2019 上海市大学生网络安全大赛 RE部分WP
这次比赛就做了这一道逆向题，看到队友的WP，下面的对v10的加密方式为RC4，从我提取的v4数组就能够察觉出这是CR4了，自己傻乎乎的用OD调试，跟踪数据半天才做出来，还是见得的少了... ...下面有几篇不错的RC4的文章：
C语言实现：https://zhoujianshi.github.io/articles/2016/RC4%E5%8A%A0%E5%AF%86%E7%AE%97%E6%B3%95/index.html
Python实现：https://specters.top/2019/03/01/Python%E5%AE%9E%E7%8E%B0RC4/
文件下载：https://www.lanzous.com/i76vcne
准备
获取信息
- 32位文件
IDA打开
int __cdecl main(int argc, const char **argv, const char **envp) { _DWORD *v4; // [esp+1Ch] [ebp-A0h] int v5; // [esp+20h] [ebp-9Ch] int v6; // [esp+24h] [ebp-98h] int v7; // [esp+28h] [ebp-94h] int v8; // [esp+2Ch] [ebp-90h] char v9; // [esp+30h] [ebp-8Ch] int v10; // [esp+50h] [ebp-6Ch] int v11; // [esp+70h] [ebp-4Ch] sub_402620(); v5 = 1701148529; v6 = 101; v7 = 0; v8 = 0; v4 = malloc(0x408u); puts("Plz solve the puzzle:"); sub_40ED00("%32s", &v9); if ( (unsigned __int8)sub_401C70(&v9) && (sub_401B60((int)&v10, &v9), sub_401850(v4, (int)&v5, strlen((const char *)&v5)), sub_4018D0(v4, &v10, 8), (unsigned __int8)sub_401950(&v10)) ) { sub_401BA0(&v9, (int)&v11); sub_40ED20("Congrats!\n%s\n", &v11); } else { puts("Failed!"); } return 0; }
通过分析，要得到flag，则需要if中的条件成立
if ( (unsigned __int8)sub_401C70(&v9) && (sub_401B60((int)&v10, &v9), sub_401850(v4, (int)&v5, strlen((const char *)&v5)), sub_4018D0(v4, &v10, 8), (unsigned __int8)sub_401950(&v10)) )
代码分析
打开sub_401C70(&v9)函数
int __cdecl sub_401C70(char *a1) { char *v2; // edx if ( strlen(a1) != 16 ) return 0; v2 = a1; while ( (unsigned __int8)(*v2 - 58) > 38u && (unsigned __int8)(*v2 - 48) <= 54u )// 输入的字符串每个字符都是'a'~'f'或者'0'~'9'之间 { if ( ++v2 == a1 + 16 ) // 遍历到字符串末尾结束，输入字符串长度为16 return 1; } return 0; }
暴力解v10
因为&&符号后面是( func1, func2, func3)形式的判断条件，因此我先直接看func3即可（func3决定条件是否成立），打开sub_401950(&v10)函数
bool __cdecl sub_401950(_BYTE *a1) { _BYTE *v1; // ecx bool result; // al v1 = a1; while ( 2 ) { switch ( *v1 ) { case 0: dword_40F028 &= dword_40F038; dword_40F02C *= dword_40F028; goto LABEL_4; case 1: if ( !dword_40F02C ) goto LABEL_6; dword_40F028 /= dword_40F02C; dword_40F024 += dword_40F034; goto LABEL_4; case 2: dword_40F030 ^= dword_40F034; dword_40F03C += dword_40F020; goto LABEL_4; case 3: dword_40F03C -= dword_40F030; dword_40F030 &= dword_40F024; goto LABEL_4; case 4: dword_40F034 *= dword_40F020; dword_40F02C -= dword_40F038; goto LABEL_4; case 5: dword_40F020 ^= dword_40F02C; dword_40F038 -= dword_40F03C; goto LABEL_4; case 6: if ( !dword_40F03C ) goto LABEL_6; dword_40F034 |= dword_40F024 / dword_40F03C; dword_40F024 /= dword_40F03C; goto LABEL_4; case 7: dword_40F038 += dword_40F028; dword_40F034 |= dword_40F024; goto LABEL_4; case 8: dword_40F020 *= dword_40F02C; dword_40F030 -= dword_40F03C; goto LABEL_4; case 9: dword_40F028 += dword_40F034; dword_40F02C ^= dword_40F030; LABEL_4: if ( ++v1 != a1 + 8 ) continue; result = (dword_40F038 == 231) + (dword_40F034 == 14456) + (dword_40F030 == 14961) + (dword_40F02C == -13264) + (dword_40F028 == 16) + (dword_40F024 == 104) + (dword_40F020 == -951) == 7; if ( dword_40F03C != -239 ) goto LABEL_6; break; default: LABEL_6: result = 0; break; } return result; } }
