![re_[SWPU201919]ReverseMe](/imgs/citynight1.jpg)
一道加密过程花里胡哨的逆向题
1
exe文件，无壳。直接拖入IDA中，根据关键字符串可定位至主要函数中:
int __thiscall sub_EC2810(void *this)
{
int v1; // eax
int v2; // ecx
int v3; // eax
int v4; // esi
signed int v5; // edi
unsigned int v6; // kr00_4
int *v7; // ecx
int *v8; // ecx
int *v9; // ecx
__int128 *v10; // edx
unsigned int v11; // edi
int v12; // eax
int v13; // eax
bool v14; // cf
unsigned __int8 v15; // al
unsigned __int8 v16; // al
unsigned __int8 v17; // al
const char *v18; // edx
int v19; // eax
int v20; // eax
int v22; // [esp-14h] [ebp-D8h]
int v23; // [esp-10h] [ebp-D4h]
int *input; // [esp+24h] [ebp-A0h]
int len_of_input; // [esp+34h] [ebp-90h]
unsigned int v26; // [esp+38h] [ebp-8Ch]
__int128 v27; // [esp+3Ch] [ebp-88h]
__int128 v28; // [esp+4Ch] [ebp-78h]
int v29; // [esp+5Ch] [ebp-68h]
__int128 v30; // [esp+60h] [ebp-64h]
__int128 v31; // [esp+70h] [ebp-54h]
int v32; // [esp+80h] [ebp-44h]
__int64 v33; // [esp+84h] [ebp-40h]
int v34; // [esp+8Ch] [ebp-38h]
__int16 v35; // [esp+90h] [ebp-34h]
int v36; // [esp+94h] [ebp-30h]
int v37; // [esp+98h] [ebp-2Ch]
int v38; // [esp+9Ch] [ebp-28h]
int v39; // [esp+A0h] [ebp-24h]
int v40; // [esp+A4h] [ebp-20h]
int v41; // [esp+A8h] [ebp-1Ch]
int v42; // [esp+ACh] [ebp-18h]
int v43; // [esp+B0h] [ebp-14h]
int v44; // [esp+C0h] [ebp-4h]
len_of_input = 0;
v26 = 15;
LOBYTE(input) = 0;
v44 = 0;
LOBYTE(v44) = 1;
v1 = print(this, "Please input your flag: ");
sub_EC3050(v1);
scan(&dword_EF0068, &input);
v34 = 1413701433;
v33 = qword_EEB9A0;
v35 = 70;
if ( len_of_input == 32 )
{
v40 = -1173078761;
v41 = 494076752;
v42 = -1811652486;
v43 = 688582768;
v5 = 0;
v30 = 0i64;
v32 = 0;
v31 = 0i64;
v6 = strlen((const char *)&v33);
do
{
v7 = (int *)&input;
if ( v26 >= 0x10 )
v7 = input;
*((_BYTE *)v7 + v5) ^= *((_BYTE *)&v33 + v5 % v6);// 循环异或
++v5;
}
while ( v5 < 32 );
v8 = (int *)&input;
v4 = (int)input;
if ( v26 >= 0x10 )
v8 = input;
v29 = 0;
v27 = 0i64;
v28 = 0i64;
*(_QWORD *)&v30 = *(_QWORD *)v8;
*((_QWORD *)&v30 + 1) = *((_QWORD *)v8 + 1);
*(_QWORD *)&v31 = *((_QWORD *)v8 + 2);
*((_QWORD *)&v31 + 1) = *((_QWORD *)v8 + 3);
complex_encrypt(v22, v23, 256, (unsigned int)&v30, (unsigned int)&v27);
v36 = 0xF80F37B3;
v37 = 0x5DAEBCBC;
v9 = &v36;
v38 = 0x864D5ABA;
v10 = &v27;
v39 = 0xD3629744;
v11 = 28;
v40 = 0x1624BA4F;
v41 = 0x1A729F0B;
v42 = 0x266D6865;
