CrackMe的分析

crackme：http://re-xe.com/?p=111

反汇编代码：

0040110C  |> /0FBE840D 48FF>/movsx   eax, byte ptr [ebp+ecx-B8]      ;  分别取用户名的每个字符
00401114  |. |41            |inc     ecx                             ;  当前是第几个字符，从1开始
00401115  |. |33C1          |xor     eax, ecx                        ;  字符值和字符位置异或，存入eax中
00401117  |. |03D8          |add     ebx, eax                        ;  累加
00401119  |. |3B4D D8       |cmp     ecx, dword ptr [ebp-28]
0040111C  |.^\75 EE         \jnz     short 0040110C
0040111E  |.  6BC0 06       imul    eax, eax, 6                      ;  eax = eax * 6 结果乘6，eax是最后一个字母异或后的结果
00401121  |.  C1E3 07       shl     ebx, 7                           ;  累加的和左移7位
00401124  |.  03C3          add     eax, ebx                         ;  eax和ebx相加
00401126  |.  8945 C8       mov     dword ptr [ebp-38], eax
00401129  |.  FF75 C8       push    dword ptr [ebp-38]
0040112C  |.  68 38B44000   push    0040B438                         ;  ASCII "%lX"
00401131  |.  8D8D 80FEFFFF lea     ecx, dword ptr [ebp-180]
00401137  |.  51            push    ecx
00401138  |.  E8 873D0000   call    00404EC4                         ;  将注册码转换为十六进制字符串
0040113D  |.  83C4 0C       add     esp, 0C
00401140  |.  8D85 80FEFFFF lea     eax, dword ptr [ebp-180]
00401146  |.  50            push    eax                              ; /String2
00401147  |.  8D95 E4FEFFFF lea     edx, dword ptr [ebp-11C]         ; |
0040114D  |.  52            push    edx                              ; |String1
0040114E  |.  E8 339C0000   call    <jmp.&KERNEL32.lstrcmpA>         ; \lstrcmp 将真码和假码比较
00401153  |.  85C0          test    eax, eax
00401155  |.  75 0D         jnz     short 00401164
00401157  |.  68 3CB44000   push    0040B43C                         ; /Text = "Congratulations! IF this number comes *FROM YOUR* keygen, Write a tutorial dude ;)."
0040115C  |.  56            push    esi                              ; |hWnd
0040115D  |.  E8 289B0000   call    <jmp.&USER32.SetWindowTextA>     ; \SetWindowTextA
00401162  |.  EB 18         jmp     short 0040117C
00401164  |>  68 90B44000   push    0040B490                         ; /Text = "This serial is *NOT* Valid!! Try again... : UNREGISTERED"
00401169  |.  56            push    esi                              ; |hWnd
0040116A  |.  E8 1B9B0000   call    <jmp.&USER32.SetWindowTextA>     ; \SetWindowTextA
0040116F  |.  EB 0B         jmp     short 0040117C
00401171  |>  68 C9B44000   push    0040B4C9                         ; /Text = "Name must contain more than 4 chars and less than 50 chars !!"
00401176  |.  56            push    esi                              ; |hWnd
00401177  |.  E8 0E9B0000   call    <jmp.&USER32.SetWindowTextA>     ; \SetWindowTextA
0040117C  |>  5F            pop     edi
0040117D  |.  5E            pop     esi
0040117E  |.  5B            pop     ebx
0040117F  |.  8BE5          mov     esp, ebp
00401181  |.  5D            pop     ebp
00401182  \.  C3            retn

算法分析及注册机：(VS2010 + XP Sp3 测试通过)

#include <iostream>
#include <string>

using namespace std;

int main(void)
{
    char username[100];
    char key[100];

    cout << "username    :" ;
    cin >> username;
    

    int length = strlen(username);
    
    
    if (length <= 4 || length >= 50) 
    {
        cout << "Name must contain more than 4 chars and less than 50 chars !!";
        cin.get();
        cin.get();
        return 0;
    }

    cout << "Key is      :";

    int value = 0;
    int sum = 0;


    for (int i =0;i<length;i++)
    
        value = (username[i]) ^ (i+1);  //1. 将每个字符和该字符所在位置异或
        sum += value;                    //2. 异或的结果累加
    }

    value *= 6;                         //3. 最后一个字符和该字符所在的位置异或,结果乘以6
    sum = (sum << 7);                   //4. 步骤2中计算出的值向左移动7位

    sum += value;                       //5. 步骤3和步骤4的值累加


    itoa(sum,key,16);                   //6. 步骤5的值转换为16进制
    
    char *upkey;
    upkey = _strupr(_strdup(key));     //7. 将16进制字符串转换为大写

    cout << upkey;

    cin.get();
    cin.get();
    
    return 0;
}