通过分析，这就是将v10的每一位进行switch选择，对一组数据进行更改，最后要满足要求，才能返回1。
因此，得到v10的值成为了这道题的关键，又因为v10的每个数都是0~9之间，且长度为8，我直接暴力求解。
#include <bits/stdc++.h> #pragma warning(disable:4996) #define _DWORD unsigned int #define _BYTE char using namespace std; bool __cdecl sub_401950(_BYTE* a1) { _BYTE* v1; // ecx bool result; // al int val_0 = 0x0A; int val_1 = 0x8A; int val_2 = 0x1A1; int val_3 = 0x12A; int val_4 = 0x269; int val_5 = 0x209; int val_6 = 0x68; int val_7 = 0x39F; int val_8 = 0x2C8; v1 = a1; while (2) { switch (*v1-48) { case 0: val_3 &= val_7; val_4 *= val_3; goto LABEL_4; case 1: if (!val_4) goto LABEL_6; val_3 /= val_4; val_2 += val_6; goto LABEL_4; case 2: val_5 ^= val_6; val_8 += val_1; goto LABEL_4; case 3: val_8 -= val_5; val_5 &= val_2; goto LABEL_4; case 4: val_6 *= val_1; val_4 -= val_7; goto LABEL_4; case 5: val_1 ^= val_4; val_7 -= val_8; goto LABEL_4; case 6: if (!val_8) goto LABEL_6; val_6 |= val_2 / val_8; val_2 /= val_8; goto LABEL_4; case 7: val_7 += val_3; val_6 |= val_2; goto LABEL_4; case 8: val_1 *= val_4; val_5 -= val_8; goto LABEL_4; case 9: val_3 += val_6; val_4 ^= val_5; LABEL_4: if (++v1 != a1 + 8) continue; result = (val_7 == 231) + (val_6 == 14456) + (val_5 == 14961) + (val_4 == -13264) + (val_3 == 16) + (val_2 == 104) + (val_1 == -951) == 7; if (val_8 != -239) goto LABEL_6; break; default: LABEL_6: result = 0; break; } return result; } } int main() { char* s = (char*)malloc(10); for (long i = 10000000; i < 100000000; ++i) { sprintf(s, "%08ld", i); //cout << s << endl; if (sub_401950(s)) { cout << i << endl; system("PAUSE"); } } system("PAUSE"); return 0; }
得到v10的值
v9的十六进制加密
得到v10值后，不能获得有关输入字符串的相关信息，打开前面的函数sub_401B60(&v10, &v9)
int __cdecl sub_401B60(int a1, _BYTE *a2) { _BYTE *v2; // ebx int i; // esi char v4; // ST08_1 _BYTE *v5; // ST00_4 int result; // eax v2 = a2; for ( i = a1; *v2; result = sub_40ED40(v5, "%02X", v4) ) { v4 = i; v5 = v2; v2 += 2; ++i; } return result; }
能够猜测到，v9加密为十六进制，存储到v10中，(如果v9="0123456789abcdef",则v10=0x0123456789abcdef。通过后面OD调试，也能够发现。）
v4数组生成
打开sub_401850(v4, &v5, strlen((const char *)&v5))函数
int __cdecl sub_401850(_DWORD *a1, int a2, int a3) { int v3; // eax int v4; // edi unsigned int *v5; // ebx int v6; // ecx int result; // eax unsigned int v8; // esi char v9; // dl unsigned int *v10; // edx v3 = 0; v4 = (int)(a1 + 2); *a1 = 0; a1[1] = 0; do { *(_DWORD *)(v4 + 4 * v3) = v3; ++v3; } while ( v3 != 256 ); v5 = a1 + 2; v6 = 0; LOBYTE(result) = 0; do { v8 = *v5; v9 = *(_BYTE *)(a2 + v6++) + *v5; result = (unsigned __int8)(v9 + result); v10 = (unsigned int *)(v4 + 4 * result); *v5 = *v10; *v10 = v8; if ( v6 >= a3 ) v6 = 0; ++v5; } while ( v5 != a1 + 258 ); return result; }
这里在生成V4数组，我从OD中提取出来
