测试文件：https://www.lanzous.com/i9la55a
准备
获取信息：
- 32位文件
- UPX壳
IDA分析
UPX脱壳后，IDA打开
int sub_804887C()
{
int v0; // eax
int result; // eax
int v2; // ecx
unsigned int v3; // et1
unsigned int i; // [esp+Ch] [ebp-FCh]
unsigned int v5; // [esp+10h] [ebp-F8h]
unsigned int v6; // [esp+14h] [ebp-F4h]
int v7; // [esp+1Ah] [ebp-EEh]
int v8; // [esp+1Eh] [ebp-EAh]
int v9; // [esp+22h] [ebp-E6h]
int v10; // [esp+26h] [ebp-E2h]
__int16 v11; // [esp+2Ah] [ebp-DEh]
char v12[30]; // [esp+2Ch] [ebp-DCh]
int v13; // [esp+4Ah] [ebp-BEh]
int v14; // [esp+4Eh] [ebp-BAh]
int v15; // [esp+52h] [ebp-B6h]
int v16; // [esp+56h] [ebp-B2h]
int v17; // [esp+5Ah] [ebp-AEh]
int v18; // [esp+5Eh] [ebp-AAh]
int v19; // [esp+62h] [ebp-A6h]
int v20; // [esp+66h] [ebp-A2h]
int v21; // [esp+6Ah] [ebp-9Eh]
int v22; // [esp+6Eh] [ebp-9Ah]
int v23; // [esp+72h] [ebp-96h]
int v24; // [esp+76h] [ebp-92h]
__int16 v25; // [esp+7Ah] [ebp-8Eh]
char v26; // [esp+7Ch] [ebp-8Ch]
unsigned int v27; // [esp+FCh] [ebp-Ch]
v27 = __readgsdword(0x14u);
v7 = ‘galF‘;
v8 = ‘ihT{‘;
v9 = ‘_a_s‘;
v10 = ‘galF‘;
v11 = ‘}‘;
v5 = strlen(&v7);
v13 = ‘8D8E‘;
v14 = ‘19DB‘;
v15 = ‘A178‘;
v16 = ‘65E1‘;
v17 = ‘F35F‘;
v18 = ‘9884‘;
v19 = ‘F286‘;
v20 = ‘4169‘;
v21 = ‘2FA2‘;
v22 = ‘F8BA‘;
v23 = ‘A7DE‘;
v24 = ‘5DFC‘;
v25 = ‘E‘;
printf("Please input your flag:");
read(0, &v26, 256);
if ( strlen(&v26) != 21 )
exit(0);
v0 = sub_8048AC2((int)&v26);
strcpy((int)v12, v0);
v6 = length(v12);
for ( i = 0; i < v6; ++i ) // 异或操作
v12[i] ^= *((_BYTE *)&v7 + i % v5);
sub_8048E24(v12, v6, &v7, v5);
if ( !strcmp(v12, &v13) )
exit(0);
printf("right!");
result = 0;
v3 = __readgsdword(0x14u);
v2 = v3 ^ v27;
if ( v3 != v27 )
sub_806FA00(v2);
return result;
}
很多未知函数，我们能够猜出是什么函数，在代码中已改。
代码分析
在代码中，有两处函数我们不清楚作用。第一处
v0 = sub_8048AC2((int)&v26);
函数传入参数是我们输入的字符串，打开函数后
int __cdecl sub_8048AC2(int a1)
{
int v2; // [esp+8h] [ebp-20h]
int v3; // [esp+Ch] [ebp-1Ch]
int v4; // [esp+10h] [ebp-18h]
int v5; // [esp+18h] [ebp-10h]
int v6; // [esp+1Ch] [ebp-Ch]
v5 = length(a1);
if ( v5 % 3 )
v2 = 4 * (v5 / 3 + 1);
else
v2 = 4 * (v5 / 3);
v6 = sub_80597A0(v2 + 1);
*(_BYTE *)(v2 + v6) = 0;
v3 = 0;
v4 = 0;
while ( v2 - 2 > v3 )
{
*(_BYTE *)(v6 + v3) = a0123456789Abcd[(unsigned __int8)(*(_BYTE *)(v4 + a1) >> 2)];
*(_BYTE *)(v6 + v3 + 1) = a0123456789Abcd[16 * (*(_BYTE *)(v4 + a1) & 3) | (unsigned __int8)(*(_BYTE *)(v4 + 1 + a1) >> 4)];
*(_BYTE *)(v6 + v3 + 2) = a0123456789Abcd[4 * (*(_BYTE *)(v4 + 1 + a1) & 0xF) | (unsigned __int8)(*(_BYTE *)(v4 + 2 + a1) >> 6)];
