用C语言写了个超难的Crackme,欢迎高手来破

iTruth

这个程序如果输入错误会直接退出,输入正确会弹出对话框提示你破解成功了
如果哪位大佬破解完觉得简单请轻喷..

C语言编写 无壳 只有写出注册机才算破解成功



现在时间到了,还没有人能写出注册机所以放出源码
第一次发这种帖子Keygenme没控制好难度还请各位大佬海涵

iTruth

我再追加一组正确的用户名和序列号来给你们做参考
username:----涉嫌联系方式----
regcode:+z0GJdRGsw13sepb

crack5

[Asm] 纯文本查看 复制代码

char* charmap = "CbcdefghGjkl+mYopLr6tOvwxyzA/BaD0FiHIJKqMNuPQRSTUV7XnZ12345sW98E";


void *__cdecl date(int a1, void *Name, int len)
{
	unsigned int p; 
	int k; 
	int i; 

	k = (*(DWORD *)a1 >> 3) & 0x3F;
	p = 64 - k;
	*(DWORD *)a1 += 8 * len;
	if (*(DWORD *)a1 < (unsigned int)(8 * len))
		++*(DWORD *)(a1 + 4);
	*(DWORD *)(a1 + 4) += (unsigned int)len >> 29;
	if (len < p)
	{
		i = 0;
	}
	else
	{
		memcpy((void *)(k + 16 + a1 + 8), Name, p);
		Transform((int *)(a1 + 8), a1 + 24);
		for (i = 64 - k; len >= (unsigned int)(i + 64); i += 64)
			Transform((int *)(a1 + 8), (int)Name + i);
		k = 0;
	}
	return memcpy((void *)(k + 16 + a1 + 8), (char *)Name + i, len - i);
}

unsigned int __cdecl Encode(int a1, int a2, unsigned int a3)
{
	unsigned int result; // eax
	unsigned int i; // [esp+8h] [ebp-8h]
	int v5; // [esp+Ch] [ebp-4h]

	v5 = 0;
	for (i = 0; ; i += 4)
	{
		result = i;
		if (i >= a3)
			break;
		*(BYTE *)(a1 + i) = *(DWORD *)(4 * v5 + a2);
		*(BYTE *)(i + 1 + a1) = *(WORD *)(4 * v5 + a2) >> 8;
		*(BYTE *)(i + 2 + a1) = *(DWORD *)(4 * v5 + a2) >> 16;
		*(BYTE *)(i + 3 + a1) = *(DWORD *)(4 * v5++ + a2) >> 24;
	}
	return result;
}
unsigned int __cdecl Final(DWORD *a1, int a2)
{
	int v2;
	char v4; 
	int v5; 
	unsigned int v6; 

	v6 = 0;
	v5 = 0;
	v6 = (*a1 >> 3) & 0x3F;
	if (v6 > 0x37)
		v2 = 120 - v6;
	else
		v2 = 56 - v6;
	v5 = v2;
	Encode((int)&v4, (int)a1, 8u);
	Update((int)a1, &PADDING, v5);
	Update((int)a1, &v4, 8);
	return Encode(a2, (int)(a1 + 2), 0x10u);
}

void 初始化(Name)
{

pi = malloc(0x10u);
               *j = 0;
	j[1] = 0;
	j[2] = 1732584193;
	j[3] = -271733879;
	j[4] = -1732584194;
	result = a1;
	a1[5] = 271733878;

NameLen = strlen(Name)

date(&j,Name,NameLen)
Final(&j, pi);
}



if(NameLen> 7 && NameLen <= 30)
{
temp = (char *)malloc(0x20u);
 for(int i = null, i <= 5 ; i++)
   {
     LOC = 4;
     LOC [i+1*4] = getLoc(temp);
   {
  
  	for (j = 0; j <= 15; ++j)
                key = (char *)malloc(0x11u);
	memset(key, 0, 0x11u);
			key[j] = charmap[abs(
			(signed int)(*(char *)(LOC[6] + j) ^ ((*(char *)(LOC[3] + j) & (*(char *)(LOC[2] + j) | *(char *)(LOC[1] + j) ^ 0xFFFFF123))
				- *(char *)(LOC[4] + j))
				* *(char *)(LOC[5] + j))
				% 64)];
   }


}
}
}



全是手打 电脑没装vs

iTruth

huzpsb 发表于 2020-2-22 21:35
[mw_shl_code=cpp,true]for ( j = 0; j

我将会在后天中午公布源代码，在此期间只要有人写出注册机我都会直接公布

huzpsb

[C++] 纯文本查看 复制代码

for ( j = 0; j <= 15; ++j )
      v6[j] = charmap[abs(
                        (signed int)(*(char *)(v8[6] + j) ^ ((*(char *)(v8[3] + j) & (*(char *)(v8[2] + j) | *(char *)(v8[1] + j) ^ 0xFFFFF123))
                                                           - *(char *)(v8[4] + j))
                                                          * *(char *)(v8[5] + j))
                      % 64)];

Nice Try.
不过我想知道magicube.h是啥...
User:----涉嫌联系方式----
Pass:xC7NIMhhKMEyuBtX
又:我刚开始以为用户名好像必须是邮箱,现在看来好像不是...

crack5

User:123456789
Pass:o3jeO4Y29BJXDwUf

iTruth

huzpsb 发表于 2020-2-22 21:35
[mw_shl_code=cpp,true]for ( j = 0; j

还有，用户名不一定是邮箱

居客

直接爆破不算吧！

居客

crack5 发表于 2020-2-22 21:41
User:123456789
Pass:o3jeO4Y29BJXDwUf

大佬能讲一下破解思路？

crack5

crack5 发表于 2020-2-22 22:05
[mw_shl_code=asm,true]char* charmap = "CbcdefghGjkl+mYopLr6tOvwxyzA/BaD0FiHIJKqMNuPQRSTUV7XnZ12345sW ...

- - 我说我没C++ 其他都可以贴出来  IDA 一看就能看出来了