v43 = 0x67C86BBA;
while ( 1 )
{
v12 = *v9;
if ( *v9 != *(_DWORD *)v10 )
break;
++v9;
v10 = (__int128 *)((char *)v10 + 4);
v14 = v11 < 4;
v11 -= 4;
if ( v14 )
{
v13 = 0;
goto LABEL_19;
}
}
v14 = (unsigned __int8)v12 < *(_BYTE *)v10;
if ( (_BYTE)v12 != *(_BYTE *)v10
|| (v15 = *((_BYTE *)v9 + 1), v14 = v15 < *((_BYTE *)v10 + 1), v15 != *((_BYTE *)v10 + 1))
|| (v16 = *((_BYTE *)v9 + 2), v14 = v16 < *((_BYTE *)v10 + 2), v16 != *((_BYTE *)v10 + 2))
|| (v17 = *((_BYTE *)v9 + 3), v14 = v17 < *((_BYTE *)v10 + 3), v17 != *((_BYTE *)v10 + 3)) )
{
v13 = -v14 | 1;
}
else
{
v13 = 0;
}
LABEL_19:
if ( v13 )
v18 = "Try again!\r\n";
else
v18 = "Congratulations! I always knew you could do it.";
v19 = print(v9, v18);
sub_EC3050(v19);
system("pause");
}
else
{
v3 = print(v2, "Try again!\r\n");
sub_EC3050(v3);
system("pause");
v4 = (int)input;
}
if ( v26 >= 0x10 )
{
v20 = v4;
if ( v26 + 1 >= 0x1000 )
{
v4 = *(_DWORD *)(v4 - 4);
if ( (unsigned int)(v20 - v4 - 4) > 0x1F )
sub_ECAFF7(v26 + 36);
}
sub_EC64DE(v4);
}
return 0;
}可以看出69到76行这一部分是一个循环异或，将输入的字符串与”SWPU_2019_CTF”进行循环异或，然后将异或后的字符串放入v30，再传入complex_encrypt()，加密结果通过v27带出，与v36进行比较。
2
那么再来看一下complex_encrypt()函数:
void __cdecl complex_encrypt(int a1, int a2, int a3, unsigned int a4_v30, unsigned int a5_v27)
{
int v5; // ecx
unsigned int v6; // ebx
int v7; // esi
unsigned int v8; // edi
unsigned int v9; // esi
unsigned int v10; // edx
unsigned int v11; // esi
_DWORD *v12; // ecx
unsigned int v13; // eax
int v14; // ebx
unsigned int v15; // esi
__m128i v16; // xmm0
__m128i v17; // xmm1
__m128i v18; // xmm0
__m128i v19; // xmm1
__m128i v20; // xmm0
__m128i v21; // xmm1
__m128i v22; // xmm0
__m128i v23; // xmm1
unsigned int v24; // ebx
int v25; // eax
unsigned int v26; // esi
int v27; // edi
int v28; // ecx
signed int v29; // [esp+20h] [ebp-20h]
int v30; // [esp+24h] [ebp-1Ch]
_DWORD *v31; // [esp+28h] [ebp-18h]
int v32; // [esp+2Ch] [ebp-14h]
int v33; // [esp+30h] [ebp-10h]
int v34; // [esp+34h] [ebp-Ch]
int v35; // [esp+38h] [ebp-8h]
v6 = a5_v27;
v7 = v5;
v8 = (unsigned int)(a3 + 31) >> 5;
v30 = 4 * v8;
v31 = (_DWORD *)sub_ECB048(4 * v8);
v32 = -1839987866;
v33 = 120;
v34 = -1839987866;
v35 = 120;
sub_EC2270(v7, &v32);
sub_EC20E0();
sub_EC2150();
sub_EC1F80();
v29 = 0;
if ( v8 )
{
do
{
dword_EF0D94 = 2 * dword_EF0DA4 ^ (unsigned __int16)(dword_EF0DBC ^ 2 * dword_EF0DA4);
