攻防世界--IgniteMe

测试文件：https://adworld.xctf.org.cn/media/task/attachments/fac4d1290e604fdfacbbe06fd1a5ca39.exe

1.准备

获取信息：

• 32位文件

2.IDA打开

打开main函数

 1 int __cdecl main(int argc, const char **argv, const char **envp)
 2 {
 3   void *v3; // eax
 4   int v4; // edx
 5   void *v5; // eax
 6   int result; // eax
 7   void *v7; // eax
 8   void *v8; // eax
 9   void *v9; // eax
10   size_t i; // [esp+4Ch] [ebp-8Ch]
11   char v11[4]; // [esp+50h] [ebp-88h]
12   char v12[28]; // [esp+58h] [ebp-80h]
13   char v13; // [esp+74h] [ebp-64h]
14 
15   v3 = (void *)sub_402B30(&unk_446360, "Give me your flag:");// 这两段代码直接理解成printf即可。下面的代码同样如此
16   sub_4013F0(v3, (int (__cdecl *)(void *))sub_403670);
17   sub_401440((int)&dword_4463F0, v4, (int)v12, 127);
18   if ( strlen(v12) < 30 && strlen(v12) > 4 )
19   {
20     strcpy(v11, "EIS{");
21     for ( i = 0; i < strlen(v11); ++i )
22     {
23       if ( v12[i] != v11[i] )                   // flag前四位为"ESI{"
24       {
25         v7 = (void *)sub_402B30(&unk_446360, "Sorry, keep trying! ");
26         sub_4013F0(v7, (int (__cdecl *)(void *))sub_403670);
27         return 0;
28       }
29     }
30     if ( v13 == 125 )
31     {
32       if ( sub_4011C0(v12) )
33         v9 = (void *)sub_402B30(&unk_446360, "Congratulations! ");
34       else
35         v9 = (void *)sub_402B30(&unk_446360, "Sorry, keep trying! ");
36       sub_4013F0(v9, (int (__cdecl *)(void *))sub_403670);
37       result = 0;
38     }
39     else
40     {
41       v8 = (void *)sub_402B30(&unk_446360, "Sorry, keep trying! ");
42       sub_4013F0(v8, (int (__cdecl *)(void *))sub_403670);
43       result = 0;
44     }
45   }
46   else
47   {
48     v5 = (void *)sub_402B30(&unk_446360, "Sorry, keep trying!");
49     sub_4013F0(v5, (int (__cdecl *)(void *))sub_403670);
50     result = 0;
51   }
52   return result;
53 }

3.代码分析

看到第32行代码

if ( sub_4011C0(v12) )
    v9 = (void *)sub_402B30(&unk_446360, "Congratulations! ");

这里通过sub_4011C0(v12)传入输入的flag来判断真假，打开函数

 1 bool __cdecl sub_4011C0(char *a1)
 2 {
 3   size_t v2; // eax
 4   signed int v3; // [esp+50h] [ebp-B0h]
 5   char v4[32]; // [esp+54h] [ebp-ACh]
 6   int v5; // [esp+74h] [ebp-8Ch]
 7   int v6; // [esp+78h] [ebp-88h]
 8   size_t i; // [esp+7Ch] [ebp-84h]
 9   char v8[128]; // [esp+80h] [ebp-80h]
10 
11   if ( strlen(a1) <= 4 )
12     return 0;
13   i = 4;
14   v6 = 0;
15   while ( i < strlen(a1) - 1 )
16     v8[v6++] = a1[i++];
17   v8[v6] = 0;
18   v5 = 0;
19   v3 = 0;
20   memset(v4, 0, 0x20u);
21   for ( i = 0; ; ++i )
22   {
23     v2 = strlen(v8);
24     if ( i >= v2 )
25       break;
26     if ( v8[i] >= 97 && v8[i] <= 122 )
27     {
28       v8[i] -= 32;
29       v3 = 1;
30     }
31     if ( !v3 && v8[i] >= 65 && v8[i] <= 90 )
32       v8[i] += 32;
33     v4[i] = byte_4420B0[i] ^ sub_4013C0(v8[i]);
34     v3 = 0;
35   }
36   return strcmp("GONDPHyGjPEKruv{{pj]X@rF", v4) == 0;
37 }

这里i是从4开始即flag的第四位开始，对flag的操作分为了两部分：

1. 第26行代码~第32行代码，将flag中的大写字母转小写，小写字母转大写。
2. 第33行代码对每位字符进行异或操作。

第一步不必多说，第二步byte_4420B0数组从文件中提取出来

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0D 13 17 11 02 01 20 1D  0C 02 19 2F 17 2B 24 1F
1E 16 09 0F 15 27 13 26  0A 2F 1E 1A 2D 0C 22 04

 sub_4013C0(v8[i])函数为

int __cdecl sub_4013C0(int a1)
{
  return (a1 ^ 0x55) + 72;
}

最后得到的字符串V4为

GONDPHyGjPEKruv{{pj]X@rF

因此我们只需要逆向操作还原flag即可

4.脚本获取

n = 28
val1 = [0x0D,0x13,0x17,0x11,0x02,0x01,0x20,0x1D,0x0C,0x02,0x19,0x2F,0x17,0x2B,
        0x24,0x1F,0x1E,0x16,0x09,0x0F,0x15,0x27,0x13,0x26,0x0A,0x2F,0x1E,0x1A,
        0x2D,0x0C,0x22,0x04]
v4 = "GONDPHyGjPEKruv{{pj]X@rF"
v8 = ""
flag = ""

for i in range(len(v4)):
    v8 += chr(((ord(v4[i]) ^ val1[i]) - 72) ^ 0x55)

for i in range(len(v8)):
    if ord(v8[i]) >= 97 and ord(v8[i]) <= 122:
        flag += chr(ord(v8[i]) - 32)
    elif ord(v8[i]) >= 65 and ord(v8[i]) <= 90:
        flag += chr(ord(v8[i]) + 32)
    else:
        flag += v8[i]

print('EIS{'+flag+'}')

5.get flag！

EIS{wadx_tdgk_aihc_ihkn_pjlm}