Misc·进阶区

1、base64÷4

image-20201130082729032

根据名称，推测为base16

下载附件，为txt文档，如下图所示。

image-20201130083601474

在线base16解码

image-20201130083706648

得到flag

flag{E33B7FD8A3B841CA9699EDDBA24B60AA}

2、wireshark-1

image-20201130083817579

由题知，本体为数据包分析

下载附件

image-20201130083934079

用wireshark打开

image-20201130084024614

追踪tcp流

image-20201130084127415

找到flag

flag{ffb7567a1d4f4abdffdb54e022f8facd}

3、reverseMe

image-20201130084413801

下载附件，得到一张图片

image-20201130084502220

由题知，大概反转就是flag。

使用win10自带的图片编辑器就可以完成反转，效果如下

image-20201130084816126

得到flag

flag{4f7548f93c7bef1dc6a0542cf04e796e}

4、Aesop_secret

image-20201130090157234

下载附件，如下

Aesop

猜测是将每一帧拼起来

使用Photoshop最终得到

Aesop1

本以为就此结束，flag是ISCC，但是不对

将原图用winhex打开，发现文件尾部有如下可疑的编码

image-20201130090617720

初步判断，应该是base64

复制出去解码

image-20201130090937049

解码失败

结合题目，以及之前拼图获得的ISCC，推测为aes加密，

尝试解码，key为ISCC

image-20201130091250504

解码获得一串短一点的字符串，推测为二次aes加密，继续解密

image-20201130091428833

得到flag

flag{DugUpADiamondADeepDarkMine}

5、a_good_idea

image-20201130092439634

下载附件，为一个图片，如下

a_very_good_idea

使用winhex查看，发现如下

image-20201130092605420

文件尾部有压缩包，怀疑是隐藏文件、

使用foremost分解

得到一个隐藏的压缩包

image-20201130092704278

解压后得到两张图片和一个txt

image-20201130092906215

文档内部显示

image-20201130092941617

让我们从像素中寻找秘密

用Stegsolve拼合图片，发现显示很奇怪，效果如下

image-20201130093949240

用ps调整参数

将曝光度调整到最高，得到一张二维码

image-20201130094135935

使用QR Research扫描

image-20201130094327967

得到flag

NCTF{m1sc_1s_very_funny!!!}

6、can_has_stdio?

