寻找屏幕录像专家注册机
各位大佬，在某处发现一张截图为屏幕录像专家注册机截图，其注册码为400位（不是常见的注册码），现在急需，求有注册机的大佬分享，感谢！！！外链:https://ang.lanzous.com/b04g57tuh 密码:d54g 这个早就有了的吧，只是具体忘了是哪个版本了 放弃这个想法吧 https://www.chinapyg.com/forum.php?mod=viewthread&tid=119276
那个注册机就是论坛大神们一起做的吧，无缘相见的。。还是直接用论坛的成品凑合下吧。
PS的毫无痕迹啊 哈哈 为什么非要鸡？论坛有可以用的版本，再不济就自己咔嚓{:lol:} 不破不立 发表于 2020-7-4 09:53
为什么非要鸡？论坛有可以用的版本，再不济就自己咔嚓
之前友坛上别人发的。
#include <iostream>
#include <Windows.h>
#include <time.h>
#include <cstring>
using namespace std;
int main()
{
char st = { 0 }; //用户名
char jqm_a = { 0 }; //机器码
char jqm = { 0 }; //机器码前20位
cout << "请输入你的用户名：";
cin >> st;
cout << "请输入你的机器码:";
cin >> jqm_a;
int q;
memcpy(jqm, jqm_a, 20);
cout << "取机器码的前20位来算序列号" << jqm << endl;
q = jqm;
jqm = jqm;
jqm = q;
q = jqm;
jqm = jqm;
jqm = q;
q = jqm;
jqm = jqm;
jqm = q;
while (1)
{
char mb = { 0 }; //运算的出的字符串
int eax_3;
int edi_3 = 0;
for (int i = 0; i < 20; i++)
{
eax_3 = st ^ jqm;
eax_3 = eax_3 * i;
eax_3 = eax_3 + edi_3;
edi_3 = eax_3;
}
edi_3 = edi_3 + 0x3039;
itoa(edi_3, mb, 10); //mb=得到那5个秘钥字符，
char str_z; //存放真正的字符串
char str; //序列号
char str1 = { 0 }; //序列号后5个字符
int esi = 0;
int eax = 0;
int ecx = 0;
int edi;
int arg_3 = 0x2D; //序列号长度-5
BYTE dl;
const char* str2 = "1234567890";
int edi_33 = edi_3;
srand((unsigned int)time(0));
while (1)
{
for (int i = 0; i < 5; i++) //生成序列号前10个字符
itoa((mb + 0x14 - 9 - (1 / 2 + i)), str_z + (i * 2), 10);
for (int i = 0; i <= 9; i++)
str = str_z;
int eax_4;
int ecx_4;
int edx_4;
edi_3 = edi_33;
edi_3 = edi_3 + 0x4d44;
eax_4 = int(edi_3 * 3.14 * 0.1594896331738427110);
edi_3 = eax_4;
ecx_4 = 0x186a0;
edx_4 = eax_4 % ecx_4;
itoa((edx_4 % 10 + 0x41) - 9 - 11 / 2 + 9, str + 10, 10);
for (int i = 12; i < 50; i++)
str = str2;
int g; //交换
g = str;
str = str;
str = g;
g = str;
str = str;
str = g;
g = str;
str = str;
str = g;
for (int i = 0; i <= 4; i++)
str1 = str;
esi = 0;
eax = 0;
ecx = 0;
while (1)
{
if (ecx == 0xB26D)
{
edi = 0;
while (edi < arg_3)
{
ecx++;
dl = 0x80;
do {
BYTE ah;
ah = WORD(eax) / 256;
if (ah >= 0x80)
{
eax = eax + eax;
WORD x = (WORD(eax) ^ 0x1021);
eax = eax / 65536;
eax = eax * 65536;
eax = eax + x;
ecx++;
}
else
eax = eax + eax;
ecx++;
if (str & dl)
{
WORD x = (WORD(eax) ^ 0x1021);
eax = eax / 65536;
eax = eax * 65536;
eax = x + eax;
}
dl = dl / 2;
} while (dl != 0);
esi++;
edi++;
}
}
ecx++;
if (ecx >= 0x186A0)
break;
}
if (WORD(eax) == atof(str1))
break;
}
int g; //生成第11,12个字符
char str_1 = { 0 };
for (int i = 0; i <= 40; i++)
str_1 = str;
g = str_1;
str_1 = str_1;
str_1 = g;
g = str_1;
str_1 = str_1;
str_1 = g;
g = str_1;
str_1 = str_1;
str_1 = g;
char str_2 = { 0 }; //产生的那20个字符
char str_3 = { 0 };
int zj = 0;
int edx_1 = 0;
int ebx_1 = 0;
int eax_1 = 0;
for (int i = 0; i < 40; i = i + 2)
{
ebx_1++;
str_3 = str_1;
str_3 = str_1;
eax_1 = atof(str_3);
edx_1 = ebx_1;
edx_1 = edx_1 / 2;
eax_1 = eax_1 + edx_1;
eax_1 = eax_1 + 9;
str_2 = eax_1;
ebx_1++;
}
int zjz = 0;
for (int i = 0; i < 19; i++)
zjz = zjz + str_2;
zjz = zjz % 10;
zjz = zjz + 0x30;
if (str_2 == zjz || str_2 >= 0x41)
{
cout << "序列号为：";
for (int i = 0; i <= 49; i++)
cout << str;
cout << endl;
break;
}
}
return 0;
}
编译报错如下： 1>C:\Users\Administrator\source\repos\PmlxzjKeygen\PmlxzjKeygen\PmlxzjKeygen.cpp(84,4): error C4996: 'itoa': The POSIX name for this item is deprecated. Instead, use the ISO C and C++ conformant name: _itoa. See online help for details.
解决办法： 打开项目属性，点击C/C++ ——> 高级 ——> 禁用特定警告 ——> 编辑中加入4996 本帖最后由 wl4592641 于 2020-7-4 10:18 编辑
楼下！！！！！！！