dword_EF0D78 = (dword_EF0DB8 << 16) | ((unsigned int)dword_EF0DAC >> 15);
v9 = (dword_EF0D70 << 16) | ((unsigned int)dword_EF0D90 >> 15);
dword_EF0D8C = (dword_EF0D68 << 16) | ((unsigned int)dword_EF0D84 >> 15);
dword_EF0D7C = (dword_EF0D70 << 16) | ((unsigned int)dword_EF0D90 >> 15);
v31[v29] = v9 ^ sub_EC2150();
sub_EC1F80();
++v29;
}
while ( v29 < (signed int)v8 );
v6 = a5_v27;
}
v10 = 0;
if ( v8 )
{
v11 = a4_v30;
if ( v8 < 0x10 || v6 <= a4_v30 + v30 - 4 && v6 + v30 - 4 >= a4_v30 )
{
v12 = v31;
}
else
{
v12 = v31;
if ( v6 > (unsigned int)&v31[v8 - 1] || v6 + v30 - 4 < (unsigned int)v31 )
{
v13 = a4_v30 + 16;
v12 = v31;
v14 = v6 + 32;
v15 = (unsigned int)v31 - a4_v30;
do
{
v16 = *(__m128i *)(v13 - 16);
v13 += 64;
v14 += 64;
v17 = _mm_xor_si128(*(__m128i *)&v31[v10], v16);
v18 = *(__m128i *)(v13 - 64);
*(__m128i *)(v14 - 96) = v17;
v19 = _mm_xor_si128(*(__m128i *)(v15 + v13 - 64), v18);
v20 = *(__m128i *)(v13 - 48);
*(__m128i *)(a5_v27 - a4_v30 + v13 - 64) = v19;
v15 = (unsigned int)v31 - a4_v30;
v21 = _mm_xor_si128(*(__m128i *)((char *)v31 + v14 - a5_v27 - 64), v20);
v22 = *(__m128i *)(v13 - 32);
*(__m128i *)(v14 - 64) = v21;
v23 = *(__m128i *)&v31[v10 + 12];
v10 += 16;
*(__m128i *)(v14 - 48) = _mm_xor_si128(v23, v22);
}
while ( v10 < (v8 & 0xFFFFFFF0) );
v6 = a5_v27;
v11 = a4_v30;
}
}
if ( v10 < v8 )
{
v24 = v6 - a4_v30;
v25 = v11 + 4 * v10;
v26 = (unsigned int)v12 - a4_v30;
v27 = v8 - v10;
do
{
v28 = *(_DWORD *)(v26 + v25); // v12 + 4 * v10
v25 += 4;
*(_DWORD *)(v24 + v25 - 4) = *(_DWORD *)(v25 - 4) ^ v28;
--v27;
}
while ( v27 );
}
}
sub_ECAE05(v31);
}第一眼看上去非常复杂，而且还用到了xmm寄存器。但仔细分析后就可以发现106行以前的部分都是在固定地生成一张表，真正的加密在106以后的一小部分，其实是就是把输入的字符串与表进行了一次异或。通过动调可以获得这个表。
3
于是可以写出脚本:
table = [0xca3e0c86, 0x19aed798, 0xa66b77e2, 0xb077a16a, 0x05379169, 0x307bf97a, 0x104b5a43, 0x28d47d86]
res = [0xF80F37B3, 0x5DAEBCBC, 0x864D5ABA, 0xD3629744, 0x1624BA4F, 0x1A729F0B, 0x266D6865, 0x67C86BBA]
c = []
for i in range(8):
c.append(res[i] ^ table[i])
b = ""
for i in c:
temp = hex(i)
b += temp[8:10]
b += temp[6:8]
b += temp[4:6]
b += temp[2:4]
blist = []
for i in range(len(b)):
if i % 2 == 0:
temp = b[i:i+2]
blist.append(int(temp, 16))
a = "SWPU_2019_CTF"
for i in range(len(blist)):
print(chr(blist[i] ^ ord(a[i%len(a)])), end="")运行后得到flag{Y0uaretheB3st!#@_VirtualCC}
4
灵活运用动调。
另外，感谢做题时robot-r sf在一旁提出了一些指导，差点就强行分析了一波complex_encrypt()。。。