Hgame 2023
[HGAME 2023 week1]test your IDA
用ida打开即可
[HGAME 2023 week1]encode
查壳
32位windows
加密函数
将输入的字节转高位和低位进行加密，后与byte_403000进行比较
解密脚本：
这里采取了爆破
enc = [8, 6, 7, 6, 1, 6, 13, 6, 5, 6, 11, 7, 5, 6, 14, 6, 3, 6, 15, 6, 4, 6, 5, 6, 15, 5, 9, 6, 3, 7, 15, 5, 5, 6, 1, 6, 3, 7, 9, 7, 15, 5, 6, 6, 15, 6, 2, 7, 15, 5, 1, 6, 15, 5, 2, 7, 5, 6, 6, 7, 5, 6, 2, 7, 3, 7, 5, 6, 15, 5, 5, 6, 14, 6, 7, 6, 9, 6, 14, 6, 5, 6, 5, 6, 2, 7, 13, 7]
flag = ''
for j in range(0, len(enc), 2):for i in range(32, 126):if i & 0xF == enc[j] and (i >> 4) & 0xF == enc[j+1]:flag += chr(i)
print(flag)
[HGAME 2023 week2]before_main
查壳：
64位linux无壳
找到主加密程序
大概是个base64加密
点开sub_12EB
直接点开里面的table表查看，按a换成字符串
可以发现是替换了table表
但拿去解密是错的
那应该就有可能是base64加密本身逻辑不同，或者有别的table表
观察加密函数，没有问题，那应该就是有别的table表了
交叉引用a0cxwsoemvjq4zd
可以看到确实有别的引用了这个table表
跟进
便可以看到别的table表，拿去解密即可
从网上抄的映射脚本：
import base64 # 导入base64模块用于解密s1 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/' # 标准表
s2 = 'qaCpwYM2tO/RP0XeSZv8kLd6nfA7UHJ1No4gF5zr3VsBQbl9juhEGymc+WTxIiDK' # base64换表
en_text = 'AMHo7dLxUEabf6Z3PdWr6cOy75i4fdfeUzL17kaV7rG=' # 密文map_text = '' # 用于存放密文通过换表映射回标准表的字符
for i in en_text:if (i != '='): # 注意密文中存在等号的情况下，不需要替换！idx = s2.index(i) # 获取每个密文的字符在换表中的索引map_text += s1[idx] # 取出标准表中的该索引的字符，就是正常base64加密的密文else:map_text += i
print(map_text) # 可以先看看标准表base64加密的密文
print(base64.b64decode(map_text)) # 直接使用提供的base64解密函数，获得明文，就是flag
疑问
为什么会是调用这个table表，可以看到程序运行的时候同步运行了init函数，我们跟进init函数即可找到刚刚的table表
[HGAME 2023 week1]easyasm
打开附件是一个txt文本，里面储存了一个用汇编代码写的加密程序
可以看到这里有个异或和加密程序
其实可以直接猜是异或0x33
enc = [0x5b,0x54,0x52,0x5e,0x56,0x48,0x44,0x56,0x5f,0x50,0x3,0x5e,0x56,0x6c,0x47,0x3,0x6c,0x41,0x56,0x6c,0x44,0x5c,0x41,0x2,0x57,0x12,0x4e]
flag = ''
for i in enc:flag += chr(i ^ 0x33)
print(flag)
得到flag
但还是得学着看汇编的hh
我自己的理解：
; void __cdecl enc(char *p)
.text:00401160 _enc proc near ; CODE XREF: _main+1B↑p //定义函数enc
.text:00401160
.text:00401160 i = dword ptr -4 //定义变量
.text:00401160 Str = dword ptr 8
.text:00401160
.text:00401160 push ebp
.text:00401161 mov ebp, esp //将基址指针ebp
.text:00401163 push ecx
.text:00401164 mov [ebp+i], 0
.text:0040116B jmp short loc_401176 //跳转到loc_401176 开始循环
.text:0040116D ; ---------------------------------------------------------------------------
.text:0040116D
.text:0040116D loc_40116D: ; CODE XREF: _enc+3B↓j
.text:0040116D mov eax, [ebp+i] //将i加载到eax中
.text:00401170 add eax, 1 //将i值加1
