找出四位数的所有吸血鬼数字（JAVA）

/**
* 找出四位数的所有吸血鬼数字
* 吸血鬼数字是指位数为偶数的数字，可以由一对数字相乘而得到，而这对数字各包含乘积的一半位数的数字，其中从最初的数字中选取的数字可以任意排序.
* 以两个0结尾的数字是不允许的。
* 例如下列数字都是吸血鬼数字
1260=21*60
1827=21*87
2187=27*81
…
* 比较笨的低效率的做法： 遍历所有四位数， 每生成一个四位数的时候，
* 在双重循环遍历两位数，在两位数的内层循环中判断是否与最外层循环的四位数相等。 如果相等把这些数字都存放到数组，进行排序后比较
* 两组数字，如果相等那么输出这个数就是要找的数字；

*/

了解下这个英文参考:吸血鬼数字

An important theoretical result found by Pete Hartley:
1. If x·y is a vampire number then x·y == x+y (mod 9) Proof: Let mod be the binary modulo operator and d(x) the sum of the decimal
digits of x.
2. It is well-known that d(x) mod 9 = x mod 9, for all x.
Assume x·y is a vampire.
3.Then it contains the same digits as x and y，and in particular d(x·y) = d(x)+d(y).
4.This leads to:
(x·y) mod 9 = d(x·y) mod 9 = (d(x)+d(y)) mod 9 = (d(x) mod 9 + d(y) mod 9) mod 9 = (x mod 9 + y mod 9) mod 9 = (x+y) mod 9

The solutions to the congruence are (x mod 9, y mod 9) in {(0,0),
(2,2), (3,6), (5,8), (6,3), (8,5)} Only these cases (6 out of 81) have
to be tested in a vampire search based on testing x·y for different
values of x and y.
public class Exercise10 {

JAVA实现：

    public static void main(String[] args) {
        // TODO 自动生成的方法存根
        int num1, num2, result, i, j, 
            count = 0;
        int[] predata = new int[4];
        int[] lastdata = new int[4];

        for(num1 = 10; num1 < 99; num1++)
            for(num2 = num1; num2 < 99; num2++){
                result = num1 *num2;
                count = 0;
                if(((num1 * num2) %9) !=((num1 + num2) %9))
                    continue;
                predata[0] = num1 / 10;
                predata[1] = num1 % 10;
                predata[2] = num2 / 10;
                predata[3] = num2 % 10;
                lastdata[0] = result / 1000;
                lastdata[1] = (result % 1000) / 100;
                lastdata[2] = (result % 1000 % 100) / 10;
                lastdata[3] = (result % 1000 % 100 % 10);
                for(i = 0; i < 4 ; i++)
                    for(j = 0; j < 4; j++){
                        if(predata[i] == lastdata[j]){
                            count++;
                            predata[i] = -1;
                            lastdata[j] = -2;
                        }
                    }
                if(count == 4)
                    System.out.println(num1 +" * " + num2 + " = "+ result);
            }

    }

}