BUUCTF-RE-reverse3

没有信息

丢进IDA

 1 __int64 __cdecl main_0()
 2 {
 3   int v0; // eax
 4   const char *v1; // eax
 5   size_t v2; // eax
 6   int v3; // edx
 7   __int64 v4; // ST08_8
 8   signed int j; // [esp+DCh] [ebp-ACh]
 9   signed int i; // [esp+E8h] [ebp-A0h]
10   signed int v8; // [esp+E8h] [ebp-A0h]
11   char Dest[108]; // [esp+F4h] [ebp-94h]
12   char Str; // [esp+160h] [ebp-28h]
13   char v11; // [esp+17Ch] [ebp-Ch]
14 
15   for ( i = 0; i < 100; ++i )
16   {
17     if ( (unsigned int)i >= 0x64 )
18       j____report_rangecheckfailure();
19     Dest[i] = 0;
20   }
21   sub_41132F("please enter the flag:");
22   sub_411375("%20s", &Str);
23   v0 = j_strlen(&Str);
24   v1 = (const char *)sub_4110BE((int)&Str, v0, (int)&v11);
25   strncpy(Dest, v1, 0x28u);
26   v8 = j_strlen(Dest);
27   for ( j = 0; j < v8; ++j )
28     Dest[j] += j;
29   v2 = j_strlen(Dest);
30   if ( !strncmp(Dest, Str2, v2) )
31     sub_41132F("rigth flag!\n");
32   else
33     sub_41132F("wrong flag!\n");
34   HIDWORD(v4) = v3;
35   LODWORD(v4) = 0;
36   return v4;
37 }

首先找到的信息就是Str2，Str2中存储的是flag变换后的字符串

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

第24行就是在对原str进行操作

然后把str的值给Dest

进入24行的函数

 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 }

在该函数的后半部分，Dst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+/='可知，这个函数应该是base64的加密函数，因此只需要解密即可。

解题脚本如下：

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)

详情参考：https://www.cnblogs.com/Mayfly-nymph/p/11465643.html