[C++知识库]2018年九月份安恒杯月赛之NewDriver

2018年九月份安恒杯月赛之NewDriver

这题是复现的，用于研究 RC4 加密，首先拿到附件：

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照例扔入 exeinfope 中查看信息：

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32 位无壳，照例扔入 IDA32 中查看信息，有 main 函数看 main 函数：（这次就用老版7.0 IDA 来分析算了）




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分析第一个自定义函数：


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分析第二个自定义函数：


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分析第三个自定义函数：


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所以总的流程是，用户输入------>base64加密------>RC4数据初始化并且加密-------->进行比较
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第一个解法，RC4解密后，Base64变表解密：
这里替换了密文和密钥，需要注意的是密钥哪里由于C语言的字符串末尾补0特性，导致key多了一个0x00，结果就大不同了，所以我们只能用16进制的没有补齐0x00的数据才行。

#include <stdio.h>

typedef struct _RC4INFO
{
    unsigned char s_box[256]; 
    unsigned char t_box[256];
}RC4_INFO,*PRC4_INFO;

void rc4_init(PRC4_INFO prc4,unsigned char key[], unsigned int keylen)
{
    int i = 0;
    int j = 0;
    unsigned char tmp;
    if (prc4 == NULL)
    {
        return;
    }
    //init sbox an KeyBox(Tbox)
    for (i = 0; i < 256; i++)
    {
        prc4->s_box[i] = i;
        prc4->t_box[i] = key[i % keylen];
    }

    //swap sbox
    for (i = 0; i < 256; i++)
    {
        j = (j + prc4->s_box[i] + prc4->t_box[i]) % 256;
        tmp = prc4->s_box[i];
        prc4->s_box[i] = prc4->s_box[j];
        prc4->s_box[j] = tmp;
    }
}
void rc4_crypt(
    unsigned char data[],       //要加密的数据
    unsigned int datalen,       //要加密的数据长度
    unsigned char key[],        //加密数据所用的Key
    unsigned int keylen)        //加密数据所用的key长度
{
    int dn = 0;  //data[n]的意思
    int i = 0;
    int j = 0;   // i j 分别用于交换sbox[i] 和 sbox[j]
    int t = 0;   //t = s[i] + s[j]
    unsigned char tmp;
    RC4_INFO rc4;
    rc4_init(&rc4, key, keylen);

    for (dn = 0; dn < datalen; dn++)
    {
        i = (i + 1) % 256;
        j = (j + rc4.s_box[i]) % 256;

        //swap
        tmp = rc4.s_box[i];
        rc4.s_box[i] = rc4.s_box[j];
        rc4.s_box[j] = tmp;
        //得到T下标用于交换
        t = (rc4.s_box[i] + rc4.s_box[j]) % 256;
        data[dn] ^= rc4.s_box[t];
    }
}
void EntryBuffer(unsigned char data[],unsigned int datalen)
{/*
    unsigned char key[] = {
  0x66, 0x6C, 0x61, 0x67, 0x7B, 0x74, 0x68, 0x69, 0x73, 0x5F, 
  0x69, 0x73, 0x5F, 0x6E, 0x6F, 0x74, 0x5F, 0x74, 0x68, 0x65, 
  0x5F, 0x66, 0x6C, 0x61, 0x67, 0x5F, 0x68, 0x61, 0x68, 0x61, 
  0x68, 0x61, 0x7D,0x00
};
*/
	//unsigned char key[] ="flag{this_is_not_the_flag_hahaha}";
    unsigned char key[] = {
  0x66, 0x6C, 0x61, 0x67, 0x7B, 0x74, 0x68, 0x69, 0x73, 0x5F, 
  0x69, 0x73, 0x5F, 0x6E, 0x6F, 0x74, 0x5F, 0x74, 0x68, 0x65, 
  0x5F, 0x66, 0x6C, 0x61, 0x67, 0x5F, 0x68, 0x61, 0x68, 0x61, 
  0x68, 0x61, 0x7D
};

	printf("\n\n%x\n\n",key[0]);
    rc4_crypt(data, datalen, key, sizeof(key) / sizeof(key[0]));
    printf("\n%d\n",sizeof(key) / sizeof(key[0]));
}
int main()
{
   unsigned char Hell[] = {
  0x20, 0xC3, 0x1A, 0xAE, 0x97, 0x3C, 0x7A, 0x41, 0xDE, 0xF6, 
  0x78, 0x15, 0xCB, 0x4B, 0x4C, 0xDC, 0x26, 0x55, 0x8B, 0x55, 
  0xE5, 0xE9, 0x55, 0x75, 0x40, 0x3D, 0x82, 0x13, 0xA5, 0x60, 
  0x13, 0x3B, 0xF5, 0xD8, 0x19, 0x0E, 0x47, 0xCF, 0x5F, 0x5E, 
  0xDE, 0x9D, 0x14, 0xBD
};
    printf("pData=%s\n\n", Hell);
    //EntryBuffer((unsigned char*)Hell,sizeof(Hell)/sizeof(Hell[0])); //加密
    //printf("pData=%s\n\n", Hell);
    EntryBuffer((unsigned char*)Hell,sizeof(Hell)/sizeof(Hell[0])); //在调用一次就是解密了
    printf("pData=%s\n\n", Hell);
    return 0;
}

