关于搜狗在线测评上面的题目详解

前几天在投递了搜狗Java开发工程师。需要搞一个在线测评，出了一道题目，是程序填空题

要求观察程序，补全decode，输出一句话：

public  class  Test  {

public  static  void  encode(byte[]  in,  byte[]  out,  int  password)
{
int  len  =  in.length;

int  seed  =  password  ^  0x1963f9c;
for  (int  i  =  0  ;  i  <  len;  ++i)  {
byte  a  =  (byte)(  (  in[i]  ^  seed  )  >>>  2  );
byte  b  =  (byte)(  (  (  ((int)in[i])  <<  13  )  ^  seed  )  >>>  (13-6)  );
a  &=  0x3f;
b  &=  0xc0;
out[i]  =  (byte)(a  |  b);
seed  =  (((seed  <<  7)  ^  seed  ^  in[i])  +  5393887);
}
}

public  static  void  decode(byte[]  in,  byte[]  out,  int  password)
{
int  len  =  in.length;

int  seed  =  password  ^  0x1963f9c;
for  (int  i  =  0  ;  i  <  len;  ++i)  {
byte a = (byte)(in[i] & 0x3f);
byte b = (byte)(in[i] & 0xc0);
a = (byte)(((a << 2) ^ seed) & 0xfc);
b = (byte)((((((int)b) << (13-6)) ^ seed) >>13) & 0x03);
out[i]  =  (byte)(a  |  b);
seed  =  (((seed  <<  7)  ^  seed  ^  out[i])  +  5393887);

}
}
public  static  void  main(String  []  args)  throws  Exception
{
int  password  =  0x99b90094;
byte[]  buf1  =  {113,  87,  1,  -29,  55,  -5,  65,  13,  -85,  78,  50,  23,  -33,  -29,  73,  117,  -97,  106,  77,  -17,  -107,  -62,  -110,  -126,  -15,  32,  -48,  -128,  };
byte[]  buf2  =  new  byte[buf1.length];
decode(buf1,  buf2,  password);
System.out.println(new  String(buf2,  "GBK"));
}

}

思前想后，发现规律。这就是一系列的移位，我们要做的就是反过来再移一次。先看看encode方法，然后再去实现decode。方法如下：

public  static  void  decode(byte[]  in,  byte[]  out,  int  password)
{
int  len  =  in.length;

int  seed  =  password  ^  0x1963f9c;
for  (int  i  =  0  ;  i  <  len;  ++i)  {
byte a = (byte)(in[i] & 0x3f);
byte b = (byte)(in[i] & 0xc0);
a = (byte)(((a << 2) ^ seed) & 0xfc);
b = (byte)((((((int)b) << (13-6)) ^ seed) >>13) & 0x03);
out[i]  =  (byte)(a  |  b);
seed  =  (((seed  <<  7)  ^  seed  ^  out[i])  +  5393887);

}
}

最后输出的结果是：听歌从搜狗音乐搜索开始！！！