.text:00401173 mov [ebp+i], eax //更新i的值
.text:00401176
.text:00401176 loc_401176: ; CODE XREF: _enc+B↑j
.text:00401176 mov ecx, [ebp+Str] //将str移动给ecx
.text:00401179 push ecx ; Str //将str压栈
.text:0040117A call _strlen //用strlen获取刚刚压栈的str长度
.text:0040117F add esp, 4
.text:00401182 cmp [ebp+i], eax //比较i 与 eax大小
.text:00401185 jge short loc_40119D //如果i大于eax则跳转loc_40119D
.text:00401187 mov edx, [ebp+Str] //将str地址赋值edx
.text:0040118A add edx, [ebp+i] //将str的地址+i的地址，即可实现对str数组元素的获取
.text:0040118D movsx eax, byte ptr [edx] //将edx地址中的元素转byte赋值到eax寄存器
.text:00401190 xor eax, 33h //将eax与0x33异或，将异或结果储存到了eax寄存器中的低地址al中，他只是对eax中的内存进行操作
.text:00401193 mov ecx, [ebp+Str] //将异或后的值赋值给ecx
.text:00401196 add ecx, [ebp+i] //再次加载str的第i个数
.text:00401199 mov [ecx], al //将加密后的结果移动到ecx中
.text:0040119B jmp short loc_40116D //跳转到loc_40116D函数
.text:0040119D ; ---------------------------------------------------------------------------
.text:0040119D
.text:0040119D loc_40119D: ; CODE XREF: _enc+25↑j
.text:0040119D mov esp, ebp //回复栈指针
.text:0040119F pop ebp //恢复基址指针
.text:004011A0 retn //结束函数
.text:004011A0 _enc endp
Input: your flag
Encrypted result: 0x5b,0x54,0x52,0x5e,0x56,0x48,0x44,0x56,0x5f,0x50,0x3,0x5e,0x56,0x6c,0x47,0x3,0x6c,0x41,0x56,0x6c,0x44,0x5c,0x41,0x2,0x57,0x12,0x4e
[HGAME 2023 week1]easyenc
die查壳
64位win
定位主程序
对v7数组以byte类型进行操作
编写脚本
#include<stdio.h>
void main() {int v8[10], v3, v5;v8[0] = 167640836;v8[1] = 11596545;v8[2] = -1376779008;v8[3] = 85394951;v8[4] = 402462699;v8[5] = 32375274;v8[6] = -100290070;v8[7] = -1407778552;v8[8] = -34995732;v8[9] = 101123568;for (int i = 0; i < 41; i++){v5 = (*((char*)v8 + i) + 86) ^ 0x32;printf("%c", v5);}
}
[HGAME 2023 week1]a_cup_of_tea
查壳
64win无壳
main函数
可以看到是接受参数Buf1和v10来加密然后与buf2进行比较
没看到哪里是v10的输入，但v10和buf1的地址相近，所以v10应该也是buf1的输入
tea加密
魔改delta的tea加密，进行四轮
返回看函数接收，接受了一个m128i_i32地址
smmword表示一个十六字节字符串
小端序储存，因此可以推测这就是tea加密的key，当然也可以用动调来看tea加密中变量的赋值
编写脚本
#include <iostream>
using namespace std;
int main()
{unsigned int enc[8] = { 778273437,-1051836401, -1690714183,1512016660,1636330974,1701168847,-1626976412,0x236A43F6 };unsigned int key[4] = { 0x12345678, 0x23456789, 0x34567890, 0x45678901 };int i, j;for (i = 0;i < 8;i += 2) {unsigned int delta = 0x543210DD;int sum = -0x543210DD * 32;for (j = 0; j < 32; j++) {enc[i + 1] -= ((enc[i] << 4) + key[2]) ^ (enc[i] + sum) ^ ((enc[i] >> 5) + key[3]);enc[i] -= ((enc[i + 1] << 4) + key[0]) ^ (enc[i + 1] + sum) ^ ((enc[i + 1] >> 5) + key[1]);sum += delta;}}printf("%s", enc);return 0;
