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