一个有趣的crack注册机

多个checkbox，选中一部分当作密码。直接上注册机原码然后简单说下道理。

// ChechkBoxReverse.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include "windows.h"

DWORD getGroupSize(int All, int SelectItem);//组合数个数
void addToArry(char *p, char *Positions, char count);
void GetGroupMain(char *Group, char *Ppositions, int GroupMaxItem, int SelectItems, int CurrentPosition, int CurrentIdex);//组合集合数组，集合最大值，排列个数,集合默认从0开始
char checkID[] = { 0x16, 0x49, 0x5e, 0x15, 0x27, 0x26, 0x21, 0x25, 0x1d, 0x59, 0x53, 0x37, 0x31, 0x48, 0x5d, 0x0c, 0x61, 0x52, 0x4d };
bool isFirst = true;
int _tmain(int argc, _TCHAR* argv[])
{
    DWORD CheckSum[20] = {0};
    int length = strlen(checkID);
    for (int i = 0; i < length-1; i++)
    {
        CheckSum[i] = (DWORD)(checkID[i])*(i + 1)*checkID[i + 1];
    }
    DWORD ResualtSum = 0xf35466 / 0x4d;
    for (int i = 1; i <= 17; i++)
    {
        DWORD GroupCount = getGroupSize(17, i);
        char *Group = (char *)malloc(i*GroupCount);
        GetGroupMain(Group, NULL, 16, i, -1, 0);

        /*
        for (int j = 0; j < GroupCount; j++)
        {
            for (int k = 0; k < i; k++)
            {
                printf_s("%d  ", *(char *)(j*i + k + Group));
            }
        }
        printf_s("
");
        int a = 1;
        */
        
        for (int j = 0; j < GroupCount; j++)
        {
            DWORD tempSum = 0;
            for (int k = 0; k < i; k++)
            {
                tempSum = tempSum + CheckSum[*(char *)(j*i + k +Group)];
            }
            if (tempSum == ResualtSum)
            {
                int counts = 0;
                for (int l = 0; l < i; l++)
                {
                    printf_s("%d
", *(char *)(j*i + l + Group));
                }
                printf_s("count=%d,index is
", i);
                system("pause");
                return 0;
            }
        }
        
        isFirst = true;
        free(Group);
    }
    system("pause");
    return 0;
}
void GetGroupMain(char *Group,char *Ppositions ,int GroupMaxItem, int SelectItems,int CurrentPosition,int CurrentIdex)//组合集合数组，集合最大值，排列个数,集合默认从0开始
{
    
    for (int i = CurrentPosition + 1; i <= GroupMaxItem-SelectItems+1+CurrentIdex; i++)
    {
        if (CurrentIdex == SelectItems - 1)
        {
            if (CurrentIdex == 0)
            {
                char a = i;
                addToArry(Group, &a, SelectItems);
            }
            else
            {
                *(Ppositions + CurrentIdex) = i;
                addToArry(Group, Ppositions, SelectItems);
            }
        }
        else
        {
            char* PointPositions = (char*)malloc(CurrentIdex + 2);
            if (CurrentIdex!=0)
            memcpy(PointPositions, Ppositions, CurrentIdex);
            *(PointPositions + CurrentIdex) = i;
            GetGroupMain(Group, PointPositions, GroupMaxItem, SelectItems, i, CurrentIdex + 1);
        }
    }
    free(Ppositions);
    return;
}

char* CurrentOffset=0;
void addToArry(char *p, char *Positions,char count)
{
    if (isFirst)
    CurrentOffset = p;
    isFirst = false;
    for (int i = 0; i < count; i++)
    {
        char a = *(Positions + i);
        *(CurrentOffset++) = a;
    }
}
DWORD getGroupSize(int All, int SelectItem)//组合数个数
{
    UINT64 sum = 1;
    for (int i = All; i> All - SelectItem; i--)
    {
        sum = sum *i;
    }
    for (int i = SelectItem; i >= 1; i--)
    {
        sum = sum /i;
    }
    return sum;
}

每个checkbox都有一个ID对应，然后索引从0开始遍历，遇到选中的box择将他的ID乘下个box再乘下个的索引，一直到最后连加起来再将最后的和做一个简单运算，再与一个固定值比较，正确则完成。

那么通过逆向简单运算得到最后的和。再假设所有box被选中算出值，这个问题就变成了先这么多值中选出一些相加，看最后和是否等于一个值。

于是这个算法的主要部分在于怎么选出这几个值。于是我在GetGroupMain这个函数中列出了不同个数的所有可能索引取值，最后去试。主要思路就是这样。用了一个递归算法选和组合数公式。在free运用上要比较注意，指针在不停的传递，肯定有更好的办法，但是不细想了。