多维数组在Java中的Math.Random – 分别限制每列的数字

huangapple
go评论99阅读模式
英文:

Multi-Dimensional Arrays in Java Math.Random - Limiting numbers seperately for each column

问题

目前，我正在学习编程的过程中。我的目标是开发一个随机生成乘法问题的软件应用程序。为了实现这一目标，我希望将多维数组的第一列中生成的随机数限制在最大值为5，第二列限制在最大值为10。我已经在限制 Math.random 的范围到5或其他指定数字方面取得了一些进展，但不幸的是，这种限制适用于数组中的每一列。

以下是我的代码：

  1. package exercises;
  3. public class TimeTable {
  4.     public static void main(String[] args) {
  5.         int[][] nums = new int[10][2];
  7.         for (int i = 0; i < 10; i++) {
  8.             for (int j = 0; j < 2; j++) {
  9.                 nums[i][j] = (int) (Math.random() * 5 + 1);
  10.             }
  11.         }
  13.         for (int i = 0; i < 10; i++) {
  14.             System.out.println(nums[i][0] + " x " + nums[i][1] + " = ");
  15.         }
  16.     }
  17. }

输出结果为：

  1. 3 x 2 = 
  2. 1 x 1 = 
  3. 5 x 4 = 
  4. 2 x 4 = 
  5. 5 x 3 = 
  6. 1 x 4 = 
  7. 5 x 1 = 
  8. 2 x 5 = 
  9. 3 x 3 = 
  10. 5 x 1 =

从上面的示例中可以看出，乘法问题两边的数字都受到了最大值为5的限制。然而，我的目标是将第一列的最大值限制为5，同时允许第二列达到最大值10。这样可以使孩子在练习乘法时专注于特定的数字。

我已经尝试通过查看文档和搜索在StackOverflow上以前提出的问题来找到解决方案，但是尚未找到可行的解决方案。感谢您的帮助。

英文:

Currently, I am in the process of learning how to code. My goal is to develop a software application that generates multiplication questions randomly. In order to achieve this, I would like to restrict the random numbers generated in the first column of the multi-dimensional array to a maximum of 5, and to a maximum of 10 in the second column. I have made some progress in limiting the range of Math.random to 5 or any specified number, but unfortunately, this limitation applies to every column in the array.

Here is my code:

  1. package exercises;
  3. public class TimeTable {
  4.     public static void main(String[] args) {
  5.         int[][] nums = new int[10][2];
  7.         for (int i = 0; i &lt; 10; i++) {
  8.             for (int j = 0; j &lt; 2; j++) {
  9.                 nums[i][j] = (int) (Math.random() * 5 + 1);
  10.             }
  11.         }
  13.         for (int i = 0; i &lt; 10; i++) {
  15.             System.out.println(nums[i][0] + &quot; x &quot; + nums[i][1] + &quot; = &quot;);
  17.         }
  18.     }
  19. }

and the output is :

  1. 3 x 2 = 
  2. 1 x 1 = 
  3. 5 x 4 = 
  4. 2 x 4 = 
  5. 5 x 3 = 
  6. 1 x 4 = 
  7. 5 x 1 = 
  8. 2 x 5 = 
  9. 3 x 3 = 
  10. 5 x 1 =

From the previous example, it can be observed that the numbers on both sides of the multiplication questions are restricted to a maximum of 5. However, my objective is to limit the first column to a maximum of 5 while allowing the second column to reach a maximum of 10. This will enable children to focus on specific numbers while practicing multiplication.

I have attempted to find solutions by reviewing documentation and searching for previously asked questions on StackOverflow, but I have yet to find a viable solution. Thank you for your assistance.

答案1

得分: 2

不需要第二个循环，因为它们并不相同：

  1.     public static void main(String[] args) {
  2.         int[][] nums = new int[10][2];
  4.         for (int i = 0; i < 10; i++) {
  5.             nums[i][0] = (int) (Math.random() * 5 + 1);
  6.             nums[i][1] = (int) (Math.random() * 10 + 1);
  7.         }
  9.         for (int i = 0; i < 10; i++) {
  10.             System.out.println(nums[i][0] + " x " + nums[i][1] + " = ");
  11.         }
  12.     }

此外，最好创建一个显式的Random对象，因为在这种情况下，您可以访问一个更易读的API，无需转换为int，并且在创建随机生成器时可以通过玩弄种子来创建相同的序列：

  1. import java.util.Random;
  3. public class TimeTable {
  4.     public static void main(String[] args) {
  5.         int[][] nums = new int[10][2];
  6.         Random random = new Random(/* 在这里可以放置一个种子以获取相同的序列 */);
  8.         for (int i = 0; i < 10; i++) {
  9.             nums[i][0] = random.nextInt(5) + 1;
  10.             nums[i][1] = random.nextInt(10) + 1;
  11.         }
  13.         for (int i = 0; i < 10; i++) {
  14.             System.out.println(nums[i][0] + " x " + nums[i][1] + " = ");
  15.         }
  16.     }
  17. }
英文:

You don't need a second loop, since they are not doing the same:

  1.     public static void main(String[] args) {
  2.         int[][] nums = new int[10][2];
  4.         for (int i = 0; i &lt; 10; i++) {
  5.             nums[i][0] = (int) (Math.random() * 5 + 1);
  6.             nums[i][1] = (int) (Math.random() * 10 + 1);
  7.         }
  9.         for (int i = 0; i &lt; 10; i++) {
  10.             System.out.println(nums[i][0] + &quot; x &quot; + nums[i][1] + &quot; = &quot;);
  11.         }
  12.     }

Also it's better to create an explicit Random object, because in that case you get access to a more readable API, that does not require casting to int, and you can also create identical sequences by playing around with the seed when creating the random generator:

  1. import java.util.Random;
  3. public class TimeTable {
  4.     public static void main(String[] args) {
  5.         int[][] nums = new int[10][2];
  6.         Random random = new Random(/* you can put a seed here to get identical sequences*/);
  8.         for (int i = 0; i &lt; 10; i++) {
  9.             nums[i][0] = random.nextInt(5) + 1;
  10.             nums[i][1] = random.nextInt(10) + 1;
  11.         }
  13.         for (int i = 0; i &lt; 10; i++) {
  14.             System.out.println(nums[i][0] + &quot; x &quot; + nums[i][1] + &quot; = &quot;);
  15.         }
  16.     }
  17. }

答案2

得分: 1

以下是翻译好的内容：

这里有一个简单的方法来实现它。首先将数组的索引列更改为首列，然后是行。您不需要使用循环来填充数组。格式化的打印语句有助于保持'='对齐。

  • Random.ints 参数为 count, minValue, maxValue (exclusive)
  1. public class TimeTable {
  2.     public static void main(String[] args) {
  3.         int[][] nums = new int[2][]; // 长度由随机数确定
  4.         Random r = new Random();
  5.         nums[0] = r.ints(10, 1, 6).toArray(); // 1 到 5 包括 5
  6.         nums[1] = r.ints(10, 1, 11).toArray(); // 1 到 10 包括 10
  7.         for (int i = 0; i < nums[0].length; i++) {
  8.             System.out.printf("%d x %-2d = %n", nums[0][i], nums[1][i]);
  9.         }
  10.     }
  11. }

打印结果

  1. 3 x 2  = 
  2. 1 x 3  = 
  3. 4 x 10 = 
  4. 1 x 9  = 
  5. 3 x 5  = 
  6. 5 x 4  = 
  7. 1 x 1  = 
  8. 2 x 4  = 
  9. 2 x 10 = 
  10. 1 x 1  =
英文:

Here is an easy way to do it. Change your array to index columns first, then rows. You don't need a loop to fill the arrays. The formatted print statement helps keep the '=' aligned.

  • Random.ints arguments are count, minValue, maxValue (exclusive)
  1. public class TimeTable {
  2.     public static void main(String[] args) {
  3.         int[][] nums = new int[2][]; // length determined by Random
  4.         Random r = new Random();
  5.         nums[0] = r.ints(10, 1, 6).toArray(); // 1 thru 5 inclusive
  6.         nums[1] = r.ints(10, 1, 11).toArray();// 1 thru 10 inclusive
  7.         for (int i = 0; i &lt; nums[0].length; i++) {
  8.             System.out.printf(&quot;%d x %-2d = %n&quot;, nums[0][i], nums[1][i]);
  9.         }
  10.     }
  11. }

prints

  1. 3 x 2  = 
  2. 1 x 3  = 
  3. 4 x 10 = 
  4. 1 x 9  = 
  5. 3 x 5  = 
  6. 5 x 4  = 
  7. 1 x 1  = 
  8. 2 x 4  = 
  9. 2 x 10 = 
  10. 1 x 1  = 
  12. </details>

huangapple
  • 本文由 发表于 2023年3月7日 02:49:36
  • 转载请务必保留本文链接：https://go.coder-hub.com/75654700.html
  • java
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定