}
用python的ctypes库
from ctypes import *
from binascii import *
from Crypto.Util import *
key = [0x12345678, 0x23456789, 0x34567890, 0x45678901]
enc = [0x2E63829D, 0xC14E400F, 0x9B39BFB9, 0x5A1F8B14, 0x61886DDE, 0x6565C6CF, 0x9F064F64, 0x236A43F6]
for i in range(0, len(enc), 2):v0 = c_uint32(enc[i])v1 = c_uint32(enc[i+1])delta = 0x543210DDsum = c_uint32(-delta * 32)r = 32for j in range(r):v1.value -= (sum.value + v0.value) ^ (key[2] + 16 * v0.value) ^ (key[3] + (v0.value >> 5))v0.value -= (sum.value + v1.value) ^ (key[0] + 16 * v1.value) ^ (key[1] + (v1.value >> 5))sum.value += deltaenc[i] = v0.valueenc[i+1] = v1.valueprint(a2b_hex(hex(enc[i])[2:].encode()).decode()[::-1],end='')print(a2b_hex(hex(enc[i+1])[2:].encode()).decode()[::-1], end='')# print(number.long_to_bytes(enc[i]).decode()[::-1],end="")# print(number.long_to_bytes(enc[i+1]).decode()[::-1],end="")
#也可以用Crypto.Util库的long_to_bytes打印
[HGAME 2023 week2]stream
python编译的程序，使用pyinstxtractor-master来拆包
stream.pyc反编译
import base64def gen(key):s = list(range(256))j = 0for i in range(256):j = (j + s[i] + ord(key[i % len(key)])) % 256tmp = s[i]s[i] = s[j]s[j] = tmpi = j = 0data = []for _ in range(50):i = (i + 1) % 256j = (j + s[i]) % 256tmp = s[i]s[i] = s[j]s[j] = tmpdata.append(s[(s[i] + s[j]) % 256])return datadef encrypt(text, key):result = ''for c, k in zip(text, gen(key)):result += chr(ord(c) ^ k)result = base64.b64encode(result.encode()).decode()return resulttext = input('Flag: ')
key = 'As_we_do_as_you_know'
enc = encrypt(text, key)
if enc == 'wr3ClVcSw7nCmMOcHcKgacOtMkvDjxZ6asKWw4nChMK8IsK7KMOOasOrdgbDlx3DqcKqwr0hw701Ly57w63CtcOl':print('yes!')return None
None('try again...')
很简单的python程序写的异或加密
首先在加密函数中把k打印出来
再编写脚本
解密脚本：
from base64 import *
key = [213, 242, 54, 127, 156, 227, 172, 100, 212, 1, 130, 92, 20, 189, 115, 12, 15, 228, 186, 225, 227, 75, 200, 119, 171, 11, 152, 15, 89, 160, 116, 157, 194, 226, 72, 147, 65, 74, 92, 21, 136, 193, 152, 94, 17, 178, 205, 195, 87, 145]
enc = 'wr3ClVcSw7nCmMOcHcKgacOtMkvDjxZ6asKWw4nChMK8IsK7KMOOasOrdgbDlx3DqcKqwr0hw701Ly57w63CtcOl'
tmp = b64decode(enc.encode()).decode()
flag = ''
for i in range(len(tmp)):flag += chr(ord(tmp[i]) ^ key[i])
print(flag)[HGAME 2023 week2]math
查壳：
64位linux无壳
分析
是一个简单的矩阵相乘方程
from z3 import *
