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《Thinking in Java》P75
练习10：（5）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
写一个程序，找出4位数的所有吸血鬼数字（Dan Forhan推荐）。
我的方案1：穷举四位数得到从左到右的 a1 a2 a3 a4 四个数字，进行排列组合，满足 n = (a1*10 + a2)*(a3*10 + a4)或类似条件，就打印出来。
好处是时间复杂度为N，但是四位数穷举本身比较次数就很高。
import java.util.StringTokenizer;/*** Created by sherry on 17/2/14.*/
public class vampire {public static void main(String[] args){for(int n = 1000;n <= 9999;n++){int a1 = n/1000;int a2 = n/100 - a1*10;int a3 = n/10 - a2*10 - a1*100;int a4 = n - a3*10 - a2*100 - a1*1000;if((a1*10 + a2)*(a3*10 + a4) == n){System.out.print(n + " ");}if((a2*10 + a1)*(a3*10 + a4) == n){System.out.print(n + " ");}if((a1*10 + a2)*(a4*10 + a3) == n){System.out.print(n + " ");}if((a2*10 + a1)*(a4*10 + a3) == n){System.out.print(n + " ");}if((a1*10 + a3)*(a2*10 + a4) == n){System.out.print(n + " ");}if((a3*10 + a1)*(a2*10 + a4) == n){System.out.print(n + " ");}if((a1*10 + a3)*(a4*10 + a2) == n){System.out.print(n + " ");}if((a3*10 + a1)*(a4*10 + a2) == n){System.out.print(n + " ");}if((a1*10 + a4)*(a2*10 + a3) == n){System.out.print(n + " ");}if((a4*10 + a1)*(a2*10 + a3) == n){System.out.print(n + " ");}if((a1*10 + a4)*(a3*10 + a2) == n){System.out.print(n + " ");}if((a4*10 + a1)*(a3*10 + a2) == n){System.out.print(n + " ");}}}
}方案2：方案1的简洁版
public class VampireNumbers { static int a(int i) {return i/1000;}static int b(int i) {return (i%1000)/100;}static int c(int i) {return ((i%1000)%100)/10;}static int d(int i) {return ((i%1000)%100)%10;}static int com(int i, int j) {return (i * 10) + j;}static void productTest (int i, int m, int n) {if(m * n == i) System.out.println(i + " = " + m + " * " + n);} public static void main(String[] args) { for(int i = 1001; i < 9999; i++) { productTest(i, com(a(i), b(i)), com(c(i), d(i)));productTest(i, com(a(i), b(i)), com(d(i), c(i)));productTest(i, com(a(i), c(i)), com(b(i), d(i)));productTest(i, com(a(i), c(i)), com(d(i), b(i)));productTest(i, com(a(i), d(i)), com(b(i), c(i)));productTest(i, com(a(i), d(i)), com(c(i), b(i)));productTest(i, com(b(i), a(i)), com(c(i), d(i)));productTest(i, com(b(i), a(i)), com(d(i), c(i)));productTest(i, com(b(i), c(i)), com(d(i), a(i)));productTest(i, com(b(i), d(i)), com(c(i), a(i)));productTest(i, com(c(i), a(i)), com(d(i), b(i)));productTest(i, com(c(i), b(i)), com(d(i), a(i)));} }
}
```方案3：另一种思路，通过先穷举两位数的方式，再比较四位数结果。时间效率为N²，但实际比较次数有改善。参考博客http://www.cnblogs.com/vincent-hv/archive/2013/04/19/3030031.html<div class="se-preview-section-delimiter"></div>
import java.util.StringTokenizer;
import java.util.Arrays;
/**
* Created by sherry on 17/2/14.
*/
public class vampire {
public static void main(String[] args){int count = 0;String[] ar_str1, ar_str2;for(int x = 10;x < 100;x++) {int from = Math.max(1000 / x, x);int to = Math.min(10000 / x, 100);for (int y = from; y < to; y++) {int integrity = x * y;//(integrity - x - y) % 9的由来：//x = 10*a1 + a2, y = 10*a3 + a4//integrity = 1000*a1 + 100*a2 + 10*a3 + a4//integrity - x - y = 990*a1 + 99*a2 + 9*a3 = 9*(110*a1 + 11*a2 + a3)//通过上式保证 x*y = x和y元素的组合四位数(integrity)if (integrity % 100 == 0 || (integrity - x - y) % 9 != 0) {continue;}ar_str1 = String.valueOf(integrity).split("");ar_str2 = (String.valueOf(x) + String.valueOf(y)).split("");Arrays.sort(ar_str1);Arrays.sort(ar_str2);if (Arrays.equals(ar_str1, ar_str2)) {// 排序后比较,为真则找到一组System.out.println( x + "*" + y + "=" + integrity);count++;}}}System.out.print("总数 = " + count + " ");
}
}
“`
运行结果：
问题在于Arrays.sort()之后，无法区别 80*60 = 6880 和 60*80 = 6880,所以只有7行结果。
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