BUU

BUU

LSB

010看了一下没发现有什么问题

丢到stegsolve,rgb调到0的时候,是张二维码

image-20241209221820579

image-20241209221855330

被嗅探的流量

找就完了

image-20241210190719549

rar

image-20241210193906460

image-20241210193931602

qr

image-20241210194009393

image-20241210194103908

镜子里面的世界

image-20241210200150364

key就是flag..无语 一开始没看到

 Hey I think we can write safely in this file without any one seeing it. Anyway, the secret key is: st3g0_saurus_wr3cks... .....

爱因斯坦

image-20241210200545871

binwalk分离得到加密的 zip 文件

image-20241210200714226

image-20241210201800447

解压得到flag

ningen

image-20241210203644671

一张jpg图片

又是压缩包

image-20241210202346915

image-20241210202530767

提示4位密码

image-20241210203316905

image-20241210203343154

小明的保险箱

image-20241210203625862

image-20241210203817756

有东西是 rar压缩包 修改后缀暴力破解

image-20241210212708616

image-20241210212742236

easycap

打开就是flag

image-20241210205212587

隐藏的钥匙

image-20241210205431452

image-20241210205523269

解码得到flag

另外一个世界

image-20241210211315108

一串二进制,转成ascii码,得到flag

image-20241210211746309

数据包中的线索

image-20241210212056197

image-20241210212046039

神秘龙卷风

image-20241210212931320

image-20241210213003559

使用工具 https://www.splitbrain.org/services/ook

image-20241210213535627

FLAG

image-20241210214431985

发现是zip文件解压缩

image-20241210214724821

假如给我三天光明

根据图片上的提示对照盲文找到密码

image-20241211194227588

image-20241211194602020

image-20241211194839931

对应密码为

kmdonowg

得到一个wav文件,摩斯密码

image-20241211195127585

-.-. - ..-. .-- .--. . .. ----- ---.. --... ...-- ..--- ..--.. ..--- ...-- -.. --..

image-20241211195456422

后门查杀

D盾扫一下

image-20241211202659896

webshell后门

image-20241211204311099

image-20241211204657854

来首歌吧

摩斯密码

..... -... -.-. ----. ..--- ..... -.... ....- ----. -.-. -... ----- .---- ---.. ---.. ..-. ..... ..--- . -.... .---- --... -.. --... ----- ----. ..--- ----. .---- ----. .---- -.-.

