• BUUCTF--reverse3


    测试文件:https://buuoj.cn/files/aa4f6c7e8d5171d520b95420ee570e79/a9d22a0e-928d-4bb4-8525-e38c9481469e.rar?token=eyJ0ZWFtX2lkIjpudWxsLCJ1c2VyX2lkIjoxOTAzLCJmaWxlX2lkIjoxNTl9.XXCjMA.QBNzEL8Ujfk_QRbQYa6XZ3dYjQ4

    1.准备

    获得信息

    1. 32位文件

    2.IDA打开

    将main函数反编译为C语言代码

    int __cdecl main(int argc, const char **argv, const char **envp)
    {
      int v3; // ebx
      int v4; // edi
      int v5; // esi
    
      return main_0(v3, v4, v5);
    }

    打开main_0函数

     1 __int64 __usercall main_0@<edx:eax>(int a1@<ebx>, int a2@<edi>, int a3@<esi>)
     2 {
     3   int v3; // eax
     4   const char *v4; // eax
     5   size_t v5; // eax
     6   int v6; // edx
     7   __int64 v7; // ST08_8
     8   signed int j; // [esp+DCh] [ebp-ACh]
     9   signed int i; // [esp+E8h] [ebp-A0h]
    10   signed int v11; // [esp+E8h] [ebp-A0h]
    11   char Dest[108]; // [esp+F4h] [ebp-94h]
    12   char Str; // [esp+160h] [ebp-28h]
    13   char v14; // [esp+17Ch] [ebp-Ch]
    14 
    15   for ( i = 0; i < 100; ++i )
    16   {
    17     if ( (unsigned int)i >= 0x64 )
    18       j____report_rangecheckfailure(a1, a2, a3);
    19     Dest[i] = 0;
    20   }
    21   sub_41132F("please enter the flag:");
    22   sub_411375("%20s", &Str);
    23   v3 = j_strlen(&Str);
    24   v4 = (const char *)sub_4110BE((int)&Str, v3, (int)&v14);
    25   strncpy(Dest, v4, 0x28u);
    26   v11 = j_strlen(Dest);
    27   for ( j = 0; j < v11; ++j )
    28     Dest[j] += j;
    29   v5 = j_strlen(Dest);
    30   if ( !strncmp(Dest, Str2, v5) )
    31     sub_41132F("rigth flag!
    ");
    32   else
    33     sub_41132F("wrong flag!
    ");
    34   HIDWORD(v7) = v6;
    35   LODWORD(v7) = 0;
    36   return v7;
    37 }

    2.1代码分析

    第24行代码,第27~28行代码在对Dest进行一些列变换

    从第30行代码分析得到,Str2中存储的就是flag变换过后的字符串,打开Str2

    .data:0041A034 ; char Str2[]
    .data:0041A034 Str2            db 'e3nifIH9b_C@n@dH',0 ; DATA XREF: _main_0+142↑o