0x00, 0x00, 0x71, 0x12, 0x62, 0x31, 0x4D, 0x97, 0x14, 0x0D, 0xED, 0xA3, 0xD6, 0xFC, 0xF1, 0x3B, 0x3C, 0x33, 0xB5, 0x22, 0xA2, 0x1A, 0x17, 0x1D, 0x98, 0x91, 0x06, 0x2A, 0x8B, 0x23, 0xE6, 0x55, 0x46, 0x3A, 0x65, 0x28, 0x30, 0x39, 0xD4, 0x0C, 0x01, 0x2D, 0x25, 0x10, 0x09, 0x8F, 0x6A, 0x3F, 0x44, 0xD8, 0x6D, 0xC5, 0xA6, 0x72, 0x07, 0x83, 0x40, 0xC6, 0x8E, 0x1F, 0x77, 0x61, 0x96, 0x4A, 0x08, 0xFE, 0x53, 0x5A, 0xA1, 0xDF, 0xB6, 0x67, 0x66, 0x5C, 0x57, 0xB8, 0xD3, 0x11, 0x52, 0x21, 0xCC, 0x56, 0x2E, 0xC2, 0x88, 0xAA, 0xF9, 0x20, 0x7A, 0x6F, 0x4E, 0x76, 0xE8, 0xC1, 0xD5, 0xBD, 0xCE, 0x9E, 0x38, 0x95, 0x50, 0xF2, 0x9F, 0xB2, 0x9A, 0x0B, 0x47, 0x16, 0x60, 0xBF, 0xFD, 0x92, 0x35, 0x89, 0xDA, 0xFF, 0x9B, 0xBA, 0x13, 0xAB, 0xF4, 0x79, 0x87, 0xAC, 0x8C, 0x73, 0x84, 0xB3, 0x0E, 0xC8, 0x26, 0xA5, 0xE7, 0x15, 0xE9, 0xC3, 0x69, 0x70, 0xE0, 0x68, 0x42, 0x81, 0xCD, 0xEB, 0xDE, 0x7D, 0xEF, 0xD0, 0x24, 0x00, 0xF0, 0x41, 0xA0, 0xEE, 0x05, 0x94, 0x85, 0xBB, 0x43, 0x02, 0xF7, 0xC0, 0xD1, 0x1B, 0x7F, 0x5B, 0xEC, 0xF6, 0x2B, 0x1E, 0xE2, 0x27, 0xFB, 0x78, 0x54, 0x58, 0xE4, 0x32, 0xDB, 0xB7, 0xC7, 0x90, 0x7C, 0xF8, 0x5D, 0x5F, 0x63, 0xBE, 0x2C, 0x0A, 0xDD, 0x9C, 0x75, 0x19, 0xC4, 0xA8, 0x86, 0x36, 0xBC, 0x8D, 0xD7, 0x7B, 0xB4, 0x5E, 0x3E, 0xA7, 0xB1, 0xE1, 0x59, 0x82, 0xB9, 0xAE, 0xD9, 0x7E, 0xAF, 0xCF, 0x9D, 0xF5, 0xFA, 0x48, 0x4F, 0xA9, 0x6C, 0x64, 0x6E, 0x49, 0x4B, 0x6B, 0x29, 0x45, 0xE5, 0x04, 0xA4, 0x4C, 0x34, 0x80, 0xD2, 0x3D, 0xE3, 0x99, 0x37, 0xDC, 0x93, 0xC9, 0xCA, 0xCB, 0xEA, 0xB0, 0x0F, 0x03, 0x8A, 0xF3, 0x51, 0x1C, 0xAD, 0x74, 0x18, 0x2F
v10的加密与解密
打开 sub_4018D0(v4, &v10, 8)函数
_DWORD *__cdecl sub_4018D0(_DWORD *a1, _BYTE *a2, int a3) { int v3; // edx int v4; // ecx int v5; // esi _BYTE *v6; // ebx int v7; // edi unsigned int *v8; // eax int v9; // edx _DWORD *v10; // ebp _DWORD *v11; // ST00_4 unsigned int v12; // ebp _DWORD *result; // eax v3 = *a1; v4 = a1[1]; v5 = (int)(a1 + 2); if ( a3 > 0 ) { v6 = a2; v7 = *a1; do { v7 = (unsigned __int8)(v7 + 1); v8 = (unsigned int *)(v5 + 4 * v7); v9 = *v8; v4 = (unsigned __int8)(*v8 + v4); v10 = (_DWORD *)(v5 + 4 * v4); v11 = v10; v12 = *v10; *v8 = v12; *v11 = v9; *v6++ ^= *(_DWORD *)(v5 + 4 * (unsigned __int8)(v9 + v12)); } while ( v6 != &a2[a3] ); v3 = v7; } result = a1; *a1 = v3; a1[1] = v4; return result; }
这里利用v4数组，对v10的每一位进行异或运算。
这个函数执行后，就是最后一个函数，刚刚我们暴力解出了61495072，v10 = '\x06\x01\x04\x09\x05\x00\x07\x02'，因此我们能够对v10的每一位进行异或，得到v10加密前的值。
本来是准备利用提取出的v4，代入函数计算出v10的值，但是没成功，哈哈哈！后面发现v4的值实际上是固定的，不会受到输入的影响并且我只需要知道*v6异或的值即可。因此，我在OD中调试，记录异或的值。
0x7c, 0x0AB, 0x2D, 0x91, 0x2F, 0x98, 0x0ED, 0xA9
我们能够写出脚本
num = [0x7c, 0x0AB, 0x2D, 0x91, 0x2F, 0x98, 0x0ED, 0xA9] v10 = '\x06\x01\x04\x09\x05\x00\x07\x02' flag = [] n = 0 for i in v10: flag.append(hex(ord(i) ^ num[n])) n = n + 1 print(flag)
['0x7a', '0xaa', '0x29', '0x98', '0x2a', '0x98', '0xea', '0xab']
组合起来，得到v9经过十六进制加密后的v10=0x7aaaa29982a98eaab
因此我们能够得到v9="7aaa29982a98eaab"，输入到程序中得到flag。
get flag！
flag{5cb92582-66a8-e5b7-d3bf-3b99df8ac7f0}