v12 = [63998, 33111, 67762, 54789, 61979, 69619, 37190, 70162, 53110, 68678, 63339, 30687, 66494, 50936, 60810, 48784, 30188, 60104, 44599, 52265, 43048, 23660, 43850, 33646, 44270]
v10 = [126, 225, 62, 40, 216, 253, 20, 124, 232, 122, 62, 23, 100, 161, 36, 118, 21, 184, 26, 142, 59, 31, 186, 82, 79]
x = [Int(f"x{i}") for i in range(25)]
s = Solver()
for i in range(5):for j in range(5):tmp = 0for k in range(5):tmp += x[5 * i + k] * v10[5 * k + j]s.add(v12[5 * i + j] == tmp)
if s.check() == sat:result = s.model()print(result)
[HGAME 2023 week3]kunmusic
用dnspy打开dll文件
定位关键函数
这里看入口点
在资源处保存文件
转换文档脚本
with open(r'D:\CTF\Reverse\Hgame\Hgame2023\[HGAME 2023 week3]kunmusic\kmusic\data', 'rb') as file:str = file.read()
result = [0] * len(str)
for i in range(len(str)):result[i] = (str[i] ^ 104)
output = bytes(result)
with open(r'D:\CTF\Reverse\Hgame\Hgame2023\[HGAME 2023 week3]kunmusic\kmusic\output.txt', 'wb') as file:file.write(output)
丢入dnspy反编译
找到关键加密函数
z3求解
from z3 import *
s = Solver()
num = [BitVec(f'num{i}',8) for i in range(13)]
array = [132, 47, 180, 7, 216, 45, 68, 6, 39, 246, 124, 2, 243, 137, 58, 172, 53, 200, 99, 91, 83, 13, 171, 80, 108, 235, 179, 58, 176, 28, 216, 36, 11, 80, 39, 162, 97, 58, 236, 130, 123, 176, 24, 212, 56, 89, 72]s.add(num[0] + 52296 + num[1] - 26211 + num[2] - 11754 + (num[3] ^ 0xA114) + num[4] * 63747 + num[5] - 52714 + num[6] - 10512 + num[7] * 12972 + num[8] + 45505 + num[9] - 21713 + num[10] - 59122 + num[11] - 12840 + (num[12] ^ 0x525F) == 12702282 )
s.add( num[0] - 25228 + (num[1] ^ 0x50DB) + (num[2] ^ 0x1FDE) + num[3] - 65307 + num[4] * 30701 + num[5] * 47555 + num[6] - 2557 + (num[7] ^ 0xBF9F) + num[8] - 7992 + (num[9] ^ 0xE079) + (num[10] ^ 0xE052) + num[11] + 13299 + num[12] - 50966 == 9946829 )
s.add( num[0] - 64801 + num[1] - 60698 + num[2] - 40853 + num[3] - 54907 + num[4] + 29882 + (num[5] ^ 0x3506) + (num[6] ^ 0x533E) + num[7] + 47366 + num[8] + 41784 + (num[9] ^ 0xD1BA) + num[10] * 58436 + num[11] * 15590 + num[12] + 58225 == 2372055 )
s.add( num[0] + 61538 + num[1] - 17121 + num[2] - 58124 + num[3] + 8186 + num[4] + 21253 + num[5] - 38524 + num[6] - 48323 + num[7] - 20556 + num[8] * 56056 + num[9] + 18568 + num[10] + 12995 + (num[11] ^ 0x995C) + num[12] + 25329 == 6732474 )
