一道CTF的逆运算
本帖最后由 小朋友呢 于 2020-6-7 13:54 编辑已经逆出来了算法和验证key，但是算法逆运算过程，实在是太难没有搞出来，来求助下大佬们
#include "stdafx.h"
#include <string.h>
#include <stdlib.h>
#include <Windows.h>
char key = {
0x16, 0xB6, 0xEC, 0x52, 0xA1, 0x5D, 0x9C, 0xA6, 0x0C, 0x44, 0x8C, 0xE5, 0xE1, 0xA5, 0x02, 0x50,
0x00, 0xBB, 0x74, 0x4B, 0x40, 0x70, 0x22, 0x35, 0x96, 0xA3, 0x0B, 0x2A, 0xF1, 0x6A, 0xC4, 0x1E,
0xA6, 0xED, 0xDD, 0xE8, 0x3C, 0x01, 0x71, 0x4F, 0x9C, 0xFE, 0x62, 0x66, 0x7F, 0xB1, 0x29, 0x95
};
int input_flag()
{
void* Dst = (void*)calloc(0x30, 1);
int* Buf1 = (int*)calloc(0x30, 1);
printf("please input string:");
scanf("%s", &Dst);
for (int i = 0; i < 6; ++i)
{
int o1 = *((int*)&Dst + i * 2);
int o2 = *((int*)&Dst + i * 2 + 1);
for (int j = 0; j < 64; ++j)
{
if (o2 >= 0)
{
_asm{
mov cl, 0x1
mov eax,
mov edx,
shld edx, eax, cl
shl eax, cl
mov, eax
mov, edx
}
}
else{
_asm{
mov cl, 0x1
mov eax,
mov edx,
shld edx, eax, cl
shl eax, cl
mov, eax
mov, edx
}
o1 = o1 ^ 0xEEFF0102;
o2 = o2 ^ 0xAABBCCDD;
}
}
Buf1 = o1;
Buf1 = o2;
}
if (!memcmp(&Buf1, key,0x30u))
puts("great~");
else
puts("wrong");
}
int _tmain(int argc, _TCHAR* argv[])
{
input_flag();
system("pause");
return 0;
}
俩字：爆破
想逆洗洗睡吧https://cdn.jsdelivr.net/gh/hishis/forum-master/public/images/patch.gif
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