第三题石像病毒

考点：异常处理

引入结构体

struct INP
{
  BYTE unknown1[8];
  BYTE key[28];
  BYTE hash[28];
  BYTE unknown2[8];
  BYTE output[48];
  BYTE input[4008];
};

分析主要逻辑

逻辑比较简单

输入长度：32
AES Key：md5("Enj0y_1t_4_fuuuN")

下面主要分析j_aesEncrypt

找到相似项目：lmshao/AES: Advanced Encryption Standard

手动导入结构体及符号

两处魔改

patch try部分，看到KeyExpansion

交换Sbox[113]与sbox[163]

int sub_652330()
{
  sbox[113] ^= sbox[163];
  sbox[163] ^= sbox[113];
  sbox[113] ^= sbox[163];
  return 1;
}

前段偶数次循环，不改变sbox

后段奇数次循环，交换sbox

相同办法分析shiftRows

程序将ROF修改为ROR

代码如下

int shiftRows(uint8_t (*state)[4]) {
    uint32_t block[4] = {0};
 
    /* i: row */
    for (int i = 0; i < 4; ++i) {
        LOAD32H(block[i], state[i]);
        block[i] = ROR32(block[i], 8*i);
        STORE32H(block[i], state[i]);
    }
 
    return 0;
}

修改逆函数及逆s盒

unsigned char inv_S[256] = {
        0x52, 0x09, 0x6a, 0xd5, 0x30, 0x36, 0xa5, 0x38, 0xbf, 0x40, 0x71, 0x9e, 0x81, 0xf3, 0xd7, 0xfb,
        0x7c, 0xe3, 0x39, 0x82, 0x9b, 0x2f, 0xff, 0x87, 0x34, 0x8e, 0x43, 0x44, 0xc4, 0xde, 0xe9, 0xcb,
        0x54, 0x7b, 0x94, 0x32, 0xa6, 0xc2, 0x23, 0x3d, 0xee, 0x4c, 0x95, 0x0b, 0x42, 0xfa, 0xc3, 0x4e,
        0x08, 0x2e, 0xa1, 0x66, 0x28, 0xd9, 0x24, 0xb2, 0x76, 0x5b, 0xa2, 0x49, 0x6d, 0x8b, 0xd1, 0x25,
        0x72, 0xf8, 0xf6, 0x64, 0x86, 0x68, 0x98, 0x16, 0xd4, 0xa4, 0x5c, 0xcc, 0x5d, 0x65, 0xb6, 0x92,
        0x6c, 0x70, 0x48, 0x50, 0xfd, 0xed, 0xb9, 0xda, 0x5e, 0x15, 0x46, 0x57, 0xa7, 0x8d, 0x9d, 0x84,
        0x90, 0xd8, 0xab, 0x00, 0x8c, 0xbc, 0xd3, 0x0a, 0xf7, 0xe4, 0x58, 0x05, 0xb8, 0xb3, 0x45, 0x06,
        0xd0, 0x2c, 0x1e, 0x8f, 0xca, 0x3f, 0x0f, 0x02, 0xc1, 0xaf, 0xbd, 0x03, 0x01, 0x13, 0x8a, 0x6b,
        0x3a, 0x91, 0x11, 0x41, 0x4f, 0x67, 0xdc, 0xea, 0x97, 0xf2, 0xcf, 0xce, 0xf0, 0xb4, 0xe6, 0x73,
        0x96, 0xac, 0x74, 0x22, 0xe7, 0xad, 0x35, 0x85, 0xe2, 0xf9, 0x37, 0xe8, 0x1c, 0x75, 0xdf, 0x6e,
        0x47, 0xf1, 0x1a, 0xa3, 0x1d, 0x29, 0xc5, 0x89, 0x6f, 0xb7, 0x62, 0x0e, 0xaa, 0x18, 0xbe, 0x1b,
        0xfc, 0x56, 0x3e, 0x4b, 0xc6, 0xd2, 0x79, 0x20, 0x9a, 0xdb, 0xc0, 0xfe, 0x78, 0xcd, 0x5a, 0xf4,
        0x1f, 0xdd, 0xa8, 0x33, 0x88, 0x07, 0xc7, 0x31, 0xb1, 0x12, 0x10, 0x59, 0x27, 0x80, 0xec, 0x5f,
        0x60, 0x51, 0x7f, 0xa9, 0x19, 0xb5, 0x4a, 0x0d, 0x2d, 0xe5, 0x7a, 0x9f, 0x93, 0xc9, 0x9c, 0xef,
        0xa0, 0xe0, 0x3b, 0x4d, 0xae, 0x2a, 0xf5, 0xb0, 0xc8, 0xeb, 0xbb, 0x3c, 0x83, 0x53, 0x99, 0x61,
        0x17, 0x2b, 0x04, 0x7e, 0xba, 0x77, 0xd6, 0x26, 0xe1, 0x69, 0x14, 0x63, 0x55, 0x21, 0x0c, 0x7d};
 
// ···
 
int invShiftRows(uint8_t (*state)[4]) {
    uint32_t block[4] = {0};
 
    /* i: row */
    for (int i = 0; i < 4; ++i) {
        LOAD32H(block[i], state[i]);
        block[i] = ROF32(block[i], 8*i);
        STORE32H(block[i], state[i]);
    }
 
    return 0;
}

这里要注意到对比结果在sub_6523D0被修改

简单赋值

int sub_6523D0()
{
  byte_65B200[16] = 0xF4;
  byte_65B200[17] = 0xF2;
  return 1;
}

实际上执行点位于GMul，通过相同的try catch处理方式

执行解密

#include <cstdio>
#include "aes.h"
 
 
int main()
{
    uint8_t key[16] = { 0x2F, 0x65, 0xB1, 0xFF, 0x31, 0xED, 0x86, 0xD0, 0x9A, 0x28, 0x5C, 0x0F, 0x40, 0x48, 0x05, 0x9D};
    uint8_t ct[32] = {
            0x57, 0x7C, 0xF5, 0x6D, 0x56, 0x96, 0x77, 0x45, 0xB0, 0xBD, 0xA1, 0xC7, 0x89, 0xA5, 0xAB, 0xDC,
            0xF2, 0xF4, 0x4B, 0xFE, 0xBE, 0xF5, 0xF5, 0x5C, 0x4D, 0x30, 0x42, 0x0F, 0x2B, 0x3B, 0xE6, 0xCB
    };
    ct[16] = 0xF4;
    ct[17] = 0xF2;
 
    uint8_t plain[33] = {0};
 
    aesDecrypt(key, 16, ct, plain, 32);
    for (int i = 0; i < 32; ++i) {
        printf("%02x ", plain[i]);
    }
    printf("\n");
    printf("%s\n", plain);
    return 0;
}