image-20201130094449560

附件解压后得到一个无后缀文件，如下

image-20201130094612009

使用010打开后，内部显示如下

image-20201130094700860

经过查询得知该编码是BrainFuck编码

BrainFuck介绍

在线解码

image-20201130100738108

得到flag

flag{esolangs_for_fun_and_profit}

7、János-the-Ripper

image-20201130101043507

附件解压后获得

image-20201130103423192

使用010打开查看，显示如下：

image-20201130103508788

根据文件头判断应该是个zip压缩包，并且其中有flag.txt文件

更改文件后缀为zip

image-20201130103604701

尝试解压，发现需要密码

image-20201130103647971

怀疑伪加密

尝试修改，失败

使用zipperello暴力破解

image-20201130104240350

爆出解压密码为fish

解压文件

image-20201130104354297

得到flag

flag{ev3n::y0u::bru7us?!}

8、快乐游戏题

image-20201130105159581

下载附件提示报毒，于是在专用测试软件的虚拟中下载运行

image-20201130105442648

玩个游戏直接通关就给flag

*

有点简单

UNCTF{c783910550de39816d1de0f103b0ae32}

9、Banmabanma

image-20201130110853652

附件下载，解压后是应该png图片

斑马斑马

一眼看去，确认是条形码

尝试用手机扫码，尝试无果

使用在线解码网站，解码成功，显示如下

image-20201130111035914

得到flag

flag{TENSHINE}

10、stage1

image-20201130120808569

附件下载，是个图片

3600c13125fe4443aeef3c55b9c1357b

拉入stegsolve

发现有二维码

image-20201130122143441

使用QR Research扫描后得到

image-20201130122423873

初步断定该字符串为十六进制数，放入在线解码

image-20201130122601206

解码出来的推测为.pyc文件，将原16进制字符串通过winhex保存为.pyc文件，效果如下：

image-20201130122819671

进行反编译

image-20201130122846849

运行即可出现flag

image-20201130123504109

得到flag

AlphaLab

11、Hear-with-your-Eyes

image-20201130134127726

附件为一段音频，如下

image-20201130134220992

拖入Audacity进行分析

image-20201130134309839

更改显示为频谱图

image-20201130134403803

显示如下

image-20201130134432976

得到flag

flag{e5353bb7b57578bd4da1c898a8e2d767}

12、再见李华

image-20201130140012186

附件解压后获得一张图片

mail2LiHua

直观上的感觉是MD5

尝试解密

image-20201130143717384

解密无果

winhex查看该图片，发现图片中有一个key.txt

image-20201130144510567

使用foremost分解

分离出一个压缩包，内含一个加密txt文件

image-20201130144737999

推测密码与题目有关，且密码尾部为LiHua

爆破,使用ARCHPR进行暴力破解，选择掩码，推测LiHua前应该有四位密码(根据1000是四位联想到)，勾选大写、小写字母及数字

image-20201201141621708

开始爆破！

image-20201201141657538

成功得到解压密码

15CCLiHua

解压文件，得到flag

image-20201201141828927

Stay hungry, Stay foolish.

顺带一提，前面给的MD5值是这个解压密码MD5加密后的前16位，或许应该爆破。

13、something_in_image

image-20201201142300310

附件下载下来是一个没有后缀的文件

image-20201201143640992

用winhex打开，搜索flag

image-20201201143705039

得到flag

Flag{yc4pl0fvjs2k1t7T}

14、simple_transfer

image-20201201143856706

附件是一个数据包文件

image-20201201144119840

winhex分析后发现其中有pdf文件

image-20201201150521889

使用foremost分离，获得pdf文件

image-20201201150631860

打开后显示图像数据不充分

image-20201201150709845

里面是一片空白，实际上是由一张显示不出来的图片挡住了flag

image-20201201150855016

得到flag

HITB{b3d0e380e9c39352c667307d010775ca}

15、适合作为桌面

image-20201202093717222

附件下载是一张图片

1

跟上方stage1的附件一样，推测解法差不多

拉入stegsolve

发现有二维码

image-20201202093922751

使用QR Research扫描二维码

image-20201202094000226

得到一串十六进制数

粘入winhex

image-20201202094107259

保存为.pyc文件

image-20201202094050271

反编译

image-20201202094140079

稍稍修改，运行

image-20201202094313901

得到flag

flag{38a57032085441e7}

16、神奇的Modbus

image-20201202133043854

附件是一个数据包

使用wireshark打开，并搜索flag、key、ctf等关键字

在搜索ctf时出现flag

image-20201202133213928

得到flag，格式为sctf{}且中间要补上一个o

sctf{Easy_Modbus}

17、MISCall

image-20201202133538651

附件下载是一个没有后缀的文件

image-20201202133604779

使用winhex打开

image-20201202133808630

看不出来是什么文件

使用文件格式分析器分析一波

image-20201202133840397

发现是.BZC压缩文件，更改后缀

使用bandzip解压

发现flag.txt

image-20201202134139067

打开查看

image-20201202134158155

提示没有东西，继续前进。。。

发现.git文件夹内部也没有flag

网上查询后得知本题考察 .git目录利用、git stash命令

将此文件夹拖入kali

image-20201202140112710

查看提交的commit history

git log

image-20201202140141745

查看堆栈中的 stash 列表

git stash list

image-20201202140238633

显示文件

git stash show

image-20201202140312053

重新进行储存 复原文件得到s.py文件

git stash apply

image-20201202140349576

运行s.py文件

python s.py

image-20201202140415030

得到flag

NCN4dd992213ae6b76f27d7340f0dde1222888df4d3

18、Reverse-it

image-20201202142242672

附件为一个无后缀的文件

使用winhex打开后看不出来是啥

同时使用binwalk等文件分析器也分析不出来文件格式

根据题意

怀疑文件的十六进制源码需要逆转

恰好文件头和文件为逆转后欸话jpg格式一样

str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000003000100A3000002303A31343A30313022323A30303A31303032364001446F62656020586F647F63786F6070254C656D656E6473710000000840000001000000084000000000000000A0000001000000040009678C80000004100000020002310270000009100000020001310000020001000000030008210A6000000100000005000B11026000000100000005000A110000010001000000030002110700080000000A200D4D40000669687542D001EFF00008400840010101000649464A401000EFF8DFF '

str_2 = str_1[::-1]

print(str_2)