s.add( num[0] - 42567 + num[1] - 17743 + num[2] * 47827 + num[3] - 10246 + (num[4] ^ 0x3F9C) + num[5] + 39390 + num[6] * 11803 + num[7] * 60332 + (num[8] ^ 0x483B) + (num[9] ^ 0x12BB) + num[10] - 25636 + num[11] - 16780 + num[12] - 62345 == 14020739 )
s.add( num[0] - 10968 + num[1] - 31780 + (num[2] ^ 0x7C71) + num[3] - 61983 + num[4] * 31048 + num[5] * 20189 + num[6] + 12337 + num[7] * 25945 + (num[8] ^ 0x1B98) + num[9] - 25369 + num[10] - 54893 + num[11] * 59949 + (num[12] ^ 0x3099) == 14434062 )
s.add( num[0] + 16689 + num[1] - 10279 + num[2] - 32918 + num[3] - 57155 + num[4] * 26571 + num[5] * 15086 + (num[6] ^ 0x59CA) + (num[7] ^ 0x5B35) + (num[8] ^ 0x3FFD) + (num[9] ^ 0x5A85) + num[10] - 40224 + num[11] + 31751 + num[12] * 8421 == 7433598 )
s.add( num[0] + 28740 + num[1] - 64696 + num[2] + 60470 + num[3] - 14752 + (num[4] ^ 0x507) + (num[5] ^ 0x89C8) + num[6] + 49467 + num[7] - 33788 + num[8] + 20606 + (num[9] ^ 0xAF4A) + num[10] * 19764 + num[11] + 48342 + num[12] * 56511 == 7989404 )
s.add( (num[0] ^ 0x7132) + num[1] + 23120 + num[2] + 22802 + num[3] * 31533 + (num[4] ^ 0x9977) + num[5] - 48576 + (num[6] ^ 0x6F7E) + num[7] - 43265 + num[8] + 22365 + num[9] + 61108 + num[10] * 2823 + num[11] - 30343 + num[12] + 14780 == 3504803 )
s.add( num[0] * 22466 + (num[1] ^ 0xDABF) + num[2] - 53658 + (num[3] ^ 0xB838) + (num[4] ^ 0x30DF) + num[5] * 59807 + num[6] + 46242 + num[7] + 3052 + (num[8] ^ 0x62BF) + num[9] + 30202 + num[10] * 22698 + num[11] + 33480 + (num[12] ^ 0x4175) == 11003580 )
s.add( num[0] * 57492 + (num[1] ^ 0x346D) + num[2] - 13941 + (num[3] ^ 0xBBDC) + num[4] * 38310 + num[5] + 9884 + num[6] - 45500 + num[7] - 19233 + num[8] + 58274 + num[9] + 36175 + (num[10] ^ 0x4888) + num[11] * 49694 + (num[12] ^ 0x2501) == 25546210 )
s.add( num[0] - 23355 + num[1] * 50164 + (num[2] ^ 0x873A) + num[3] + 52703 + num[4] + 36245 + num[5] * 46648 + (num[6] ^ 0x12FA) + (num[7] ^ 0xA376) + num[8] * 27122 + (num[9] ^ 0xA44A) + num[10] * 15676 + num[11] - 31863 + num[12] + 62510 == 11333836 )
s.add( num[0] * 30523 + (num[1] ^ 0x1F36) + num[2] + 39058 + num[3] * 57549 + (num[4] ^ 0xD0C0) + num[5] * 4275 + num[6] - 48863 + (num[7] ^ 0xD88C) + (num[8] ^ 0xA40) + (num[9] ^ 0x3554) + num[10] + 62231 + num[11] + 19456 + num[12] - 13195 == 13863722)
s.add(num[0]==ord('h')^array[0])
s.add(num[1]==ord('g')^array[1])
s.add(num[2]==ord('a')^array[2])
s.add(num[3]==ord('m')^array[3])
s.add(num[4]==ord('e')^array[4])
s.add(num[5]==ord('{')^array[5]) #z3多解，猜测求解
