打印十字图

来源：互联网发布：relief 人工智能编辑：程序博客网时间：2024/04/29 13:29

小明为某机构设计了一个十字型的徽标（并非红十字会啊），如下所示(可参见p1.jpg)

import java.util.Scanner;public class dibage {public static void main(String[] args) {// TODO Auto-generated method stubScanner scan = new Scanner(System.in);int line = scan.nextInt();shuzu(line);}//定义一个二维数组，赋值为 $  然后找 dian的位置 。public static void shuzu(int n) {int num = 5 + 4 * n, s = 0;char[][] arr = new char[num][num];for (int i = 0; i < arr.length; i++) {for (int j = 0; j < arr.length; j++) {arr[i][j] = '$';}}show(arr);dayin(arr);}public static void dayin(char[][] arr) {for (int i = 0; i < arr.length; i++) {for (int j = 0; j < arr.length; j++) {System.out.print(arr[i][j]);}System.out.print("\n");}}public static void show(char[][] arr) {int lain = arr.length / 2 + 1;for (int count = 0; count < lain; count += 2) {for (int i =  count; i < lain; i++) {for (int j =  count; j < lain; j++) {//对角线  if (i == 0 && j == 0 || i == j && i % 2 == 1) {arr[i][j] = '.';}//if (i == count + 1 && j > 2 + count) {arr[i][j] = '.';}if (j == count + 1 && i > 2 + count) {arr[i][j] = '.';}if (i == count - 1 && j < count) {arr[i][j] = '.';}if (j == count - 1 && i < count) {arr[i][j] = '.';}if (i == count + 1 && j < 2 + count) {arr[i][j] = '.';}if (j == count + 1 && i < 2 + count) {arr[i][j] = '.';}}}//右上角for(int i=0+count,x=0+count;i<lain;i++,x++)            {                    for(int j=lain,y=lain-2;j<arr.length-count;j++,y--)                    {                            arr[i][j] = arr[x][y];                                                    }                                                        }//左下角for (int i = lain, x=lain - 2; i < arr.length - count; i++ , x--) {for (int j = count , y = count; j < lain ;  j++ , y++) {arr[i][j]= arr[x][y] ;}}//右下角for (int i = lain, x=lain - 2; i < arr.length - count; i++ , x--) {for(int j=lain,y=lain-2;j<arr.length-count;j++,y--)                {                        arr[i][j] = arr[y][x];                                                }}}}}

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