使用脚本逆转

粘贴进新文件，保存为jpg

image-20201202144227945

保存成功

image-20201202144245339

发现图片还要反转

ausdghaiou

本想使用win10自带编辑器，进行反转，结果文件太小，不支持

image-20201202144604435

随后使用ps进行反转，得到flag

image-20201202144537035

SECCON{6in_tex7}

19、打野

image-20201203084240584

使用winhex查看

image-20201203084439440

无果

使用文件分析器分析

image-20201203084520203

是分析为bmp

使用binwalk分析

image-20201203084641877

还是bmp

使用zsteg查看是否有隐写数据

image-20201203084753474

得到flag

qwxf{you_say_chick_beautiful?}

20、倒立屋

image-20201203091301477

使用winhex查看

image-20201203091339856

无果

使用binwalk和分拣分析器均无果

尝试stegsolve

image-20201203091448425

同样无果

使用zsteg查看文件

image-20201203091520887

发现疑似flag

IsCc_2019

多次提交后无果

网上查询后得知，flag提交要结合题目翻转。。。。。。

flag{9102_cCsI}

21、2017_Dating_in_Singapore

image-20201203101308091

下载附件为一个日历

image-20201203101345125

将之前的题目描述保存到记事本中

使用-明显就是分割符

手动处理一下

image-20201203101750874

一共十二行，对应12个月

daliy

得到flag

HITB{CTFFUN}

22、Erik-Baleog-and-Olaf

image-20201203105658132

附件是一个没有后缀的文件

image-20201203105730363

使用winhex查看

image-20201203105751034

是一个png图片文件，且文件尾部有个链接

image-20201203105849745

将附件添加后缀png

stego100

打开隐藏链接也是一个图片

image-20201203110020892

使用beyond compare比较两张图片

image-20201203105624478

得到一个二维码

二维码

保存下来使用QR Research扫描试试

image-20201203110352809

得到flag

flag{#justdiffit}

23、Ditf

image-20201203111958440

附件解压是一个png图片

使用binwalk题区出一个rar加密压缩包

image-20201203112124617

image-20201203112152106

尝试暴力破解失败

继续从图片上寻找线索

使用一系列隐写查看工具过了一遍均无所获

最后尝试使用winhex修改图片高度

image-20201203112454115

得到解压密码

image-20201203112414574

解压压缩包得到一个pcapng数据包文件

image-20201203112617144

先于http中搜索各种关键字

在搜索png时，找到一个kiss.png的记录，其中有一段可以代码

image-20201203113453015

尝试将字符串base64解码

image-20201203113535137

得到flag

flag{Oz_4nd_Hir0_lov3_For3ver}

24、心仪的公司

image-20201203114649278

附件解压后是一个pcapng数据包

image-20201203114741224

使用strings命令

image-20201203115441709

得到flag

fl4g:{ftop_Is_Waiting_4_y}

25、就在其中

image-20201228091711433

附件解压后得到一个数据包

image-20201228091753268

拉入winhex分析

可以看到它似乎是传输某些文档的流量包，推测flag与传输的文档有关

追踪tcp流，发现有个类似传输文件目录的东西，其中的.key文件需要重视，含flag文档很有可能是需要解密。

image-20201228091954744

发现私钥

image-20201228091914673

之后还在里面发现了一个压缩包文件以及pdf文件的ascii码，但是都复原不了

使用binwalk扫一遍

image-20201228092707410

提取出来很多奇奇怪怪的东西

其中有一个key.txt文件

image-20201228092823332

打开显示乱码

image-20201228092919100

推测使用先前获得的私钥解密

先将之前的私钥保存为test.key

再使用openssl

openssl rsautl -decrypt -in key.txt -inkey test.key -out flag.txt

其中 -in 为要解密的加密文档 -inkey 为密钥 -out 为输出文档

image-20201228093232333

打开flag.txt

image-20201228093308385

得到flag

flag{haPPy_Use_0penSsI}

26、Miscellaneous-200

image-20201228162720481

附件是一个文档

image-20201228162810950

推测是像素点

导入PIL库，使用脚本

from ast import literal_eval as make_tuple
from PIL import Image

f = open('flag.txt', 'r')
corl = [make_tuple(line) for line in f.readlines()]
f.close()
img0 = Image.new('RGB', (270, 270), '#ffffff')
k = 0
for i in range(246):
    for j in range(246):
        img0.putpixel([i, j], corl[k])
        k = k + 1
img0.save("result.png")

生成图片

image-20201228162923953

得到flag

flag{ youc@n'tseeme }

27、