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取主机名() 和 luochen（) 运算
第一组运算 j = 申请空白字节集(256)
依次次转入 1~256的字节集
{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136,137,138,139,140,141,142,143,144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160,161,162,163,164,165,166,167,168,169,170,171,172,173,174,175,176,177,178,179,180,181,182,183,184,185,186,187,188,189,190,191,192,193,194,195,196,197,198,199,200,201,202,203,204,205,206,207,208,209,210,211,212,213,214,215,216,217,218,219,220,221,222,223,224,225,226,227,228,229,230,231,232,233,234,235,236,237,238,239,240,241,242,243,244,245,246,247,248,249,250,251,252,253,254,255}
第二组 运算 J = 申请空白字节集(256)
int c;
const char* temp = "luochen";
for (int i = 0; i <=256 i++)
{
c = c+1;
J[i] =temp[c];
if(c = strlen(temp))
{
j ＝ 0;
}
}
第三组运算 没去逆大概就是
for (int i = 0; i <=256 i++)
{
k ＝ k ＋ 位与 (空白字节 [i], 255) ＋ 位与 (空白字节2 [i], 255)
}
然后在和机器码运算 这里猜的！！
最后得到 机器码的运算的结果 在用Base64编码 校验编辑框1.内容是否等于 编码后运算的机器码 如果等于 就进行下一个 注册码的运算
注册码相对于比较简单
就是取主机名 和 取MAC 地址 注意他这个取MAC 是 PIC去取或者通过 注册表取 或 WQL
得到的MAC地址 大概是这样的 40:50:60:70:80:90:10 然后机器码和 MAC文本相加 在Base编码 判断是否与注册码编辑框的内容相同
校验OD 部分就没什么好说的了 就是 取标题和窗口 而已
发表于 2020-2-27 11:27
发表于 2020-2-27 12:48
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发表于 2020-2-27 12:51
