Question

对于我所在的课程，我们有一个程序要写，要求玩家猜测x和y坐标，直到找到宝藏。该项目由2个名为Treasure.java和TreasureHunt.java的文件组成，它们使用Treasure类来玩游戏。我到目前为止的代码是......

package treasurehunt;

public class Treasure {

public int POSITIONS_PER_ROW = 20;
public int MAX_DISTANCE = 401;
private String name;
private int xLocation;
private int yLocation;
private Boolean found;


    public Treasure(){
     name = "";
     xLocation = 0;
     yLocation = 0;
     found = false;
    }

    public Treasure(String name){
        this.name = name;
        xLocation = 0;
        yLocation = 0;
        found = false;
    }

    public int getXLocation(){
        return xLocation;
    }

    public int getYLocation(){
        return yLocation;
    }

    public boolean isFound(){
        return found; 
    }

    public void hideTreasure(){
        xLocation = 1+(int)(Math.random()*POSITIONS_PER_ROW);
        yLocation = 1+(int)(Math.random()*POSITIONS_PER_ROW);
    }

    public int treasureStatus(int xStick , int yStick){
     if (xStick == xLocation && yStick == yLocation){
        return 0;
    } else if (xStick != xLocation || yStick != yLocation) {
        return Math.abs((yStick-yLocation)+(xStick-xLocation));           
    } else return MAX_DISTANCE;



}


}

--------------------------------------- AND ------- ------------------------------------------

package treasurehunt;

import java.util.Scanner;

public class TreasureHunt {

public static final int MAX_DISTANCE = 401;
public static final int POSITIONS_PER_ROW = 20;

public static void main(String[] args) {
    Treasure gold = new Treasure();
    Treasure diamond = new Treasure();
    char playAgain = 0;

    gold.hideTreasure();
    diamond.hideTreasure();

    Scanner keyboard = new Scanner(System.in);

    System.out.println("Treasures have been hidden.");

    do{
    System.out.println("");
    System.out.print("Enter x and y coordinates to search: ");
    int xStick = keyboard.nextInt();
    int yStick = keyboard.nextInt();

    int dist1 = diamond.treasureStatus(xStick, yStick);
    int dist2 = gold.treasureStatus(xStick , yStick);

    if (dist1 == 0 || dist1 == MAX_DISTANCE){
        System.out.println("Diamonds: FOUND!");
    }else{
        System.out.println("Diamonds: " + dist1 + " away");
    }

    if (dist2 == 0 || dist2 == MAX_DISTANCE){
        System.out.println("Gold: FOUND!");
    }else{
        System.out.println("Gold: " + dist2 + " away");
    }

    }while(!diamond.isFound() && !gold.isFound());




    do{
    System.out.println("Play again? y/n:");
    playAgain = keyboard.next().charAt(0);
    }while(diamond.isFound() && gold.isFound());

}
}

当程序正确运行时，应该看起来像这样....

Treasures have been hidden.
Enter x and y coordinates to search: 17 11
Diamonds: 2 away
Gold: 9 away

Enter x and y coordinates to search: 17 13
Diamonds: FOUND!
Gold: 7 away

Enter x and y coordinates to search: 21 13
Diamonds: FOUND!
Gold: 9 away

Enter x and y coordinates to search: 17 16
Diamonds: FOUND!
Gold: 4 away

Enter x and y coordinates to search: 17 19
Diamonds: FOUND!
Gold: 1 away

Enter x and y coordinates to search: 18 19
Diamonds: FOUND!
Gold: FOUND!

It took you 6 tries to find 2 treasures.

Play again? y/n: n

我的问题是，一旦你找到第一个宝藏，你如何防止它改变“发现！”当你猜到第二个宝藏的坐标时找不到。我现在的代码删除了“FOUND！”并在每次猜到第二个宝藏的坐标时将其更改为＃away。谢谢你的帮助！

Answer 1

尝试这样的事情（这与你的初始帖子差不多）。你需要更新你的Treasure类添加这样的setFound方法，并按如下方式更改你的hideTreasure方法。

// Set's the found 
public void setFound(boolean found) {
  this.found = found;
}

// Setting found to false on hide!
public void hideTreasure() {
  this.found = false;
  xLocation = 1 + (int) (Math.random() * POSITIONS_PER_ROW);
  yLocation = 1 + (int) (Math.random() * POSITIONS_PER_ROW);
}

// Finally update the main method to use the changes above.
public static void main(String[] args) {
    Treasure gold = new Treasure();
    Treasure diamond = new Treasure();
    char playAgain = 0;

    gold.hideTreasure();
    diamond.hideTreasure();

    Scanner keyboard = new Scanner(System.in);
    System.out.println("Treasures have been hidden.");

    try {
      do {
        System.out.println("");
        System.out
            .print("Enter x and y coordinates to search: ");
        int xStick = keyboard.nextInt();
        int yStick = keyboard.nextInt();

        if (!diamond.isFound()) {
          int dist1 = diamond.treasureStatus(xStick,
              yStick);
          if (dist1 == 0 || dist1 == MAX_DISTANCE) {
            diamond.setFound(true);
            System.out.println("Diamonds: FOUND!");
          } else {
            System.out.println("Diamonds: " + dist1
                + " away");
          }
        }

        if (!gold.isFound()) {
          int dist2 = gold.treasureStatus(xStick,
              yStick);
          if (dist2 == 0 || dist2 == MAX_DISTANCE) {
            gold.setFound(true);
            System.out.println("Gold: FOUND!");
          } else {
            System.out.println("Gold: " + dist2
                + " away");
          }
        }
        if (diamond.isFound() && gold.isFound()) {
          System.out.println("Play again? y/n:");
          playAgain = keyboard.next().charAt(0);
          if (playAgain != 'Y' && playAgain != 'y') {
            System.exit(0);
          } else {
            diamond.hideTreasure();
            gold.hideTreasure();
          }
        }
      } while (true);
    } finally {
      keyboard.close();
    }
  }

Answer 2

我认为你可以使用一个标志（int值或boolean）值。并且您可以根据找到的宝藏状态或距离改变距离来更改旗帜的值。

我认为您可以声明一个变量，如

int flag=0;

一旦你发现宝藏标志= 1;

 if( flag=0)
 {
     String pout="write distance value here"
 }

 if(flag==1)
 {
    String pout="found !!";
 }

Answer 3

你真的需要让found成为treasureStatus方法中的永久停止点。现在你有一个＆＃34;发现＆＃34;变量，但除非你在代码中使它成为一个不可改变的停止点，否则它对你没有任何好处。

在宝藏状态方法中有类似的东西...

 public int treasureStatus(int xStick , int yStick){
 if ((xStick == xLocation && yStick == yLocation) || found == true){
    return 0;
 } else if ...

这将使方法if found == true每次只需return 0，将Treasure标记为已找到。

（作为一个注释，第一次回答Stack Overflow问题，我为我可怕的格式化道歉。）

Java中的寻宝计划

3 个答案: