一道搜狗机试题的解答

在网上看到搜狗这道机试题，觉得好深奥。一时技痒，尝试做了下，运行结果比较有意思-“搜狗输入法支持各种炫酷的皮肤，彰显个性的你！！！”。运算涉及到异或、与、（无符号）移位、还有强制转型，难度并不算大。

题目如下：

 

/**
 * 搜狗机试题，根据encode方法写出decode方法
 * 
 */
public class TestDecode {

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

		int seed = password ^ 0x8c357ca5;
		for (int i = 0; i < len; ++i) {
			byte a = (byte) ((in[i] ^ seed) >>> 5);
			byte b = (byte) (((((int) in[i]) << 16) ^ seed) >>> (16 - 3));
			a &= 0x7;
			b &= 0xf8;
			out[i] = (byte) (a | b);
			seed = (seed * 3687989 ^ seed ^ in[i]);
		}
	}

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

		int seed = password ^ 0x8c357ca5;
		for (int i = 0; i < len; ++i) {
                     //填写代码
		}

	}

	public static void main(String[] args) throws Exception {
		int password = 0xe87dd9d3;
		byte[] buf1 = { 29, -16, 96, 43, -85, 25, -96, 83, 13, 66, -109, 49,
				-111, 0, 60, -101, 99, -86, -38, 86, -35, 48, 23, 83, -102, 25,
				73, -116, -101, -88, -5, 14, -14, -112, 87, -87, 2, 108, -58,
				40, 56, 12, 108, 77, 83, 38, 20, -114, };
		byte[] buf2 = new byte[buf1.length];
		decode(buf1, buf2, password);
		System.out.println(new String(buf2, "GBK"));
	}
}

 

 首先，分析加密方法。seed每次加密都有变化，解密的时候，也需要计算。计算代码为：seed = (seed * 3687989 ^ seed ^ out[i]);。 循环体内，12-17行是由明文生成密文的方法。

12行，byte a = (byte) ((in[i] ^ seed) >>> 5);将明文与种子异或右移5位取证，得到的byte的3-1位是明文8-6位的与种子8-6位的异或结果。13行，byte b = (byte) (((((int) in[i]) << 16) ^ seed) >>> (16 - 3));将明文左移至24-17位与种子24-17位异或然后右移13位，得到的byte的8-4位是是原明文的5-1位与种子24-17位异或的结果。14行，

a &= 0x7;//与00000111与运算，取低三位。15行，b &= 0xf8;// 与11111000与运算取高五位。16行，out[i] = (byte) (a | b);使用a的低三位与b的高5位合成密文。很明显，经过简单的运算，明文的8-6位生成了密文的3-1位，明文的5-1位生成了密文的8-4位。

然后，根据加密方法，逐步求解。int seedForLow3 = seed;int seedForHigh5 = seed >>> 16；byte low3 = (byte) (in[i] << 5); byte hight5 = (byte) (in[i] >>> 3);//将高地位与其种子对齐。out[i] = (byte) (((low3 ^ seedForLow3) & 0xe0) | ((hight5 ^ seedForHigh5) & 0x1f));将高低位分别与种子异或后拼合为原明文。seed = (seed * 3687989 ^ seed ^ out[i]);计算新种子

完整的代码如下：/**

 * 搜狗机试题，根据encode方法写出decode方法
 * 
 * @author SunShadow
 * 
 */
public class TestDecode {

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

		int seed = password ^ 0x8c357ca5;
		for (int i = 0; i < len; ++i) {
			// 与种子异或后右移5位，即高3位与876位异或 高三位至3-1
			byte a = (byte) ((in[i] ^ seed) >>> 5);
			// 与种子的24-17异或后取低五位 低五位至8-4
			byte b = (byte) (((((int) in[i]) << 16) ^ seed) >>> (16 - 3));
			a &= 0x7;// 取低三位 00000111
			b &= 0xf8;// 取高五位11111000
			out[i] = (byte) (a | b);
			seed = (seed * 3687989 ^ seed ^ in[i]);
		}
	}

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

		int seed = password ^ 0x8c357ca5;
		for (int i = 0; i < len; ++i) {
			int seedForLow3 = seed;
			int seedForHigh5 = seed >>> 16;
			byte low3 = (byte) (in[i] << 5); // 现在在8-6位
			byte hight5 = (byte) (in[i] >>> 3);// 现在在5-1位
			out[i] = (byte) (((low3 ^ seedForLow3) & 0xe0) | ((hight5 ^ seedForHigh5) & 0x1f));
			seed = (seed * 3687989 ^ seed ^ out[i]);// 计算种子
		}

	}

	public static void main(String[] args) throws Exception {
		int password = 0xe87dd9d3;
		byte[] buf1 = { 29, -16, 96, 43, -85, 25, -96, 83, 13, 66, -109, 49,
				-111, 0, 60, -101, 99, -86, -38, 86, -35, 48, 23, 83, -102, 25,
				73, -116, -101, -88, -5, 14, -14, -112, 87, -87, 2, 108, -58,
				40, 56, 12, 108, 77, 83, 38, 20, -114, };
		byte[] buf2 = new byte[buf1.length];
		decode(buf1, buf2, password);
		System.out.println(new String(buf2, "GBK"));
	}
}