我如何在java中解决这个问题 [英] How do I solve this in java

问题描述

实现一个输入两个整数m和n的程序，并绘制m个2个方块，每个大小为n×n，排列在m×m表中。也就是说，它绘制一个m行m列的表，每个行的元素是n×n的正方形。例如，如果输入整数为4和6，则绘制16个正方形，每个尺寸为6×6，排列在4×4表中。每个方块的周长由符号*（星号）组成，内部空间用符号o（小写字母o）填充。



输入和输出：



你的程序应该提示用户输入两个整数，m和n，在2和8之间，并绘制m 2个方格，每个尺寸n×n。



我的尝试：



有不知道，这个帽子我试过了吗？

 import java.io. *; 
 public class JavaApplication19 {
 
 static void generateSquare（int n）
 {
 int [] [] magicSquare = new int [n] [n]; 
 
 //初始化1 
的位置i = n / 2; 
 int j = n-1; 
 
 //逐个将所有值放在幻方
中（int num = 1; num< = n * n;）
 {
 if（ i ==  -  1&& j == n）//第3个条件
 {
j = n-2; 
 i = 0; 
} 
其他
 {
 //第一个条件助手如果下一个数字
 //走出正方形的右边
 if（j == n ）
j = 0; 
 
 //第一个条件助手如果下一个数字是
 //则超出正方形的上限
 if（i <0）
 i = n-1; 
} 
 
 //第二个条件
 if（magicSquare [i] [j]！= 0）
 {
 j  -  = 2; 
 i ++; 
继续; 
} 
 else 
 //设定数字
 magicSquare [i] [j] = num ++; 
 
 //第一个条件
 j ++;一世 - ; 
} 
 
 // print magic square 
 System.out.println（The Magic Square for+ n +：）; 
 System.out.println（每行或每列的总和+ n *（n * n + 1）/ 2 +：）; 
 for（i = 0; i< n; i ++）
 {
 for（j = 0; j< n; j ++）
 System.out.print（* ）; 
 System.out.println（）; 
} 
} 
 public static void main（String [] args）{
 //仅当n为奇数时才有效
 int n = 3; 
 generateSquare（n）; 
} 
 
} 

解决方案

仅回答从未答复列表中删除：由OP解决。

Implement a program that inputs two integers, m and n, and draws m 2 squares, each of the size n × n, arranged in an m × m table. That is, it draws a table with m rows and m columns, each element of which is an n × n square. For example, if the input integers are 4 and 6, it draws 16 squares, each of the size 6 × 6, arranged in a 4 × 4 table. The perimeter of each square is composed of the symbol "*" (asterisk), and the space inside is filled with the symbol "o" (lowercase letter o).

Input and output:

Your program should prompt the user to enter two integers, m and n, between 2 and 8, and draw m 2 squares, each of the size n × n.

What I have tried:

have no idea ,this hat i have tried

import java.io.*;
public class JavaApplication19 {

   static void generateSquare(int n)
    {
        int[][] magicSquare = new int[n][n];
         
        // Initialize position for 1
        int i = n/2;
        int j = n-1;
  
        // One by one put all values in magic square
        for (int num=1; num <= n*n; )
        {
            if (i==-1 && j==n) //3rd condition
            {
                j = n-2;
                i = 0;
            }
            else
            {
                //1st condition helper if next number 
                // goes to out of square's right side
                if (j == n)
                    j = 0;
                     
                //1st condition helper if next number is 
                // goes to out of square's upper side
                if (i < 0)
                    i=n-1;
            }
             
            //2nd condition
            if (magicSquare[i][j] != 0) 
            {
                j -= 2;
                i++;
                continue;
            }
            else
                //set number
                magicSquare[i][j] = num++; 
                 
            //1st condition
            j++;  i--; 
        }
  
        // print magic square
        System.out.println("The Magic Square for "+n+":");
        System.out.println("Sum of each row or column "+n*(n*n+1)/2+":");
        for(i=0; i<n; i++)
        {
            for(j=0; j<n; j++)
                System.out.print(" *");
            System.out.println();
        }
    }
    public static void main(String[] args) {
       // Works only when n is odd
        int n = 3;
        generateSquare(n);
    }
    
}

解决方案

Answered only to remove from unanswered list: solved by OP.

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