Plato Video To iPod Converter V3.46算法分析兼RSA-64实战

直接切入正题。

试炼码：
111111111111111122222222222222223333333333333333

1. PEiD：VC6编写，Kanal插件识别出了Miracl库，所以用IDA + CryptoSIG导出map文件；
2. OD载入，Loadmap上一步的map文件，根据出错提示轻松找到0040791C，跟进这个call：

00406920 /$>sub esp, 300
00406926 |.>or ecx, FFFFFFFF
00406929 |.>xor eax, eax
0040692B |.>push ebx
0040692C |.>push esi
0040692D |.>mov esi, [esp+30C]
00406934 |.>push edi
00406935 |.>mov edi, esi
00406937 |.>repne scas byte ptr es:[edi]
00406939 |.>not ecx
0040693B |.>dec ecx
0040693C |.>cmp ecx, 30 ; 注册码长度：48=3*16
0040693F |.>jnz 00406A2C
00406945 |.>xor ebx, ebx
00406947 |.>mov ecx, 3F
0040694C |.>lea edi, [esp+10D]
00406953 |.>mov [esp+10C], bl
0040695A |.>rep stos dword ptr es:[edi]
0040695C |.>stos word ptr es:[edi]
0040695E |.>stos byte ptr es:[edi]
0040695F |.>mov ecx, 3F
00406964 |.>xor eax, eax
00406966 |.>lea edi, [esp+D]
0040696A |.>mov [esp+C], bl
0040696E |.>rep stos dword ptr es:[edi]
00406970 |.>stos word ptr es:[edi]
00406972 |.>stos byte ptr es:[edi]
00406973 |.>mov ecx, 3F
00406978 |.>xor eax, eax
0040697A |.>lea edi, [esp+20D]
00406981 |.>mov [esp+20C], bl
00406988 |.>rep stos dword ptr es:[edi]
0040698A |.>stos word ptr es:[edi]
0040698C |.>stos byte ptr es:[edi]
0040698D |.>mov edi, [<&MSVCRT.strncpy>] ; MSVCRT.strncpy
00406993 |.>push 10 ; /maxlen = 10 (16.)
00406995 |.>lea eax, [esp+110] ; |
0040699C |.>push esi ; |src
0040699D |.>push eax ; |sn_1:取sn前16位
0040699E |.>call edi ; \strncpy
004069A0 |.>lea ecx, [esi+10]
004069A3 |.>push 10
004069A5 |.>lea edx, [esp+1C]
004069A9 |.>push ecx
004069AA |.>push edx ; sn_2:sn中间16位
004069AB |.>call edi
004069AD |.>add esi, 20
004069B0 |.>push 10
004069B2 |.>lea eax, [esp+228]
004069B9 |.>push esi
004069BA |.>push eax ; sn_3:sn最后16位
004069BB |.>call edi
004069BD |.>lea ecx, [esp+130]
004069C4 |.>push 0041B570 ; ASCII "C09EE4DD93F3D6D1"
004069C9 |.>push ecx ; sn_1
004069CA |.>mov [esp+148], bl
004069D1 |.>mov [esp+48], bl
004069D5 |.>mov [esp+248], bl
004069DC |.>call <mir> ; 计算出c1