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第二种解法，由于RC4末尾是异或操作，所以可以直接抽出内存中初始化后的数据来重新加密即可：
在RC4的异或运算对合性中，RC4加密解密使用同一套算法。

key=[0x66, 0x32, 0xCA, 0xA0, 0xBF, 0x98, 0x2D, 0x76, 0xF1, 0x59, 0x2A, 0x4A, 0xF4, 0x30, 0xAD, 0xD2, 0x1D, 0x02, 0xD8, 0x23, 0x89, 0x5D, 0x83, 0x38, 0x09, 0xF2, 0x74, 0x65, 0x40, 0x19, 0xC6, 0xDD, 0x18, 0xD3, 0x8F, 0x6C, 0x8B, 0xC0, 0xC5, 0x54, 0x2E, 0x81, 0x10, 0xC4, 0x26, 0x56, 0x5F, 0x53, 0x80, 0x43, 0x27, 0x62, 0xEA, 0x3D, 0xE6, 0x00, 0xE7, 0xB7, 0x50, 0x94, 0x90, 0x4C, 0x3F, 0x9D, 0x07, 0xE0, 0xA3, 0x9C, 0x4E, 0x0F, 0x9F, 0xFE, 0x5B, 0x8E, 0xDE, 0x88, 0x72, 0x2F, 0xC1, 0x67, 0x31, 0x70, 0x8D, 0xFD, 0xBE, 0x64, 0xC3, 0xBD, 0x6B, 0x7A, 0xCF, 0x0C, 0x34, 0x1F, 0x6F, 0x01, 0xF0, 0x7C, 0x5E, 0xA4, 0x1E, 0x49, 0x8C, 0x75, 0x1C, 0xE3, 0x20, 0x48, 0x28, 0x79, 0xA5, 0x7F, 0xF5, 0xEC, 0x4F, 0x78, 0x58, 0x11, 0xF7, 0xCD, 0x91, 0x13, 0xFC, 0xB8, 0x2C, 0x04, 0xEE, 0xD5, 0x08, 0x44, 0xA9, 0xE1, 0xB1, 0x42, 0x84, 0x29, 0xA7, 0x47, 0x97, 0x7E, 0xE8, 0xB3, 0x60, 0x0B, 0xF9, 0x4B, 0x3C, 0x77, 0x17, 0x03, 0x82, 0x69, 0x87, 0xD4, 0x95, 0x1A, 0x33, 0x25, 0x6E, 0xCC, 0xD6, 0xBB, 0x99, 0xB0, 0x85, 0x41, 0xB2, 0x0D, 0xDB, 0x35, 0x3B, 0x5C, 0xF8, 0xED, 0x9E, 0xA6, 0x96, 0x39, 0x63, 0x0A, 0x1B, 0x93, 0x21, 0x46, 0x12, 0xD0, 0xB4, 0x22, 0x51, 0xC9, 0x61, 0xD1, 0x2B, 0xAA, 0x45, 0x06, 0x05, 0xCE, 0xFA, 0x92, 0x68, 0xAB, 0x36, 0xDA, 0xC8, 0xE2, 0x37, 0xD9, 0xA2, 0x5A, 0xD7, 0x6A, 0xB5, 0xFF, 0xE9, 0xBA, 0x52, 0x15, 0xF6, 0xBC, 0x9A, 0xB6, 0xEF, 0x6D, 0xCB, 0x4D, 0xAE, 0xE4, 0xA1, 0xAC, 0xEB, 0x0E, 0x71, 0x7B, 0xF3, 0x24, 0xC2, 0xFB, 0x7D, 0x86, 0x55, 0xAF, 0x3A, 0xDF, 0x3E, 0x14, 0xB9, 0x9B, 0x16, 0xDC, 0x73, 0x57, 0xE5, 0xC7, 0x8A, 0xA8, 0x66, 0x6C, 0x61, 0x67, 0x7B, 0x74, 0x68, 0x69, 0x73, 0x5F, 0x69, 0x73, 0x5F, 0x6E, 0x6F, 0x74, 0x5F, 0x74, 0x68, 0x65, 0x5F, 0x66, 0x6C, 0x61, 0x67, 0x5F, 0x68, 0x61, 0x68, 0x61, 0x68, 0x61, 0x7D]
key2=[0x20, 0xC3, 0x1A, 0xAE, 0x97, 0x3C, 0x7A, 0x41, 0xDE, 0xF6, 0x78, 0x15, 0xCB, 0x4B, 0x4C, 0xDC, 0x26, 0x55, 0x8B, 0x55, 0xE5, 0xE9, 0x55, 0x75, 0x40, 0x3D, 0x82, 0x13, 0xA5, 0x60, 0x13, 0x3B, 0xF5, 0xD8, 0x19, 0x0E, 0x47, 0xCF, 0x5F, 0x5E, 0xDE, 0x9D, 0x14, 0xBD]
key3=[]
v3=0
v4=0
v6=0
s=""
#这里直接仿写IDA中第三个加密函数逻辑即可
for v3 in range(44):
  v3+=1
  v6=key[v3]
  v4=(v6+v4)%256
  key[v3]=key[v4]
  key[v4]=v6
  key3.append(key[(key[v3]+v6)%256])
for i in range(44):
  key2[i]^=key3[i]
for i in key2:
  s+=chr(i)
print s

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解毕！
敬礼！