java – 如何从控制台接受用户数据
发布时间：2020-12-15 04:30:43 所属栏目：Java 来源：网络整理
导读：我的控制台 java应用程序生成一个包含一些随机数的3×3矩阵.我想要做的是从集合中删除一些随机数,然后允许用户输入这些数字 到目前为止,我已经尝试了以下但是它不起作用 package magicsquare;import java.io.BufferedReader;import java.io.IOException;impo
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我的控制台
java应用程序生成一个包含一些随机数的3×3矩阵.我想要做的是从集合中删除一些随机数,然后允许用户输入这些数字
到目前为止,我已经尝试了以下但是它不起作用 package magicsquare;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Random;
public class MagicSquare {
/**
* @param args the command line arguments
*/
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.print("nnEnter the size of the matrix : ");
int n = Integer.parseInt(br.readLine());
if (n > 5 && n < 2) {
System.out.println("Enter a number between 2 to 5 ");
} else {
int A[][] = new int[n][n]; // Creating the Magic Matrix
int i,j,k,t;
/*Initializing every cell of the matrix with 0 */
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
A[i][j] = 0;
}
}
/* When the size of the matrix is Odd */
if (n % 2 != 0) {
i = 0;
j = n / 2;
k = 1;
while (k <= n * n) {
A[i][j] = k++;
i--; // Making one step upward
j++; // Moving one step to the right
if (i < 0 && j > n - 1) // Condition for the top-right corner element
{
i = i + 2;
j--;
}
if (i < 0) // Wrapping around the row if it goes out of boundary
{
i = n - 1;
}
if (j > n - 1) // Wrapping around the column if it goes out of boundary
{
j = 0;
}
if (A[i][j] > 0) // Condition when the cell is already filled
{
i = i + 2;
j--;
}
}
} /* When the size of the matrix is even */ else {
k = 1;
/* Filling the matrix with natural numbers from 1 till n*n */
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
A[i][j] = k++;
}
}
j = n - 1;
for (i = 0; i < n / 2; i++) {
/* swapping corner elements of primary diagonal */
t = A[i][i];
A[i][i] = A[j][j];
A[j][j] = t;
/* swapping corner elements of secondary diagonal */
t = A[i][j];
A[i][j] = A[j][i];
A[j][i] = t;
j--;
}
}
/* Printing the Magic matrix */
System.out.println("The Magic Matrix of size " + n + "x" + n + " is:");
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
//remove random element from array
Random rand = new Random();
for (i = 0; i < 4 ; i++)
{
int randomNum1 = rand.nextInt((n-1) - 0 + 1) + 0;
int randomNum2 = rand.nextInt((n-1) - 0 + 1) + 0;
BufferedReader dr = new BufferedReader(new InputStreamReader(System.in));
A[randomNum1][randomNum2] = Integer.parseInt(dr.readLine());
}
System.out.print(A[i][j] + "t");
}
System.out.println();
}
}
}
}
编辑 我想要发生的是矩阵显示时缺少一些数字.然后用户将光标放在缺失的位置,然后输入数字 解决方法
在您的代码中,您允许用户为显示的每个矩阵数选择多个数字,因此在打印一个数字之前需要多个输入.你想要这样的东西吗？
/* Printing the Magic matrix */
System.out.println("The Magic Matrix of size " + n + "x" + n + " is:");
Random rand = new Random();
for (i = 0; i < 2; i++) { //lets a user choose 2 numbers to be replaced
System.out.println("Please input a number: ");
int randomNum1 = rand.nextInt((n - 1) - 0 + 1) + 0;
int randomNum2 = rand.nextInt((n - 1) - 0 + 1) + 0;
BufferedReader dr = new BufferedReader(new InputStreamReader(System.in));
A[randomNum1][randomNum2] = Integer.parseInt(dr.readLine());
}
for (i = 0; i < n; i++) {
System.out.println("n");
for (j = 0; j < n; j++)
System.out.print(A[i][j] + "t");
}
System.out.println();
}
}
} 编辑：可能会发生相同的数字被此代码替换两次. （编辑：李大同） 【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容! |