mir标签是我自己加的，跟进去可以看见miracl库函数。
{
004067F0 >/$>sub esp, 108 ; 因为这个函数调用了3次，所以我们用了分号和sn_x(表示sn_1,sn_2,sn_3)
004067F6 |.>push ebx
004067F7 |.>push ebp
004067F8 |.>push esi
004067F9 |.>push edi
004067FA |.>push 0 ; 0
004067FC |.>push 64 ; 0x64
004067FE |.>call <mirsys>
00406803 |.>push 0
00406805 |.>mov dword ptr [eax+234], 10 ; 16进制
0040680F |.>call <_mirvar>
00406814 |.>push 0
00406816 |.>mov ebx, eax
00406818 |.>call <_mirvar>
0040681D |.>push 0
0040681F |.>mov [esp+24], eax
00406823 |.>call <_mirvar>
00406828 |.>push 0
0040682A |.>mov ebp, eax
0040682C |.>call <_mirvar>
00406831 |.>mov [esp+2C], eax
00406835 |.>mov ecx, 3F
0040683A |.>xor eax, eax
0040683C |.>lea edi, [esp+31]
00406840 |.>mov byte ptr [esp+30], 0
00406845 |.>add esp, 18
00406848 |.>rep stos dword ptr es:[edi]
0040684A |.>stos word ptr es:[edi]
0040684C |.>stos byte ptr es:[edi]
0040684D |.>mov edi, [esp+11C]
00406854 |.>or ecx, FFFFFFFF
00406857 |.>xor eax, eax
00406859 |.>xor esi, esi
0040685B |.>repne scas byte ptr es:[edi]
0040685D |.>not ecx
0040685F |.>dec ecx
00406860 |.>mov edi, ecx
00406862 |.>test edi, edi
00406864 |.>jle short 0040688A
00406866 |>>/mov eax, [esp+11C]
0040686D |.>|movsx ecx, byte ptr [esi+eax]
00406871 |.>|push ecx ; /c
00406872 |.>|call [<&MSVCRT.isxdigit>] ; \isxdigit
00406878 |.>|add esp, 4
0040687B |.>|test eax, eax
0040687D |.>|je 00406905
00406883 |.>|inc esi
00406884 |.>|cmp esi, edi
00406886 |.>\jl short 00406866
00406888 |.>test edi, edi
0040688A |>>je short 00406905
0040688C |.>mov edx, [esp+11C]
00406893 |.>push edx ; sn_1;sn_2;sn_3
00406894 |.>push ebx
00406895 |.>call <_cinstr>
0040689A |.>mov eax, [esp+128]
004068A1 |.>push eax ; n1;n1;n2
004068A2 |.>push ebp ; n
004068A3 |.>call <_cinstr>
004068A8 |.>mov edi, [esp+24]
004068AC |.>push 0041B554 ; ASCII "10001"
004068B1 |.>push edi ; e
004068B2 |.>call <_cinstr>
004068B7 |.>push ebp ; n
004068B8 |.>push ebx ; sn_1;sn_2;sn_3
004068B9 |.>call <_compare> ; sn_x < n
004068BE |.>add esp, 20
004068C1 |.>cmp eax, -1
004068C4 |.>jnz short 00406905
004068C6 |.>mov esi, [esp+10]
004068CA |.>push esi ; c
004068CB |.>push ebp ; n
004068CC |.>push edi ; e
004068CD |.>push ebx ; sn_x
004068CE |.>call <_powmod> ; c = sn_x^e (mod n)
004068D3 |.>lea ecx, [esp+28]
004068D7 |.>push 0
004068D9 |.>push ecx
004068DA |.>push esi ; c
004068DB |.>push 100
004068E0 |.>call <_big_to_bytes>
004068E5 |.>push ebx
004068E6 |.>call <_mirkill>
004068EB |.>push esi
004068EC |.>call <_mirkill>
004068F1 |.>push ebp
004068F2 |.>call <_mirkill>
004068F7 |.>push edi
004068F8 |.>call <_mirkill>
004068FD |.>add esp, 30
00406900 |.>call <_mirexit>
00406905 |>>lea edx, [esp+18] ; 深刻理解下面这个函数很重要^_^，请查阅MSDN这个函数下面的例子。
00406909 |.>push edx ; /s
0040690A |.>call [<&MSVCRT.atoi>] ; \atoi
00406910 |.>add esp, 4
00406913 |.>pop edi
00406914 |.>pop esi
00406915 |.>pop ebp
00406916 |.>pop ebx
00406917 |.>add esp, 108
0040691D \.>retn
}
上面的RSA清晰明了。

004069E1 |.>lea edx, [esp+38]
004069E5 |.>push 0041B570 ; ASCII "C09EE4DD93F3D6D1"
004069EA |.>push edx ; sn_2
004069EB |.>mov esi, eax
004069ED |.>call <mir> ; 计算出c2
004069F2 |.>mov edi, eax
004069F4 |.>lea eax, [esp+240]
004069FB |.>push 0041B55C ; ASCII "9C70A3F13DE1C2F1"
00406A00 |.>push eax ; sn_3
00406A01 |.>call <mir> ; 计算出c3
00406A06 |.>add esp, 3C
00406A09 |.>cmp esi, ebx
00406A0B >jle short 00406A2A
00406A0D |.>cmp edi, ebx
00406A0F |.>jle short 00406A2A
00406A11 |.>cmp eax, ebx
00406A13 |.>jle short 00406A2A
00406A15 |.>add edi, esi ; c1 + c2
00406A17 |.>xor ecx, ecx
00406A19 |.>cmp edi, eax ; 比较c1 + c2和c3是否相等，相等则成功
00406A1B |.>pop edi
00406A1C |.>sete cl
00406A1F |.>pop esi
00406A20 |.>mov eax, ecx
00406A22 |.>pop ebx
00406A23 |.>add esp, 300
00406A29 |.>retn
00406A2A |>>xor eax, eax
00406A2C |>>pop edi
00406A2D |.>pop esi
00406A2E |.>pop ebx
00406A2F |.>add esp, 300
00406A35 \.>retn