*(_BYTE *)(v6 + v3 + 3) = a0123456789Abcd[*(_BYTE *)(v4 + 2 + a1) & 0x3F];
v4 += 3;
v3 += 4;
}
if ( v5 % 3 == 1 )
{
*(_BYTE *)(v3 - 2 + v6) = 61;
*(_BYTE *)(v3 - 1 + v6) = 61;
}
else if ( v5 % 3 == 2 )
{
*(_BYTE *)(v3 - 1 + v6) = 61;
}
return v6;
}
.rodata:080BB9A8 a0123456789Abcd db ‘0123456789+/abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ‘,0
通过观察，这应该是一个变表的Base64加密
第二处
sub_8048E24((int)v10, v4, (int)&v5, v3)
unsigned int __cdecl sub_8048E24(int a1, unsigned int a2, int a3, int a4)
{
char v4; // ST2B_1
unsigned int result; // eax
unsigned int v6; // et1
int v7; // [esp+1Ch] [ebp-11Ch]
int v8; // [esp+20h] [ebp-118h]
unsigned int i; // [esp+24h] [ebp-114h]
char v10[256]; // [esp+2Ch] [ebp-10Ch]
unsigned int v11; // [esp+12Ch] [ebp-Ch]
v11 = __readgsdword(0x14u);
sub_8048CC2((int)v10, a3, a4);
LOBYTE(v7) = 0;
LOBYTE(v8) = 0;
for ( i = 0; i < a2; ++i )
{
v7 = (unsigned __int8)(v7 + 1);
v8 = (unsigned __int8)(v10[v7] + v8);
v4 = v10[v7];
v10[v7] = v10[v8];
v10[v8] = v4;
*(_BYTE *)(a1 + i) ^= v10[(unsigned __int8)(v10[v7] + v10[v8])];
}
v6 = __readgsdword(0x14u);
result = v6 ^ v11;
if ( v6 != v11 )
sub_806FA00();
return result;
}
这应该是一个RC4加密函数， sub_8048CC2((int)v10, a3, a4);函数是初始化函数
思路
这样整个代码的思路就很明晰了，对输入字符串进行base64加密后，再进行异或操作，最后进行RC4加密(Key = "Flag{This_a_Flag}")，并与"E8D8BD91871A1E56F53F4889682F96142AF2AB8FED7ACFD5E"比较
解密
解密网站：
http://tool.chacuo.net/cryptrc4
http://ctf.ssleye.com/rc4.html
http://tools.jb51.net/password/rc4_encode
https://gchq.github.io/CyberChef/
得到解密的字符串：23152553081a5938126a3931275b0b1313085c330b356101511f105c
# -*- coding:utf-8 -*-
from base64 import b64decode
key = ‘Flag{This_a_Flag}‘
decode_byte = ‘23152553081a5938126a3931275b0b1313085c330b356101511f105c‘
encode_base64 = ‘‘
lists = []
for i in range(len(decode_byte)//2):
lists.append(int(decode_byte[i*2:(i+1)*2],16))
# for i in range(0,len(decode_byte),2):
# lists.append(int(decode_byte[i:i+2],16))
print (lists)
for i in range(len(lists)):
encode_base64 += chr(lists[i] ^ ord(key[i%len(key)]))
t = ‘0123456789+/abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ=‘
table = ‘ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=‘
table = str.maketrans(t, table)
flag = b64decode(encode_base64.translate(table))
print(flag)
get flag!
BJD{0v0_Y0u_g07_1T!}
E8D8BD91871A1E56F53F4889682F96142AF2AB8FED7ACFD5E
原文地址：https://www.cnblogs.com/Mayfly-nymph/p/12347921.html
时间： 2024-11-09 00:35:15