新出炉的CM，易语言，求大佬指教
本帖最后由 CrazyNut 于 2018-4-27 17:50 编辑三无产品
密码纯数字 并且 唯一，获取密码算成功 = =
这个爆破太简单了，没啥意思= =
希望能获取算法的大佬给我说下怎么获取的！！！推算出密码的大佬说一下推算逻辑 谢谢{:301_971:}
源码附上就是简单的算法，还有大佬还弄了一个计算器出来
int sub_40100B()
{
void *v0; // eax
void *v1; // eax
void *v2; // eax
void *v3; // eax
void *v4; // eax
void *v5; // eax
void *v6; // eax
void *v7; // eax
void *v8; // eax
void *v9; // eax
void *v10; // eax
void *v11; // eax
double v12; // st7
double v13; // st7
int result; // eax
void *v15; //
int lpMem; //
int lpMema; //
int lpMemb; //
int lpMemc; //
int lpMemd; //
int lpMeme; //
void *lpMem_4; //
void *lpMem_4a; //
void *lpMem_4b; //
void *lpMem_4c; //
void *lpMem_4d; //
void *lpMem_4e; //
v15 = (void *)sub_4015A0(1375797249, 369164290, 8, -1);
if ( dword_4A65B0 )
sub_40158E(dword_4A65B0);
dword_4A65B0 = v15;
LOBYTE(v0) = (_BYTE)v15;
if ( !v15 )
v0 = &unk_481B5C;
v1 = (void *)sub_40159A(2, (char)v0);
lpMem_4 = v1;
if ( !v1 )
v1 = &unk_481B5C;
lpMem = sub_40159A(1, (char)v1);
if ( lpMem_4 )
sub_40158E(lpMem_4);
dword_4A65B4 = lpMem;
LOBYTE(v2) = (_BYTE)dword_4A65B0;
if ( !dword_4A65B0 )
v2 = &unk_481B5C;
v3 = (void *)sub_40159A(3, (char)v2);
lpMem_4a = v3;
if ( !v3 )
v3 = &unk_481B5C;
lpMema = sub_40159A(1, (char)v3);
if ( lpMem_4a )
sub_40158E(lpMem_4a);
dword_4A65B8 = lpMema;
LOBYTE(v4) = (_BYTE)dword_4A65B0;
if ( !dword_4A65B0 )
v4 = &unk_481B5C;
v5 = (void *)sub_40159A(3, (char)v4);
lpMem_4b = v5;
if ( !v5 )
v5 = &unk_481B5C;
lpMemb = sub_40159A(1, (char)v5);
if ( lpMem_4b )
sub_40158E(lpMem_4b);
dword_4A65BC = lpMemb;
LOBYTE(v6) = (_BYTE)dword_4A65B0;
if ( !dword_4A65B0 )
v6 = &unk_481B5C;
v7 = (void *)sub_40159A(2, (char)v6);
lpMem_4c = v7;
if ( !v7 )
v7 = &unk_481B5C;
lpMemc = sub_40159A(1, (char)v7);
if ( lpMem_4c )
sub_40158E(lpMem_4c);
dword_4A65C0 = lpMemc;
LOBYTE(v8) = (_BYTE)dword_4A65B0;
if ( !dword_4A65B0 )
v8 = &unk_481B5C;
v9 = (void *)sub_40159A(3, (char)v8);
lpMem_4d = v9;
if ( !v9 )
v9 = &unk_481B5C;
lpMemd = sub_40159A(1, (char)v9);
if ( lpMem_4d )
sub_40158E(lpMem_4d);
dword_4A65C4 = lpMemd;
LOBYTE(v10) = (_BYTE)dword_4A65B0;
if ( !dword_4A65B0 )
v10 = &unk_481B5C;
v11 = (void *)sub_40159A(3, (char)v10);
lpMem_4e = v11;
if ( !v11 )
v11 = &unk_481B5C;
lpMeme = sub_40159A(1, (char)v11);
if ( lpMem_4e )
sub_40158E(lpMem_4e);
dword_4A65C8 = lpMeme;
v12 = (double)dword_4A65B4 + (double)dword_4A65B8 * (double)dword_4A65BC + (double)dword_4A65C0 - 82.0;
if ( v12 < 0.0 )
v12 = -v12;
if ( v12 > 0.0000001 )
return sub_4014AD();
v13 = (double)dword_4A65C4 + (double)dword_4A65C8 - 108.0;
if ( v13 < 0.0 )
v13 = -v13;
if ( v13 > 0.0000001 )
result = sub_4014AD();
else
result = sub_401405();
return result;
} 我说大佬啊，你给个附件也好啊。一张图能看出啥来 云在天 发表于 2018-4-26 13:23
我说大佬啊，你给个附件也好啊。一张图能看出啥来
刚刚有点bug我重新发= = 爆破算吗？
00401484|.68 751B4800 push CM.00481B75;密码正确
旧年白白白 发表于 2018-4-26 15:11
爆破算吗？
00401484|.68 751B4800 push CM.00481B75;密码正确
这个爆破就太没意思了。、。、{:301_999:} CrazyNut 发表于 2018-4-26 15:16
这个爆破就太没意思了。、。、
哈哈哈我懒 所以走最快的路{:301_997:} 旧年白白白 发表于 2018-4-26 15:19
哈哈哈我懒 所以走最快的路
这个爆破没什么难度 0.0 试试获取密码呢 云在天 发表于 2018-4-26 13:23
我说大佬啊，你给个附件也好啊。一张图能看出啥来
哈哈，我差点笑出声，竟然云大神也有懵逼的时候 rurg 发表于 2018-4-26 16:27
int sub_40100B()
{
void *v0; // eax
{:301_971:}大佬这是啥 啊 0.0怎么搞出来的