整理一下程序验证的思路：
(1)使用n1和sn_1计算出c1，并转换结果为十进制（atoi函数）；
(2)使用n1和sn_2计算出c2，并转换结果为十进制；
(3)使用n2和sn_3计算出c3，并转换结果为十进制；
(4)判断c1 + c2和c3相等否，相等则成功，反之则失败。

经RSATool计算：
=====================
n1=p1 * q1:
C09EE4DD93F3D6D1
p1:
BCAAC583
q1:
1055D561B
d1:
325A57214FA2FBF1

n2=p2 * q2:
9C70A3F13DE1C2F1
p2:
FEED9D0B
q2:
9D190573
d2:
4E31E75F8231BB79
=====================

好了，剩下就是写注册机了，下面是注册机的源代码和几组可用的注册码：

//KeyGen for [Plato Video To iPod Converter V3.46]
//Cracked by [HappyTown]

/* RSA inside */
#include <stdio.h>
#include <stdlib.h>
#include <windows.h>
#include <time.h>
#include <miracl.h>

#define MAXLEN 500
int main()
{
    int i=0, Len=0, k=0, c=0;
    unsigned char sn[MAXLEN] = {0};
    unsigned char h[MAXLEN] = {0};
    unsigned char Table[100] = "1234";

    big n1,n2,d1,d2,bTemp,c1,c2,c3;
    miracl *mip = mirsys(0x64, 0);
    n1 = mirvar(0);
    n2 = mirvar(0);
    d1 = mirvar(0);
    d2 = mirvar(0);
    bTemp = mirvar(0);
    c1 = mirvar(0);
    c2 = mirvar(0);
    c3 = mirvar(0);

    mip->IOBASE = 16;
    cinstr(n1, "C09EE4DD93F3D6D1");
    cinstr(n2, "9C70A3F13DE1C2F1");
    cinstr(d1, "325A57214FA2FBF1");
    cinstr(d2, "4E31E75F8231BB79");

again:
    //get c1,c2,c3
    srand((unsigned int)time(NULL));
    for(i=0; i<4; i++)
    {
        c = rand() % 4;
        k = c * 2 + 0x30;
        c = c + 0x30;    //下面直接令c1 = c2，所以c3 = 2 * c1，偷懒了，呵呵
        h[i*2] = 0x33;
        h[i*2+1] = c;
        sn[i*2] = 0x33;
        sn[i*2+1] = k;
    }
    cinstr(c1, h);
    copy(c1, c2);    //令c1 = c2
    cinstr(c3, sn);

    //RSA: c1,c2
    powmod(c1, d1, n1, bTemp);
    cotstr(bTemp, sn);
    Len = lstrlen(sn);
    if(Len != 16)    //因为在cotstr转换时会把前面的0去掉，所以检测一下是否够16位
    {
        goto again;
    }
    memcpy(h, sn, lstrlen(sn));
    lstrcat(h, sn);

    //RSA:c3
    powmod(c3, d2, n2, bTemp);
    memset(sn, 0, MAXLEN);
    cotstr(bTemp, sn);
    Len = lstrlen(sn);
    if(Len != 16)
    {
        goto again;
    }
    lstrcat(h, sn);
    if(lstrlen(h) != 48)
    {
        goto again;
    }
    MessageBox(NULL, h, "Serial number: Cracked by [HappyTown] 2007-01-11", 0);

    //kill miracl
    mirkill(n1); mirkill(n2); mirkill(d1); mirkill(d2);
    mirkill(bTemp); mirkill(c1);mirkill(c2);mirkill(c3);
    mirexit();

    printf("\n\n\n\n");
    return 0;
}

几组可用的注册码：
342AE11C959F35FE342AE11C959F35FE54BFCA2DFA071050
8886DA1EF4C9EF0B8886DA1EF4C9EF0B3A9E704477BA7D31
68918DCE58A45DB468918DCE58A45DB44F11A44999044928

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【经验总结】
(1)一些常见的大数库一定要熟悉；
(2)如果想深入密码学算法，务必多动手分析这方面的CrackMe。

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【版权声明】: 本文原创于看雪技术论坛, 转载请注明作者并保持文章的完整, 谢谢!

2007年01月11日 13:56:12

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