image-20241211211652603

面具下的flag

image-20241211211817834

image-20241211211942684

压缩包 zip伪加密 修改第一个模块和第二个模块的全局标志位为偶数 解压缩出来flag.vmdk

vmdk文件可以用7z解压

VMWARE虚拟机的配置文件是.VMX格式,而.VMDK 格式则是虚拟机完成安装后生成的虚拟磁盘映像文件

7z x flag.vmdk -o./

image-20241212150039605

解压出来四个文件夹

image-20241212150146901

分别是

image-20241212150304660

解码

image-20241212150533705

image-20241212150503212

flag{N7F5_AD5_i5_funny!}

九连环

image-20241212152109584

jpg zip binwalk分离

伪加密 修改标志位 解压

image-20241212153611338

修改后缀为jpg steghide查看图片隐藏信息

image-20241212154513157

解压缩

image-20241212154558870

[BJDCTF2020]认真你就输了

image-20241212155505837

xls实际上是zip文件

image-20241212155643575

被劫持的神秘礼物

image-20241212160922000

根据提示进行拼接

image-20241212161259824

flag{1d240aafe21a86afc11f38a45b541a49}

被偷走的文件

有个 flag.rar

image-20241212161724497

binwalk分离 爆破出密码

image-20241212162922525

image-20241212162949654

[BJDCTF2020]藏藏藏

image-20241212171623317

010将其保存为zip文件

image-20241212172212968

解压缩得到flag

[GXYCTF2019]佛系青年

解压出一个png图片

010查看时jpg文件 修改后缀

010再查看zip文件 发现是zip伪加密 修改标注位 解压缩

image-20241212172601046

image-20241212173026547

[BJDCTF2020]你猜我是个啥

无法解压缩 010看一下

image-20241212173223870

交一下

刷新过的图片

根据图片的提示

F5隐写,工具F5-steganography

image-20241212203416491

发现是zip文件

但是我打开output.txt是空的 但是cat出来显示是zip 从kali里拖出来也是空的

后面就是伪加密 修改标志位 解压缩 得到flag

秘密文件

分析流量包 看到rar

foremost一下

image-20241212210151507

分离出来rar文件 爆破

image-20241212210053042

image-20241212210240208

[BJDCTF2020]鸡你太美

将副本文件丢到010 发现缺少文件头

添加gif文件头

image-20241212214204264

[BJDCTF2020]just_a_rar

image-20241212214552038

image-20241212214654398

snake

image-20241212214747077

分离文件 解压缩得到key和cipher文件

key解码 得到提示

What is Nicki Minaj's favorite song that refers to snakes?

Nicki Minaj's查有关蛇的歌

image-20241212220116656

snake不只代表蛇 还代表serpent

image-20241212220716518

得到flag

菜刀666

与菜刀有关的一般都是post

http.request.method==POST

image-20241213202216636

导出zip文件 需要密码解压

image-20241213212339492

是一张jpg图片

image-20241213212753826

得到密码

image-20241213213030130

[BJDCTF2020]一叶障目

修改图片宽高

import zlib
import struct

with open(r'C:\Users\jkx\Desktop\1.png','rb') as image_data:
    bin_data = image_data.read()
data = bytearray(bin_data[12:29])     #这段数据就是png图中IHDR段的16进制数据,不包括开始的length和最后CRC
crc32key = eval(str(bin_data[29:33]).replace(r'\x', '').replace("b'", '0x').replace("'", ''))
#理论上0xffffffff,但考虑到屏幕实际,0x0fff就差不多了
n = 4096
#高和宽一起爆破
for w in range(n):
    # q为8字节,i为4字节,h为2字节
    width = bytearray(struct.pack('>i', w))
    for h in range(n):
        height = bytearray(struct.pack('>i', h))
        for x in range(4):
            data[x+4] = width[x]
            data[x+8] = height[x]
        crc32result = zlib.crc32(data)
        if crc32result == crc32key:
            print("width:%s  height:%s" % (bytearray(width).hex(),bytearray(height).hex()))
            exit()

image-20241213215847618

image-20241213220117372

image-20241213220124264

[SWPU2019]神奇的二维码

image-20241213220310313

binwalk分离一下

image-20241214152017273

base64解码doc文件得到密码 解压缩得到mp3

-- --- .-. ... . .. ... ...- . .-. -.-- ...- . .-. -.-- . .- ... -.--

image-20241214152516591

转换为小写就是flag

[BJDCTF2020]纳尼

添加GIF文件头 得到gif

image-20241214155600335

image-20241214155610415image-20241214155611996image-20241214155614030

base64解码flag

CTF{wang_bao_qiang_is_sad}

[HBNIS2018]excel破解

xls丢进010

image-20241214165240229

[HBNIS2018]来题中等的吧

打开图片是一段摩斯密码

.- .-.. .--. .... .- .-.. .- -...

image-20241214165559510

梅花香之苦寒来

将图片里的十六进制保存出来 转换为ascii

with open('hex.txt','r') as h:
    h=h.read()
tem=''
for i in range(0,len(h),2):
    tem='0x'+h[i]+h[i+1]
    tem=int(tem,base=16)
    print(chr(tem),end='')

image-20241218171429672

得到坐标 将其转换成 gnuplot 能识别的形式

with open('ascii.txt','r')as a:
    a=a.read()
a=a.split()
tem=''
for i in range(0,len(a)):
    tem=a[i]
    tem=tem.lstrip('(')
    tem=tem.rstrip(')')
    for j in range(0,len(tem)):
        if tem[j]==',':
            tem=tem[:j]+' '+tem[j+1:]
    print(tem)

gnuplot 绘图

image-20241218172708459

image-20241218172914327

对于大量的十六进制 可以考虑将其转换成坐标 gnuplot 绘画

谁赢了比赛?