    因此,我们只需要将Str2反向变换即可得到flag。

    第27~28行代码的变换没有难度,再来看看第24行代码的函数。

     1 void *__cdecl sub_411AB0(char *a1, unsigned int a2, int *a3)
     2 {
     3   int v4; // STE0_4
     4   int v5; // STE0_4
     5   int v6; // STE0_4
     6   int v7; // [esp+D4h] [ebp-38h]
     7   signed int i; // [esp+E0h] [ebp-2Ch]
     8   unsigned int v9; // [esp+ECh] [ebp-20h]
     9   int v10; // [esp+ECh] [ebp-20h]
    10   signed int v11; // [esp+ECh] [ebp-20h]
    11   void *Dst; // [esp+F8h] [ebp-14h]
    12   char *v13; // [esp+104h] [ebp-8h]
    13 
    14   if ( !a1 || !a2 )
    15     return 0;
    16   v9 = a2 / 3;
    17   if ( (signed int)(a2 / 3) % 3 )
    18     ++v9;
    19   v10 = 4 * v9;
    20   *a3 = v10;
    21   Dst = malloc(v10 + 1);
    22   if ( !Dst )
    23     return 0;
    24   j_memset(Dst, 0, v10 + 1);
    25   v13 = a1;
    26   v11 = a2;
    27   v7 = 0;
    28   while ( v11 > 0 )
    29   {
    30     byte_41A144[2] = 0;
    31     byte_41A144[1] = 0;
    32     byte_41A144[0] = 0;
    33     for ( i = 0; i < 3 && v11 >= 1; ++i )
    34     {
    35       byte_41A144[i] = *v13;
    36       --v11;
    37       ++v13;
    38     }
    39     if ( !i )
    40       break;
    41     switch ( i )
    42     {
    43       case 1:
    44         *((_BYTE *)Dst + v7) = aAbcdefghijklmn[(signed int)(unsigned __int8)byte_41A144[0] >> 2];
    45         v4 = v7 + 1;
    46         *((_BYTE *)Dst + v4++) = aAbcdefghijklmn[((byte_41A144[1] & 0xF0) >> 4) | 16 * (byte_41A144[0] & 3)];
    47         *((_BYTE *)Dst + v4++) = aAbcdefghijklmn[64];
    48         *((_BYTE *)Dst + v4) = aAbcdefghijklmn[64];
    49         v7 = v4 + 1;
    50         break;
    51       case 2:
    52         *((_BYTE *)Dst + v7) = aAbcdefghijklmn[(signed int)(unsigned __int8)byte_41A144[0] >> 2];
    53         v5 = v7 + 1;
    54         *((_BYTE *)Dst + v5++) = aAbcdefghijklmn[((byte_41A144[1] & 0xF0) >> 4) | 16 * (byte_41A144[0] & 3)];
    55         *((_BYTE *)Dst + v5++) = aAbcdefghijklmn[((byte_41A144[2] & 0xC0) >> 6) | 4 * (byte_41A144[1] & 0xF)];
    56         *((_BYTE *)Dst + v5) = aAbcdefghijklmn[64];
    57         v7 = v5 + 1;
    58         break;
    59       case 3:
    60         *((_BYTE *)Dst + v7) = aAbcdefghijklmn[(signed int)(unsigned __int8)byte_41A144[0] >> 2];
    61         v6 = v7 + 1;
    62         *((_BYTE *)Dst + v6++) = aAbcdefghijklmn[((byte_41A144[1] & 0xF0) >> 4) | 16 * (byte_41A144[0] & 3)];
    63         *((_BYTE *)Dst + v6++) = aAbcdefghijklmn[((byte_41A144[2] & 0xC0) >> 6) | 4 * (byte_41A144[1] & 0xF)];
    64         *((_BYTE *)Dst + v6) = aAbcdefghijklmn[byte_41A144[2] & 0x3F];
    65         v7 = v6 + 1;
    66         break;
    67     }
    68   }
    69   *((_BYTE *)Dst + v7) = 0;
    70   return Dst;
    71 }

    在代码的下部分,可以看到Dst会经过aAbcdefghijklmn[]数组的变换,打开此处

    .rdata:00417B30 aAbcdefghijklmn db 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/='
    .rdata:00417B30                                         ; DATA XREF: .text:004117E8↑o
    .rdata:00417B30                                         ; .text:00411827↑o ...
    .rdata:00417B30                 db 0
    .rdata:00417B72                 align 4

    从'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/='可以看出,这个函数应该是base64的加密函数,因此只需要解密即可。

    3.脚本获取flag

    import base64
    
    str1 = 'e3nifIH9b_C@n@dH'
    x = ''
    flag = ''
    
    for j in range(0, len(str1)):
        x += chr(ord(str1[j]) - j)
    
    flag = base64.b64decode(x)
    flag = flag.decode('ASCII')
    print(flag)

    4.get flag!

    flag{i_l0ve_you}

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  • 原文地址:https://www.cnblogs.com/Mayfly-nymph/p/11465643.html
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