第三题，抛砖引玉

验证过程：
1.计算注册码的格式：xxxxxxxx-xxxxxxxx-xxxxxxxx-xxxxxxxx以及字符集合法性，见如下letter数组
2.一些移位操作计算出一个20字节的HASH值
3.对输入的用户名，后面追加C盘逻辑卷的序列号以及一个字符串"Tencent"，用变形的类SHA1算法计算一个HASH值
4.比较前后两个，相同则ok
能看到注册码的格式把注册码分为四部分，其中每一组8Byte计算出5个BYTE，对应于用户名的HASH值。
计算部分如下，用户名那部分的代码就不贴了，抠出来就是。
DWORD USER_HASH[5]; //这里记录的是用户名+VolumeSerialNumber+Tencent最终生成的hash
char letter[] = "ABCDEFGHJKMNPQRSTVWXYZ1234567890";
char reg_code[4][9] = {0};
void CalcReg()
{
    bool succ = false;
    BYTE* byte_USER_HASH = (BYTE*)USER_HASH;
    for(int i=0; i<4; i++)
    {
        int pos = i*5;
        succ = false;
        for(int a0=0; a0<32 && !succ; a0++)
        {
            for(int a1=0; a1<32 && !succ; a1++)
            {
                if ((((a0<<3)|(a1>>2))&0xFF) != byte_USER_HASH[pos])
                    continue;
                for(int a2=0; a2<32 && !succ; a2++)
                {
                    for(int a3=0; a3<32 && !succ; a3++)
                    {
                        if ((((a1<<6)|(a2<<1)|(a3>>4))&0xFF) != byte_USER_HASH[pos+1])
                            continue;
                        for(int a4=0; a4<32 && !succ; a4++)
                        {
                            if ((((a3<<4)|(a4>>1))&0xFF)!= byte_USER_HASH[pos+2])
                                continue;
                            for(int a5=0; a5<32 && !succ; a5++)
                            {
                                for(int a6=0; a6<32 && !succ; a6++)
                                {
                                    if ((((a4<<7)|(a5<<2)|(a6>>3))&0xFF) != byte_USER_HASH[pos+3])
                                        continue;
                                    for(int a7=0; a7<32; a7++)
                                    {
                                        if((((a6<<5)|a7)&0xFF) == byte_USER_HASH[pos+4])
                                        {
                                            reg_code[i][0] = letter[a0];
                                            reg_code[i][1] = letter[a1];
                                            reg_code[i][2] = letter[a2];
                                            reg_code[i][3] = letter[a3];
                                            reg_code[i][4] = letter[a4];
                                            reg_code[i][5] = letter[a5];
                                            reg_code[i][6] = letter[a6];
                                            reg_code[i][7] = letter[a7];
                                            succ = true;
                                            break;
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }
}