【阿尼亚不会CTF】第十五届CISCN创新实践能力赛—线上初赛部分writeup

爱波吉的阿尼亚

已于 2022-06-02 10:53:39 修改

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分类专栏：阿尼亚的CTF之旅文章标签： python go 网络安全

于 2022-05-30 20:36:06 首次发布

本文链接：https://blog.youkuaiyun.com/qq_45943172/article/details/125052554

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ez_usb

基于挑战码的双向认证、基于挑战码的双向认证2：

基于挑战码的双向认证3

ISO9798

ez_usb

题目描述：

简单的流量。

解题思路：

下载附件，解压得到一个流量包，如题，是USB流量分析。

观察发现主要有2.4.1、2.8.1、2.10.1三个Source的键盘输入，分别进行筛选并导出；

然后利用kali自带的tshark工具提取对应的内容；

tshark -T json -r 241.pcapng > 241.json
tshark -T json -r 281.pcapng > 2101.json
tshark -T json -r 2101.pcapng > 281.json
//用法 tshark -T json -r 数据包名称 > 要导出的文件

得到的文件json文件用Notepad++打开，其中的usb.capdata标黄部分对应的就是键盘操作；

然后通过编写Python脚本来对应键盘操作的具体内容；

#!/usr/bin/env python
# -*- coding:utf-8 -*-

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('usbdata.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))

usb.src == "2.4.1"没有发现

usb.src == "2.8.1"得到以下内容：

526172211a0700<CAP>c<CAP>f907300000d00000000000000c4527424943500300000002<CAP>a000000<CAP>02b9f9b0530778b5541d33080020000000666c61672<CAP>e<CAP>747874<CAP>b9b<CAP>a013242f3a<CAP>fc<CAP>000b092c229d6e994167c05<CAP>a7<CAP>8708b271f<CAP>fc<CAP>042ae3d251e65536<CAP>f9a<CAP>da87c77406b67d0<CAP>e6316684766<CAP>a86e844d<CAP>c81aa2<CAP>c72c71348d10c4<CAP>c<DEL>3d7b<CAP>00400700

其中，<CAP>是大写锁定，<DEL>是删除键，手工打一遍就行；

本来以为是编码的，尝试解密失败了，然后再看了看，哦！好像是rar文件头诶！？上Winhex！

保存并解压，发现有密码，1-6位爆破一下，发现不行，看来得找密码了；

usb.src == "2.10.1"得到以下内容：

35c535765e50074a

这个虽然看着不太像，但是试了试，是密码！

成功得到flag!!!

基于挑战码的双向认证、基于挑战码的双向认证2：

题目描述：

本题含有两个flag，请点击“下发赛题”，本题容器下发后的端口是ssh端口，ssh的账号密码均为：player；ssh登录上去可自行修改密码。请参考说明文档，获取flag，并在本题提交第一个获取到的flag。

请提交“基于挑战码的双向认证”题目容器内的第二个flag。

解题思路：

首先，本题解题关键——谢谢出题师傅！谢谢出题师傅！！谢谢出题师傅！！！

ssh连接，习惯性先……尝试……搜索一下……好了，两题flag到手，晚安！！

grep -r "flag{" /

基于挑战码的双向认证3

题目描述：

请点击“下发赛题”，本题容器下发后的端口是ssh端口，ssh的账号密码均为：player；ssh登录上去可自行修改密码。请参考第1题的说明文档，获取flag，并在本题提交flag。

解题思路：

老样子，本题解题关键依旧是——谢谢出题师傅！谢谢出题师傅！！谢谢出题师傅！！！

ssh连接，再来习惯性先……尝试……搜索一下……阿巴阿巴……Permission denied来了！

小问题，不要慌，弱口令trytry！！！芜湖，root密码是toor！好耶！

然后嘛，就是熟悉的……尝试……搜索一下……好了，又一个flag到手，晚安啦！！

find / -name flag2.txt

ISO9798

题目描述：

听说 ISO9798-2 很安全，但实际上是这样的吗？

解题思路：

感谢Crypto师傅大发慈悲，有生之年做出这么多Crypto，真是太开心了好叭！！！

下发赛题，给的nc连接，要我给他算个SHA256头部的四位；

正好最近看了看Go语言，发现一个github上大佬的Go语言SHA-256爆破脚本；

链接：https://github.com/hydewww/sha256-go

package main

import (
	"bytes"
	"crypto/sha256"
	"encoding/hex"
	"fmt"
	"runtime"
	"sync"
	"time"
)

var (
	chars     = []byte("ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz1234567890")
	tail      = []byte("YH864OHM84wAhpLp")
	result, _ = hex.DecodeString("96dd533dd79b6a14007dc3e2968080463a92092da555a37c67b1f17d8004b760")
	wg        sync.WaitGroup
)

func sha(s []byte) {
	for _, ch1 := range s {
		for _, ch2 := range chars {
			for _, ch3 := range chars {
				for _, ch4 := range chars {
					head := []byte{ch1, ch2, ch3, ch4}
					h := sha256.New()
					h.Write(head)
					h.Write(tail)
					if bytes.Equal(h.Sum(nil), result) {
						fmt.Println(string(head))
					}
				}
			}
		}
	}
	wg.Done()
}

func main() {
	threads := runtime.NumCPU() // 获取cpu逻辑核心数（包括超线程）
	start := time.Now()

	/* len(chars) = sum * sthreads + (sum+1) * (threads-sthreads) */
	snum := len(chars) / threads
	sthreads := threads*(1+snum) - len(chars)

	wg.Add(threads)
	for i := 0; i < threads; i++ {
		if i < sthreads {
			go sha(chars[snum*i : snum*(i+1)])
		} else {
			base := snum * sthreads
			go sha(chars[base+(snum+1)*(i-sthreads) : base+(snum+1)*(i-sthreads+1)])
		}
	}
	wg.Wait()
	end := time.Since(start)
	fmt.Println(end)
}

爆破完，发给他，又给我提了个问题，阿巴阿巴……