吸血鬼数

来源：互联网发布：java基础程序设计步骤编辑：程序博客网时间：2024/04/27 14:32

找到100000以下的所有吸血鬼数：

package com.inter;import java.util.Arrays;public class Gui {public static void main(String[] args) {int temp = 0;String tt = "";String ii = "";String jj = "";for (int i = 0; i <= 1000; i++) {for (int j = 0; j <= 1000; j++) {temp = i * j;tt = String.valueOf(temp);if (tt.length() % 2 == 0) {// 判断是偶数位的数进去jj = String.valueOf(j);ii = String.valueOf(i);if (jj.length() == ii.length()&& tt.length() / 2 == jj.length()) {// 位数判断完毕String jjii = jj + ii;int[] ji = new int[jjii.length()];for (int k = 0; k < ji.length; k++) {ji[k] = Integer.parseInt(jjii.substring(k, k + 1));}int[] te = new int[tt.length()];for (int k = 0; k < te.length; k++) {te[k] = Integer.parseInt(tt.substring(k, k + 1));}// 下面我要将两个数组排序 然后比较maoPao(ji);maoPao(te);if (Arrays.equals(te, ji)) {System.out.println(jj + "*" + ii + "=" + tt);}}}}}}public static void maoPao(int[] x) {for (int i = 0; i < x.length; i++) {for (int j = i + 1; j < x.length; j++) {if (x[i] > x[j]) {int temp = x[i];x[i] = x[j];x[j] = temp;}}}}}

我写的有大量可以优化的地方，和朋友打赌很简单，所以就顾及了些时间，现在又不想改了。

判断是不是吸血鬼数：

package com.inter;import java.util.Arrays;public class GuiTwo {public static void main(String[] args) {int n = 315900;String nn = String.valueOf(n);int[] ff = new int[nn.length()];int[] fn = new int[nn.length()];if(nn.length()%2!=0){}else{int temp = (int)Math.pow(10, nn.length()/2-1);for (int i = temp; i <= temp*10; i++) {for (int j = temp; j <= temp*10; j++) {if(i*j==n){String iijj = String.valueOf(i)+String.valueOf(j);for (int k = 0; k < nn.length(); k++) {ff[k] = Integer.parseInt(iijj.substring(k, k+1));fn[k] = Integer.parseInt(nn.substring(k, k+1));}Arrays.sort(ff);Arrays.sort(fn);if(Arrays.equals(ff, fn)){System.out.println("是吸血鬼数！！！");break;}}}}}}}