#include <iostream> #include <Windows.h> #include <time.h> #include <cstring> using namespace std; int main() { char st[21] = { 0 }; //用户名 char jqm_a[31] = { 0 }; //机器码 char jqm[21] = { 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[2]; jqm[2] = jqm[18]; jqm[18] = q; q = jqm[4]; jqm[4] = jqm[15]; jqm[15] = q; q = jqm[8]; jqm[8] = jqm[11]; jqm[11] = q; while (1) { char mb[7] = { 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[i] ^ jqm[i]; 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[51]; //存放真正的字符串 char str[51]; //序列号 char str1[6] = { 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[i] + 0x14 - 9 - (1 / 2 + i)), str_z + (i * 2), 10); for (int i = 0; i <= 9; i++) str[i] = str_z[i]; 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[i] = str2[rand() % 10]; int g; //交换 g = str[2]; str[2] = str[38]; str[38] = g; g = str[4]; str[4] = str[25]; str[25] = g; g = str[9]; str[9] = str[31]; str[31] = g; for (int i = 0; i <= 4; i++) str1[i] = str[45 + i]; 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[esi] & 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[41] = { 0 }; for (int i = 0; i <= 40; i++) str_1[i] = str[i]; g = str_1[2]; str_1[2] = str_1[38]; str_1[38] = g; g = str_1[4]; str_1[4] = str_1[25]; str_1[25] = g; g = str_1[9]; str_1[9] = str_1[31]; str_1[31] = g; char str_2[21] = { 0 }; //产生的那20个字符 char str_3[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[0] = str_1[i]; str_3[1] = str_1[i + 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[i / 2] = eax_1; ebx_1++; } int zjz = 0; for (int i = 0; i < 19; i++) zjz = zjz + str_2[i]; zjz = zjz % 10; zjz = zjz + 0x30; if (str_2[19] == zjz || str_2[19] >= 0x41) { cout << "序列号为："; for (int i = 0; i <= 49; i++) cout << str[i]; cout << endl; break; } } return 0; }
运算时间长，可见是通过逆向来的验证程序进行对比，不断爆破出正确的注册码，所以稳定性会很糟糕，看人品，一次不行多试几次。
毕竟人家软件说了，即使注册成功也会有问题，所以各位看着办吧，代码从PYG偶然发现的，吾爱破解和飘云阁这两个地方算是现在破解的主要聚集地了。
希望能有新的注册机出现......
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