if s.check() == sat:result = s.model()
num_values = [result[num[i]].as_long() for i in range(13)]flag = ''
for i in range(len(array)):flag += chr((array[i] ^ num_values[i % len(num_values)] ) & 0x7f )
print(flag)
[HGAME 2023 week2]VidarCamera
找到主加密程序
长度40的字节
uIntArr是密文
查看 m8encrypthkIa6DI函数
我改了一下变量名称（有一说一java太丑了）
可以看到这里有个很明显的xtea加密
修改了delta，round，加密过程
(func1和func2点进去查看是什么意思，其实就是简单的索引)
解密函数：
from ctypes import *
from Crypto.Util import *
enc = [0x260202FA, 0x1B451064, 0x867B61F1, 0x228033C5, 0xF15D82DC, 0x9D8430B1, 0x19F2B1E7, 0x2BBA859C, 0x2A08291D, 0xDC707918]
key = [2233, 4455, 6677, 8899]
for i in range(len(enc) - 2, -1, -1):v0 = c_uint32(enc[i])v1 = c_uint32(enc[i+1])delta = 878077251r = 33sum = c_uint32(r * delta)for j in range(r):sum.value -= deltav1.value -= (key[(sum.value >> 11) & 3] + sum.value) ^ (v0.value + ((v0.value >> 5) ^ (v0.value << 4)))v0.value -= (key[sum.value & 3] + sum.value) ^ (v1.value + ((v1.value >> 5) ^ (v1.value << 4))) ^ sum.valueenc[i] = v0.valueenc[i+1] = v1.value
for i in range(10):print(number.long_to_bytes(enc[i]).decode()[::-1], end='')
[HGAME 2023 week4]vm
点进主要加密函数
这里就是虚拟机的opcode，a1就是ip
导入结构体(根据sub_140001000函数)
struct vm
{_DWORD r[6];_DWORD ip;_DWORD esp;_BYTE zf;
};
分析vm指令
mov
有赋值操作，即mov
push
点进stack，即可看到一堆未定义的空间
将寄存器的值赋值给这些空间，即可知道这是模拟压栈操作
pop
与上面的类似，将stack中的值赋值给寄存器，模拟出栈操作(将上面的函数结合其实很容易看出来)
运算
进行了一系列运算，分别是add, sub, mul, xor, shl, shr
cmp
两个if值的比较，如果比较成功则zf为0不成功则为1，为后面的jz和jnz服务的
jmp
跳转到下一条opcode指令处，将opcode下一条指令赋值给ip
jz
JZ：jump if zero
如果if为假（即zf为0），则执行下一条指令
jnz
JNZ：jump if not zero
如果if为真（即zf不为0），则执行下一条指令
编写脚本
opcode = [ 0x00, 0x03, 0x02, 0x00, 0x03, 0x00, 0x02, 0x03, 0x00, 0x00,0x00, 0x00, 0x00, 0x02, 0x01, 0x00, 0x00, 0x03, 0x02, 0x32,0x03, 0x00, 0x02, 0x03, 0x00, 0x00, 0x00, 0x00, 0x03, 0x00,0x01, 0x00, 0x00, 0x03, 0x02, 0x64, 0x03, 0x00, 0x02, 0x03,0x00, 0x00, 0x00, 0x00, 0x03, 0x03, 0x01, 0x00, 0x00, 0x03,0x00, 0x08, 0x00, 0x02, 0x02, 0x01, 0x03, 0x04, 0x01, 0x00,0x03, 0x05, 0x02, 0x00, 0x03, 0x00, 0x01, 0x02, 0x00, 0x02,0x00, 0x01, 0x01, 0x00, 0x00, 0x03, 0x00, 0x01, 0x03, 0x00,0x03, 0x00, 0x00, 0x02, 0x00, 0x03, 0x00, 0x03, 0x01, 0x28,0x04, 0x06, 0x5F, 0x05, 0x00, 0x00, 0x03, 0x03, 0x00, 0x02,0x01, 0x00, 0x03, 0x02, 0x96, 0x03, 0x00, 0x02, 0x03, 0x00,0x00, 0x00, 0x00, 0x04, 0x07, 0x88, 0x00, 0x03, 0x00, 0x01,0x03, 0x00, 0x03, 0x00, 0x00, 0x02, 0x00, 0x03, 0x00, 0x03,0x01, 0x28, 0x04, 0x07, 0x63, 0xFF, 0xFF]
