.ascii_uppercase:
        index=grid.upper().index(i)
        flag+=string.ascii_uppercase[index]
        continue
    flag+=i
print flag

解读一下 脚本

就是遍历 enc 如果大写就到 第二个if语句 查询它在grid的位置 然后再去正序的 26个大写字母中找到对应的位置

小写字母同此

Flag: QCTF{cCgeLdnrIBCX9G1g13KFfeLNsnMRdOwf}

[DDCTF2018]第四扩展FS

鼠标右键查看属性 发现Pactera 猜测可能是压缩包密码

猜测有压缩包文件

foremost分离

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4kTwwTesD{iiFk}Duu{eh!sw{FCD{eFwiCwFie4DFw}T!1w!{}esTsi!noaseC{}{CD1ehCaDCDuDDuTa}soTT}D{!4T!wDaonuCo1nFw}DC1{FhFuheCFCnuoeuTCFknuuwDFkna{{sDnD{{khCunDDTC4wkw{u4DDDDTaa11h1F}CaahC4oi!DFu4DskDoTowawk1ueT{kaakhCDkFuFF4TwDk4Dao{e}D4wFhw{sDFCnw{DseD41heT{}TDenD!{aDeTuksT1s1Dk{D!e{4DuD{{kD{nnCi141seeksioew44TeDDD4eC{n{{Tu{DnwT1oaFCsFiDi}eF!DwCw{{neT}D4}{TuhoD{Dk!iDu1F{s4!u!hs4ha}CaaFFs{suFuDCiTDDasiDCh!CaC{{TFhDkDDuku1!iuT!hTnohFknki1Fnu1ah1Tk}11DsTDwD{oiDDeow4eF{{TDhoh!}CDCDoTDaFihCk1iFn1DnkC}DaCTwi

因为题目描述中说到频次很重要，所以尝试使用在线的词频密码分析网站，但没有发现，所以考虑另一种可能，就是不用还原成句子，只需要里面的每个词的频次。然后写个脚本跑一下：

下面这个比较自动我没跑出来

# -*- coding:utf-8 -*-
with open ('file.txt','r') as f:
    dd3 = f.read()

a = {}
for i in dd3:
    if i not in a:
        a[i]=1
    else:
        a[i] +=1

b = sorted(a.items(), lambda x, y: cmp(x[1], y[1]), reverse=True)
flag = ''
for i in b:
    flag += i[0]
print(flag)

# DCTF{x1n9shaNgbIci}

下面脚本跑完之后自己统计

import collections
f=open('file.txt','r')
print (collections.Counter(f.read()))

得到flag:DDCTF{huanwe1sik4o!}

[RoarCTF2019]forensic

附件已经失效

参:https://blog.csdn.net/m0_46481239/article/details/108873022

Beautiful_Side

foremost 分离一张一半的二维码

来这个网站上面手撸一份然后扫码即可，最后得到flag{OQWIC_4DS1A_S034S}

[XMAN2018排位赛]file

linux挂载讲解

http://c.biancheng.net/view/2859.html

下载得到一个attachment.img。可以挂载。

发现一个空的lost+found文件夹。

用 extundelete img文件 –-restore-all 命令对 attachment.img 进行恢复。(需要安装 extundelete)

apt-get install  extundelete

在同级目录下会生成一个名为RECOVERED_FILES的文件夹。flag就在恢复的文件里。

最后别忘记 umount /mnt/cdrom 取消挂载。

[MRCTF2020]摇滚DJ（建议大声播放

这个音频听起来就知道是什么了，无线电，用RX-SSTV工具，不知道怎么用的可以去这个兄弟博客BUUCTF-刷题记录-7

也可以使用QSSTV解析

flag{r3ce1ved_4n_img}

[ACTF新生赛2020]frequency

打开是一个word

显示隐藏文字有一段

备注有一段(我下载下来的文件没有)

