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! "); 32 else 33 sub_41132F("wrong flag! "); 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)