i = 0
for j in range(10000):if opcode[i] == 0xff:breakmatch opcode[i]:case 0: #movprint(f'({i})', end=' ') #地址match (opcode[i + 1]):case 1:print('mov flag[r[2]], r[1]')case 2:print(f'mov r[{opcode[i + 2]}], r[{opcode[i + 3]}]')case 3:print(f'mov r[{opcode[i + 2]}], {opcode[i + 3]}')case _:print('mov r[0]， flag[r[2]] ')i += 4case 1: #pushprint(f'({i})', end=' ')match (opcode[i + 1]):case 1:print('push r[0]')case 2:print('push r[2]')case 3:print('case r[3]')case _:print('push r[0]')i += 2case 2: #popprint(f'({i})', end=' ')match (opcode[i + 1]):case 1:print('pop r[1]')case 2:print('pop r[2]')case 3:print('pop r[3]')case _:print('pop r[0]')i += 2case 3: #运算print(f'({i})', end=' ')match (opcode[i + 1]):case 0:print(f'add r{[opcode[i + 2]]} r{[opcode[i + 3]]}')case 1:print(f'sub r{[opcode[i + 2]]} r{[opcode[i + 3]]}')case 2:print(f'mul r{[opcode[i + 2]]} r{[opcode[i + 3]]}')case 3:print(f'xor r{[opcode[i + 2]]} r{[opcode[i + 3]]}')case 4:print(f'shl r{[opcode[i + 2]]} r{[opcode[i + 3]]}')case 5:print(f'shr r{[opcode[i + 2]]} r{[opcode[i + 3]]}')i += 4case 4: #cmpprint(f'({i})', end=' ')print('cmp r[0] r[1]')i += 1case 5: #jmpprint(f'({i})', end=' ')print(f'jmp {opcode[i + 1]}')i += 2case 6: #jzprint(f'({i})', end=' ')print(f'jz {opcode[i + 1]}')i += 2case 7: #jnzprint(f'({i})', end=' ')print(f'jnz {opcode[i + 1]}')i += 2
打印出来的汇编指令（我给了一些注释）
(0) mov r[2], 0
(4) add r[2] r[3] //r[3]就是i
(8) mov r[0]， flag[r[2]] //这里的flag[r[2]]就是输入的input
(12) mov r[1], r[0]
(16) mov r[2], 50
(20) add r[2] r[3]
(24) mov r[0]， flag[r[2]] //将flag内存处偏移50 + i的值存入r[0]
(28) add r[1] r[0] //将input 与 偏移50+i的值相加
(32) mov r[2], 100
(36) add r[2] r[3]
(40) mov r[0]， flag[r[2]] //将flag内存处偏移100 + i的值存入r[0]
(44) xor r[1] r[0] //将input 与 偏移100+i的值异或，将值存入v[1]
(48) mov r[0], 8 //赋值8给r[0]
(52) mov r[2], r[1]
(56) shl r[1] r[0]
(60) shr r[2] r[0]
(64) add r[1] r[2] //将异或后的值左移八位右移八位相加存入r[1]（也就是低八位和高八位换位）
(68) mov r[0], r[1] //将运算结果保存在r[0]中
(72) push r[0] //将运算结果压入栈中
(74) mov r[0], 1
(78) add r[3] r[0]
(82) mov r[0], r[3]
(86) mov r[1], 40
(90) cmp r[0] r[1] //判断循环结束
(91) jz 95
(93) jmp 0
(95) mov r[3], 0
(99) pop r[1] //将结果出栈保存到r[1]中,所以最后一个结果先与第一个比较
(101) mov r[2], 150
(105) add r[2] r[3]
(109) mov r[0]， flag[r[2]]
(113) cmp r[0] r[1] //将运算结果与偏移150位的数据比较
(114) jnz 136
(116) mov r[0], 1
(120) add r[3] r[0]
(124) mov r[0], r[3]
(128) mov r[1], 40