strings一下 发现flag.txt

image-20241218173336826

foremost 分离出rar 爆破密码

image-20241218173615570

image-20241218173926390

Ps分析gif图片

image-20241218174208390

保存图片到 stegsolve里

image-20241218220126546

image-20241218220111312

[ACTF新生赛2020]outguess

image-20241218221126068

一开始没在在乎这个备注 社会主义核心价值观编码,绝了

image-20241218221243432

结合题目名字 用工具outguess

image-20241218223051342

-k 表示密码,-r应该是表示解密的意思,1.txt表示把解密后的内容写到1.txt

image-20241218223145903

[WUSTCTF2020]find_me

发现盲文

image-20241219221106098

解码得到flag

image-20241220172430203

穿越时空的思念

音频文件 有一段摩斯电码

image-20241220201154182

第二段

image-20241220201219265

第一段转小写就是flag{}

[GUET-CTF2019]KO

Brainfuck/Ook! Obfuscation/Encoding

直接解码

image-20250303134025448

flag{welcome to CTF}

[ACTF新生赛2020]base64隐写

base64隐写就是 给你一大段字符串,让你找flag

直接上脚本

这里是找了两个脚本,以备不时之需

import base64

def get_base64_diff_value(s1, s2):
    base64chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
    res = 0
    for i in range(min(len(s1), len(s2))):
        if s1[i] != s2[i]:
            res += abs(base64chars.index(s1[i]) - base64chars.index(s2[i]))
    return res

def solve_stego():
    with open('ComeOn!.txt', 'r', encoding='utf-8') as f:
        file_lines = f.readlines()
        bin_str = ''
        for line in file_lines:
            steg_line = line.strip()
            try:
                norm_line = base64.b64encode(base64.b64decode(steg_line)).decode('utf-8').strip()
                diff = get_base64_diff_value(steg_line, norm_line)
                print(diff)
                pads_num = steg_line.count('=')
                if diff:
                    bin_str += bin(diff)[2:].zfill(pads_num * 2)
                else:
                    bin_str += '0' * pads_num * 2
                print(goflag(bin_str))
            except base64.binascii.Error:
                print(f"Error decoding line: {line}")

def goflag(bin_str):
    res_str = ''
    for i in range(0, len(bin_str), 8):
        try:
            res_str += chr(int(bin_str[i:i + 8], 2))
        except ValueError:
            pass
    return res_str

if __name__ == '__main__':
    solve_stego()
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'''
e=d.splitlines()
binstr=""
base64="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
for i in e :
    if i.find("==")>0:
        temp=bin((base64.find(i[-3])&15))[2:]
    #取倒数第3个字符,在base64找到对应的索引数(就是编码数),取低4位,再转换为二进制字符
        binstr=binstr + "0"*(4-len(temp))+temp #二进制字符补高位0后,连接字符到binstr
    elif i.find("=")>0:
        temp=bin((base64.find(i[-2])&3))[2:] #取倒数第2个字符,在base64找到对应的索引数(就是编码数),取低2位,再转换为二进制字符
        binstr=binstr + "0"*(2-len(temp))+temp #二进制字符补高位0后,连接字符到binstr
str=""
for i in range(0,len(binstr),8):
    str=str+chr(int(binstr[i:i+8],2)) #从左到右,每取8位转换为ascii字符,连接字符到字符串
print(str) #结果是 Base_sixty_four_point_five转换为

浅浅补充一下base64隐写解法:

