请一JAVA算法逆运算，400积分

[galaxypilot]

private static void eighth(byte[] target){//[5, 4, 3, 8, 6, 0, 12, 10, 10, 2, 15]
byte abyte0[] = new byte[11];
long l3 = (target[9] << 4) + target[10]; //47
for(int i = 0; i < 9; i++)
{
long l5 = 9L * l3 + (long)i;//423
long l6;
long l7 = ((l6 = 0x7fffffffL) - 1L) / 256L;//8388607
long l8 = (l6 << 1) + 2L;//4294967296
long l9 = 16807L;//16807
long l10 = 1L;
for(int i1 = 0; (long)i1 < l5 + 12L; i1++)
{
if((l10 = (int)(l10 *= l9)) < 0L)
l10 += l8;
l10 %= l6;
}
byte byte1;
byte1 = (byte)(int)(l10 / l7);
abyte0[i] = (byte)(target[i] - byte1 & 0xf);//[0, 10, 7, 5, 0, 7, 1, 1, 11, 0, 0];
}
}
byte参数 target 实际值 =[5, 4, 3, 8, 6, 0, 12, 10, 10, 2, 15]，
经该方法运算后 abyte0 =[0, 10, 7, 5, 0, 7, 1, 1, 11, 0, 0];
现在要根据 abyte0 =[0, 10, 7, 5, 0, 7, 1, 1, 11, 0, 0] 逆向算出 target，数组最后两位可以不管。

[qybao]

你这个有信息丢失，也就是l3没法知道了，还原不了
如果能知道l3，也就是你把数组的后两位也保留，估计还可以还原

public class Sample {
    public static void main(String[] args) {
        try {
            byte[] org = {5, 4, 3, 8, 6, 0, 12, 10, 10, 2, 15};
            byte[] chg = eighth(org);
            System.out.println(Arrays.toString(chg)); //[0, 10, 7, 5, 0, 7, 1, 1, 11, 0, 0];
            byte b1 = 2, b2 = 15; //原数组的9，10位元素
            long l3 = (b1 << 4) + b2;
            byte[] target = restore(chg, l3); //原数组后两位没丢失的话，l3可以放到方法里求，如果丢失，看看通过有什么方法传参数
            System.out.println(Arrays.toString(target));
             
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    
    private static byte[] restore(byte[] src, long l3) {
        byte[] result = new byte[11];
        for (int i=0; i<9; i++) {
            long l5  = 9L * l3 + (long)i;//423
            long l6;
            long l7 = ((l6 = 0x7fffffffL) - 1L) / 256L;//8388607
            long l8 = (l6 << 1) + 2L;//4294967296
            long l9 = 16807L;//16807
            long l10 = 1L;
            for(int i1 = 0; (long)i1 < l5 + 12L; i1++)
            {
                if((l10 = (int)(l10 *= l9)) < 0L)
                    l10 += l8;
                l10 %= l6;
            }
            byte byte1;
            byte1  = (byte)(int)(l10 / l7);
            result[i] = (byte)((src[i] + byte1) & 0xf);
        }
        return result;
    }
    
    private static byte[] eighth(byte[] target){//[5, 4, 3, 8, 6, 0, 12, 10, 10, 2, 15]
        byte abyte0[] = new byte[11];
        long l3 = (target[9] << 4) + target[10]; //47
        System.out.println(l3);
        for(int i = 0; i < 9; i++)
        {
            long l5  = 9L * l3 + (long)i;//423
            long l6;
            long l7 = ((l6 = 0x7fffffffL) - 1L) / 256L;//8388607
            long l8 = (l6 << 1) + 2L;//4294967296
            long l9 = 16807L;//16807
            long l10 = 1L;
            for(int i1 = 0; (long)i1 < l5 + 12L; i1++)
            {
                if((l10 = (int)(l10 *= l9)) < 0L)
                    l10 += l8;
                l10 %= l6;
            }
            byte byte1;
            byte1  = (byte)(int)(l10 / l7);
            abyte0[i] = (byte)((target[i] - byte1) & 0xf);//[0, 10, 7, 5, 0, 7, 1, 1, 11, 0, 0];
        }
        //abyte0[9] = target[9]; //保留后两位的话
        //abyte0[10] = target[10];//还有可能还原
        return abyte0;
    }
}

[galaxypilot]

这么快，牛X， long l3 = (target[9] << 4) + target[10]; 是参数数组最后两位计算得到了， 我给的例子中[5, 4, 3, 8, 6, 0, 12, 10, 10, 2, 15] 是47 我试一下，晚点再回粘。谢谢！

[galaxypilot]

牛X，方便加个微信吗？