根据题目名字提示两端base64解密后进行词频统计得到flag

在线词频统计 http://www.aihanyu.org/cncorpus/CpsTongji.aspx

[SCTF2019]Ready_Player_One

开着飞机按W往上冲到上面去就有flag了

flag{You_Are_The_Ready_Player_One!!!For_Sure!!!}

greatescape

下载附件 打开得到一个流量包

追踪tcp流

在流18看到

右下角到流19的时候出现一个key，是一个rsa私钥格式的

把私钥保存为ssc.key

查了一下，这有可能是在向ftp服务器传送私钥，我们得到了私钥，就可以解密TLS报文

文件->首选项->Protocols->TLS->Edit->点击”+”号

然后在tcp.stream eq 80追踪TLS Stream即可发现flag

参:https://blog.csdn.net/mochu7777777/article/details/107743870

[INSHack2019]gflag

根据description得知，这是一种编程语言

通过搜索引擎搜索，得知是G语言(G-Code)

接着找到一个G语言的在线运行站：https://ncviewer.com/

把代码复制进去，点击PLOT，调整角度，放大

flag{3d_pr1nt3d_fl49}

[INSHack2018]INSanity

打开一个 md 交了就是flag…

[GUET-CTF2019]520的暗示

photo.dat，与0x33异或

data = open("photo.dat",'rb')
strs = data.read()
flag = open("1.jpg",'ab+')
for i in strs:
    flag.write(bytes([i ^ 0x33]))

得到一张运营商的定位

根据LTE可以定位基站地址http://www.gpsspg.com/bs.htm

[INSHack2018]Self Congratulation

图片左上角有一堆黑白块，按白色为0黑色为1转成二进制，得到：

00110001001
10010001100
11001101000
01101010011
01100011011
10011100000

二进制转ASCII得到：12345678

flag{12345678}

key不在这里

QR 识别一下

发现这个 m 是十进制 有flag的痕迹 102 108 97 103 十进制的flag格式

写脚本转换成字符串

from urllib import parse

s = "10210897103375566531005253102975053545155505050521025256555254995410298561015151985150375568"
flag = ""
i = 0

while(i <= len(s)):
    if int(s[i:i+3]) < 127:
        flag += chr(int(s[i:i+3]))
        i += 3
    else:
        flag += chr(int(s[i:i+2]))
        i += 2

print(parse.unquote(flag))
# flag{5d45fa256372224f48746c6fb8e33b32}

很好的色彩呃？

六个颜色可以看出来有不一样 ps看16进制 得到

把最后两位 得到

将每部分最后两位取出来，得到

6161706a6573

转字符串

flag{aapjes}

[BSidesSF2019]table-tennis

给了一个流量包，使用wireshark打开，通过查找字符串和跟踪TCP数据流并没有发现什么，以我的水平，基本上到这里就结束了。

后面查看了其他人写的writeup，跟着复现了一下。

这个流量包中除了大量的加密的TLS和TCP数据流，还有一些ICMP数据包，发现里面有HTML数据。

这里要将这些HTML数据提取出来，照着大佬使用python scapy模块写的脚本提取数据，我也依样画瓢谢了一下并且理解了脚本的意思，脚本比较简单：

from scapy.all import *
 
packets = rdpcap('out.pcapng')    #rdpcap()读取pcapng文件
 
for packet in packets:            #遍历每一个数据包
    if packet.haslayer(ICMP):            #haslayer()判断数据包的类型，此处为ICMP
        if packet[ICMP].type == 0:       #每一个ICMP的type值为0的包
            print packet[ICMP].load[-8:],      #打印每个数据包的最后8位，因为前面数据是重复的

将这段base64加密的数据空格合并 并解密就能得到flag

flag{JustAS0ngAb0utP1ngP0ng}

参:https://mochu.blog.csdn.net/article/details/109859233

[MRCTF2020]小O的考研复试

#!/usr/bin/python

a=2
for i in range(19260816):
	a = a * 10 + 2
	a%=(1e9+7)

print(a)
#577302567.0

[INSHack2017]insanity

直接交

[INSHack2019]Sanity

直接交

[INSHack2019]INSAnity

直接交

[RCTF2019]disk

winhex 其中一段

剩下看大佬文章吧 有时间再来复现

https://blog.csdn.net/mochu7777777/article/details/109841367

[CFI-CTF 2018]webLogon capture

用wireshark打开.pcapng文件 追踪http流发现邮件账号和密码

密码 url解密得到flag

[HDCTF2019]你能发现什么蛛丝马迹吗