(132) cmp r[0] r[1] //继续循环
(133) jnz 99
将输入与input内存偏移50处的值相加，后与偏移100的值异或，最后与偏移150处的值逆序（出栈压栈操作导致的）比较
编写解密脚本
enc = [18432, 61696, 16384, 8448, 13569, 25600, 30721, 63744, 6145, 20992, 9472, 23809, 18176, 64768, 26881, 23552, 44801, 45568, 60417, 20993, 20225, 6657, 20480, 34049, 52480, 8960, 63488, 3072, 52992, 15617, 17665, 33280, 53761, 10497, 54529, 1537, 41473, 56832, 42497, 51713]
enc = enc[::-1]
key2 =[155, 168, 2, 188, 172, 156, 206, 250, 2, 185, 255, 58, 116, 72, 25, 105, 232, 3, 203, 201, 255, 252, 128, 214, 141, 215, 114, 0, 167, 29, 61, 153, 136, 153, 191, 232, 150, 46, 93, 87]
key1 = [201, 169, 189, 139, 23, 194, 110, 248, 245, 110, 99, 99, 213, 70, 93, 22, 152, 56, 48, 115, 56, 193, 94, 237, 176,41, 90, 24, 64, 167, 253, 10, 30, 120, 139, 98, 219, 15, 143, 156]
flag = ''
for i in range(len(enc)):tmp1 = (enc[i] & 0xFF << 8) + ((enc[i] >> 8) & 0xFF)tmp2 = tmp1 ^ key1[i]tmp3 = tmp2 - key2[i]flag += chr(tmp3 & 0xff)
print(flag)
flag : hgame{y0ur_rever5e_sk1ll_i5_very_g0od!!}
高低位互换操作：
b = 18432
low_byte = b & 0xFF # 低八位 (masking with 0xFF)
high_byte = (b >> 8) & 0xFF # 高八位 (shift right and mask with 0xFF)
swapped_value = (low_byte << 8) + high_byte
[HGAME 2023 week3]cpp
全靠大佬wp成就TAT
进入丑的一b的main函数
甚至都没翻到哪里对input进行加密了
那就先动调
下断点动调
既然都不知道input都有多长，那就先输入一个比较长的数
一直f7步入
进入到一个不知道干啥的函数
跳出去继续f7
这次进入了一个疑似加密的函数，是一个异或函数
跳到异或部分
这里就是把eax和ecx进行异或，eax应该就是加密值，但我不知道为什么
使用ida脚本打印ecx的值
from idaapi import *
print(hex(get_reg_val("ecx")),end = ',')
0x4037a04e,0xfdda0246,0x3c6efa21,0xcf9cd9af,0x673347b9,0xdec4ee0,0x1380c4d1,0x3ab2a932,0x25d50a7,0x834a3982,0xcb6ea25f,0xa26ba4ab,0xa1c42135,0xd1063eba,0x2397fefc,0x55c7d126,0xabababab
进入比较函数
还有两位在汇编界面看
这就是一个比较函数
可以看到enc长度是40，那key应该也是40
编写汇编脚本
from struct import *
key = [0x4037a04e,0xfdda0246,0x3c6efa21,0xcf9cd9af,0x673347b9,0xdec4ee0,0x1380c4d1,0x3ab2a932,0x25d50a7,0x834a3982]
enc = [0x28, 0x50, 0xC1, 0x23, 0x98, 0xA1, 0x41, 0x36, 0x4C, 0x31, 0xCB, 0x52, 0x90, 0xF1, 0xAC, 0xCC, 0xF, 0x6C, 0x2A, 0x89, 0x7F, 0xDF, 0x11, 0x84, 0x7F, 0xE6, 0xA2, 0xE0, 0x59, 0xC7, 0xC5, 0x46, 0x5D, 0x29, 0x38, 0x93, 0xED, 0x15, 0x7A, 0xFF,]
key_LE = []
flag = []
for i in key:key_LE.extend((pack('>I', i)))
for i in range(len(enc)):print(chr(key_LE[i] ^ enc[i]), end='')
貌似是需要把enc进行小端序转换的，我这边直接把key大端序打印了，方便点
hgame{Cpp_1s_much_m0r3_dlff1cult_th4n_C}
只能说不懂的多的一b，哎哎，看wp也没看懂
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