能力值:
( LV13,RANK:370 )
2 楼
能力值:
( LV12,RANK:410 )
3 楼
没搞定。。。汗
能力值:
( LV9,RANK:250 )
4 楼
能力值:
( LV9,RANK:250 )
5 楼
00492143 |. BB 01000000 mov ebx, 1 ebx=1
00492148 |> 8BC3 /mov eax, ebx
0049214A |. B9 03000000 |mov ecx, 3
0049214F |. 99 |cdq
00492150 |. F7F9 |idiv ecx
00492152 |. 80C2 1A |add dl, 1A
00492155 |. 8855 F7 |mov [ebp-9], dl
00492158 |. 8D45 FC |lea eax, [ebp-4]
0049215B |. E8 F022F7FF |call 00404450
00492160 |. 8B55 FC |mov edx, [ebp-4]
00492163 |. 8A541A FF |mov dl, [edx+ebx-1] ; 字符读入
00492167 |. 8A4D F7 |mov cl, [ebp-9]
0049216A |. 0AD1 |or dl, cl
0049216C |. 885418 FF |mov [eax+ebx-1], dl
00492170 |. 8D45 FC |lea eax, [ebp-4]
00492173 |. E8 D822F7FF |call 00404450
00492178 |. 8BD3 |mov edx, ebx
0049217A |. 8B4D FC |mov ecx, [ebp-4]
0049217D |. 325419 FF |xor dl, [ecx+ebx-1]
00492181 |. 885418 FF |mov [eax+ebx-1], dl
00492185 |. 43 |inc ebx
00492186 |. 4E |dec esi
00492187 |.^ 75 BF \jnz short 00492148
00492189 |> 8BF7 mov esi, edi
0049218B |. 85F6 test esi, esi
0049218D |. 7E 32 jle short 004921C1
0049218F |. BB 01000000 mov ebx, 1
00492194 |> 8D43 0D /lea eax, [ebx+D] ; 14 (第一次)使固定数,然后每次运算+1
00492197 |. B9 03000000 |mov ecx, 3 ; 3 是常熟
0049219C |. 99 |cdq ; 青空 DL
0049219D |. F7F9 |idiv ecx ; / ECX
0049219F |. 80C2 10 |add dl, 10 ; DL 保存上面余数 + 10
004921A2 |. 8855 F7 |mov [ebp-9], dl
004921A5 |. 8D45 F8 |lea eax, [ebp-8]
004921A8 |. E8 A322F7FF |call 00404450
004921AD |. 8B55 F8 |mov edx, [ebp-8] ; 假码地址如 EDX
004921B0 |. 8A541A FF |mov dl, [edx+ebx-1] ; 假码每位入 DL
004921B4 |. 8A4D F7 |mov cl, [ebp-9] ; 上面预算的结果入 CL
004921B7 |. 32D1 |xor dl, cl ; 两个数做异域运算
004921B9 |. 885418 FF |mov [eax+ebx-1], dl ; dl 是结果如内存
004921BD |. 43 |inc ebx ; ebx +1
004921BE |. 4E |dec esi
004921BF |.^ 75 D3 \jnz short 00492194
能力值:
( LV9,RANK:650 )
6 楼
最初由 wangjun 发布 暴破不算! 找出注册码,写出注册机。 附件地址:www.163.com ---------------------- 用户:pediy8088@163.com ........
试了试勉强通过了
上传的附件:
能力值:
( LV9,RANK:650 )
7 楼
程序算法不知是不是这样的:
#include <string.h>
#include "stdio.h"
void main()
{
char str1[10];
int i,num[10],len;
scanf("%s",&str1);
len=strlen(str1);
if (len<7 || len>10)
printf("请输入7位到10位有效字符\n");
else
for(i=0;i<len;i++)
{
num[i]=(((i+1)%0x3+0x1a) | str1[i]) ^ (i+1);
num[i]=num[i]^((i+1+0xd)%3+0x10);
printf("%c",num[i]);
}
printf("\n");
getchar();
}
能力值:
( LV2,RANK:10 )
8 楼
procedure TForm1.Button2Click(Sender: TObject);
var
i,sum:Integer;
begin
sum:=0;
for i:=1 to length(edit2.Text) do begin
sum:=(((i+13)mod 3)+16)xor ord(edit2.Text[i]);
end;
label2.Caption:=IntToStr(sum);
end;
procedure TForm1.Button1Click(Sender: TObject);
var
i,sum,v:integer;
begin
sum:=0;
for i:=1 to length(edit1.Text) do
begin
sum:= i xor (ord(edit1.text[i]) or (i mod 3+26)) ;
end;
label1.Caption:=IntToStr(sum);
end;
能力值:
( LV2,RANK:10 )
9 楼
晚了一步
void keygen(char* name)
{
char nlen, i, seed;
nlen = strlen(name);
if(nlen < 7 || nlen > 10)
{
printf("Invalid Name. 注册名长度∈[7,10]\r\n");
return;
}
for(i=0; i<nlen; i++)
{
seed = (i+1)%3 + 0x1a;
name[i] |= seed;
name[i] = name[i]^(i+1);
seed = (i+1+0xd)%3 + 0x10;
name[i] = name[i]^seed;
}
printf("Serial:\t\"%s\"\r\n", name);
}