是在正常情况下base64解码过程中,将不影响解码结果的提取出来组成二进制串,再转换成ascii字符串

具体就是

已知:字符串"Rascals"经过 Base64 编码后变为"UmFzY2FSCW==''

解析:UmFzY2Fscw== 首先丢掉末尾的== ,得到UmFzY2Fscw,然后 每个字符查表转换为对应的6位索引

image-20250303151313631

解码过程中,会有一部分数据是会被丢弃的,当然被丢弃的这部分数据(正常编码过程中补位的0)是不会影响正常解码结果的。如果说,我们不用0补位,而是用一些其他的数据补位,而这些数据的位置就用于隐写

[SWPU2019]我有一只马里奥

运行一下给的exe得到一个txt

image-20250303152359115

根据提示查看一下ntfs交换数据流

参考NTFS文件流实践

这篇文章还有流其他的特性

image-20250303153221212

参考

“在NTFS中,主数据流指的是文件或目录的标准内容,通常对用户可见,而备用数据流(ADS)则隐藏”

“ADS没有大小限制且多个数据流可以和一个正常文件关联。ADS的内容也不仅限于text文本数据,基本上只要是二进制格式文件都可以被作为ADS备用流嵌入” (也就是文件后缀是jpg,png,asp,exe的都可以)

”宿住消失寄生消失“

image-20250303153831534

简单来说就是

我们先创建了一个宿住文件为test.txt,内容随意

image-20250303153438713

在创建一个关联数据流文件

打开cmd,在目录下echo

image-20250303153655367

在输入命令 notepad test.txt:okk.txt

image-20250303153749357

[MRCTF2020]ezmisc

image-20250303201019585

image-20250303201041914

[GXYCTF2019]gakki

jpg里藏了个rar

image-20250303202856957

爆破rar密码

image-20250303203234533

解压缩得到一段字符串

本题的解题思路就是统计里面出现的字符的个数,根据出现的次数排列即可得到flag

alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890!@#$%^&*()_+- =\\{\\}[]"
strings = open('flag.txt').read()
 
result = {}
for i in alphabet:
	counts = strings.count(i)
	i = '{0}'.format(i)
	result[i] = counts
 
res = sorted(result.items(),key=lambda item:item[1],reverse=True)
for data in res:
	print(data)
 
for i in res:
	flag = str(i[0])
	print(flag[0],end="")

image-20250303204014694

[HBNIS2018]caesar

image-20250303204549087

凯撒密码 直接上脚本

def encrypt(plaintext):
	# j即为key
    for j in range(26):
        str_list = list(plaintext)
        i = 0
        while i <len(plaintext):
            if not str_list[i].isalpha():
                str_list[i] = str_list[i]
            else:
                a = "A" if str_list[i].isupper() else "a"
                str_list[i] = chr((ord(str_list[i]) - ord(a) + j) % 26 + ord(a))
            i = i + 1
 
        print(''.join(str_list))
 
if __name__ == '__main__':
    plaintext = "gmbhjtdbftbs"
    encrypt(plaintext)

image-20250303204513933

[SUCTF2018]single dog

一张披着jpg后缀的png

image-20250303205100897

有个小问题就是89504E47是png文件头 但是file出来却是jpg

image-20250303210243995

先是binwalk了一下 发现zip 解压缩得到

image-20250303210422589

这么一堆东西

参考文章

image-20250303210802199

说是将末尾的括号去掉 换成toString()

emmmm..

image-20250303211410846

还是在线解密好使..

黑客帝国

一个txt里放着Rar

爆破一下

image-20250303211808872

得到一张不支持png格式的png

看到文件尾是FFD9 修改文件头和后缀

image-20250303213210539

得到flag

image-20250303213031680

[HBNIS2018]低个头

image-20250303213408403

无语...

image-20250303213615701

[SWPU2019]伟大的侦探

010里EBCDIC编码

image-20250303214614046

得到密码wllm_is_the_best_team!

image-20250303214705013

一堆小人..

image-20250303214800970

image-20250303215246071

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