从java课本中画图

Question

我正在研究来自 Reges、Stuart 和 Martin Stepp 的第 2 章自检问题。构建 Java 程序：返璞归真方法。我试图获得下面的输出与我的代码。我正在尝试确定 line 到 ! 到 \ 到 / 的关系以及计算 for loops 所需的数学。这不是作业，我也不需要答案，方向或指导是我所寻求的。

!!!!!!!!!!!!!!!!!!!!!!
\!!!!!!!!!!!!!!!!!!//
\\!!!!!!!!!!!!!!////
\\\!!!!!!!!!!//////
\\\\!!!!!!////////
\\\\\!!//////////

我现在的代码是：

/**
 * Created on 8/28/15.
 * Following Reges, Stuart, and Martin Stepp. Building Java Programs: A Back to Basics Approach.
 * Chapter 2 Self-Check Problems
 */
public class Ch2_SelfCheckProblems {
    public static void main(String[] args) {
        question34();

    }

    public static void  question34(){
/**
* Table
* Line 1   ! = 22  \ = 0   / = 0
* Line 2   ! = 18  \ = 2   / = 2
* Line 3   ! = 14  \ = 4   / = 4
* Line 4   ! = 10  \ = 6   / = 6
* Line 5   ! = 6   \ = 8   / = 8
* Line 6   ! = 2   \ = 10  / = 10
*/

        for (int line = 1; line <= 6; line++){
            for (int i = 1; i <= 22; i++){
//                for (int j = 1; j <= (line - 1); j++){
//                    System.out.print("\");
//                }
                System.out.print("!");
            }
            System.out.println();
        }
    }



}

Answer 1

答案直接在您的代码中写了 "table" 的评论中，您在代码中写了符号的计数。 for循环能够以任何步长增加和减少

for(int i = 22 ; i>2; i=i-4) 例如

希望这不是太多。

Answer 2

试试这个：（我手边没有编译器）

for (int line = 1; line <= 6; line++){
  for(int i = 1; i<=line-1; i++) {
    System.out.print("\");
  }
  for (int i = 1; i <= 22 - 4*(line-1); i++){
    System.out.print("!");
  }
  for(int i = 1; i<=line-1; i++) {
    System.out.print("//");
  }
  System.out.println();
}

有什么不懂的可以留言。洗耳恭听。

Answer 3

这完全取决于您希望打印的行数。如果你有 x 行（这里是 6），那么你可以打印你想要的，如下：

int lines = 6;
for (int i = lines; i > 0; i--) { //start from the top line (6), finish at the lowest line (1)
    //print backslashes
    for (int back = 0; back < (lines-i)*2; back++) { //lines-i is the difference from the top line. add two extra slashes at each new line
        System.out.print("\");
    }

    //print !s
    for (int up = 0; up < (i*4)-2; up++) {
        System.out.print("!");
    }

    //print slashes (as many as the backslashes)
    for (int forw = 0; forw < (lines-i)*2; forw++) {
        System.out.print("/");
    } 
    System.out.println();            
}

如果您总是想要 6 行，那么只需跳过 int lines = 6; 语句并将所有地方的 lines 替换为 6。

因此，在第一行，您打印 4*x-2 '!'s and 0 '\'s and '/'s。
在第二行少打印 4 个 '!'s and 2 more '\'s and '/'s。
...
在最后一行打印 2 '!'s and (x-1)*2 '\'s and '/'s.

一般来说，当给定 x 行时，您正在寻找的关系如下：

At line i, counting from 1 (lowest) to x (top), print:
'\': (x-i)*2 times
'!': (i*4)-2 times
'/': (x-i)*2 times

Answer 4

/**
 * Created on 8/28/15.
 * Following Reges, Stuart, and Martin Stepp. Building Java Programs: A Back to Basics Approach. 3rd Edition.
 * Chapter 2 Self-Check Problems Question 34 & 35
 */
public class Ch2_SelfCheckProblems {

    public static final int SIZE = 4;

    public static void main(String[] args) {
        System.out.println("Practice:");
        question34();
        System.out.println("Partially correct:");
        q34();
        System.out.println("Solution, scalable with constant:");
        solution34();

    }

    /**
     * Table 6 lines; CONSTANT = 6
     * Line 1   ! = 22  \ = 0   / = 0
     * Line 2   ! = 18  \ = 2   / = 2
     * Line 3   ! = 14  \ = 4   / = 4
     * Line 4   ! = 10  \ = 6   / = 6
     * Line 5   ! = 6   \ = 8   / = 8
     * Line 6   ! = 2   \ = 10  / = 10
     *
     * Table 4 lines; CONSTANT = 4
     * Line 1   ! = 14  \ = 0   / = 0
     * Line 2   ! = 10  \ = 2   / = 2
     * Line 3   ! = 6   \ = 4   / = 4
     * Line 4   ! = 2   \ = 6   / = 6
     */

    public static void  question34(){

        for (int line = 1; line <= 6; line++){
            for (int i = 1; i <= (22-2*(line-1)); i++){
//                for (int j = 1; j <= (line - 1); j++){
//                    System.out.print("\");
//                }
                System.out.print("!");
            }
            System.out.println();
        }
    }

    public static void q34(){
        for (int line = 1; line <= 6; line++){
            for(int i = 1; i<=line-1; i++) {
                System.out.print("\\");
            }
            for (int i = 1; i <= 22 -(4*(line-1)); i++){
                System.out.print("!");
            }
            for(int i = 1; i<=line-1; i++) {
                System.out.print("//");
            }
            System.out.println();
        }
    }

    public static void solution34(){
        for (int line = 1; line <= SIZE; line++){
            for(int i = 1; i<=((2 * line) - 2); i++){
                System.out.print("\");
            }
            for (int i = 1; i <= ( -4 * line + ( 4 * SIZE + 2 ) ); i++){
                System.out.print("!");
            }
            for(int i = 1; i<= ((2 * line) - 2); i++){
                System.out.print("/");
            }
            System.out.println();
        }
    }
}

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