Question

我目前正在对游戏拾取棒进行任务，其中代码提示用户输入游戏中的棒数，然后玩家1和玩家3选择在（1-3）棒之间选择捆绑和任何选择最后一根棍子的人赢得比赛。我目前停留在我的getWinner（）和我的pickUpSticks（）方法中。我的想法是尝试让玩家1为偶数而玩家2为奇数，因此通过将当前玩家增加1并使用if语句来查看是否为％2来决定玩家是什么，这将很容易在两者之间切换转过来吧。我遇到的问题是代码不能正常工作，而不是让赔率或平均值确定它是否为1或2，它只是增加了玩家，所以它变为0,1,2,3,4,5 ..我遇到的第二个问题是我不确定把声明放在哪里numOfSticks＆lt; 0，我现在拥有它的当前位置它允许numOfSticks变为负数但是一旦你尝试输入一个新值它会抛出异常，我希望它在它之前抛出异常允许它变为负数。下面我发布了代码，任何帮助将不胜感激。

import java.util.Scanner;

/**
 * Plays a game of Sticks over and over until the players wish to quit.
 * 
 *
 */
public class SticksGameApp 
{

    int intialNumOfSticks;
    int numOfSticks;
    int currentPlayer = 0; 


   public SticksGameApp () {
         intialNumOfSticks = 20;    
         numOfSticks = 20;  
   }

   public SticksGameApp(int startSticks) {
         intialNumOfSticks = startSticks;
         numOfSticks = startSticks;
   }

   public int getNumOfSticks() {
         return numOfSticks;

   }

   public int getCurrentPlayer() {
         return currentPlayer;
   }

   public String getPile() {
         String pile = "";

         for(int i=0;i<numOfSticks;i++){
               String c = "|";
               pile += c;
         }
         return pile;   
  }

   public boolean isOver() {
       if (numOfSticks == 0) return true;
       return false;
   }

   public int getWinner() {
   // player 1 is even, player 2 is odd
   int player1 = 1;
   int player2 = 2;
    if (currentPlayer  % 2 == 0) {
        return player1;
    }
    else if (currentPlayer % 2 == 1) {
        return player2;
    }
    else {
         return 0;
    }


   }

   public void reset() {
    currentPlayer = 0;
    numOfSticks = intialNumOfSticks;

   }

   public void pickUpSticks(int someInt)throws InvalidOptionException {
    if (someInt < 1 || someInt > 3 || numOfSticks < 0)throw new InvalidOptionException();
        currentPlayer++;
        numOfSticks = numOfSticks - someInt;


   }

   /**
     * Plays the game of Sticks over and over until the players quit.
     * 
     * @param args not used.
     */
    public static void main(String[] args)
    {       
        int numSticks;
      //Initialize a Scanner object to read from the console input stream.
        Scanner consuleInput = new Scanner(System.in);

        //Ask the user for the number sticks to play with.
        System.out.println("It's time for an exciting game of Sticks!");
        System.out.println("How many sticks would you like to start with?");
        int intialNumOfSticks = consuleInput.nextInt();

        //Create a sticks game with the inputed number of sticks.
        SticksGameApp myGame = new SticksGameApp(intialNumOfSticks);

        //Keep playing games until the players want to quit.
        String playAgain = "Y";
        while (playAgain.toUpperCase().startsWith("Y"))
        {
            //Start with a fresh game.
            myGame.reset();

            //Keep playing until the game is over.
            while (!myGame.isOver())
            {
                //Get the player's selection for the number of sticks to pick up.
                System.out.println("\n\nPlayer " + myGame.getCurrentPlayer() + " there are " + myGame.getNumOfSticks() + " stick(s) left.");
                System.out.println(myGame.getPile());
                System.out.print("How many do you want to take (1-3): ");
                numSticks = consuleInput.nextInt();

                //Try picking up that many sticks.  If there is an exception, have the player reenter.
                try
                {
                    myGame.pickUpSticks(numSticks);
                }
                catch(InvalidOptionException e)
                {
                    System.out.println("That is an invalid number of sticks!!! Try again.");
                }
            }

            //Once the game is over, display who won.
            System.out.println("Player " + myGame.getWinner() + " wins!");

            //As the user to play again.
            System.out.println("\n\nThat was exciting, would you like to play again?");
            playAgain = consuleInput.next();
        }

        //Close the scanner object.
        consuleInput.close();

    }




}


Player 0 there are 5 stick(s) left.
|||||
How many do you want to take (1-3): 2


Player 1 there are 3 stick(s) left.
|||
How many do you want to take (1-3): 1


Player 2 there are 2 stick(s) left.
||
How many do you want to take (1-3): 3


Player 3 there are -1 stick(s) left.

How many do you want to take (1-3): 2
That is an invalid number of sticks!!! Try again.


Player 3 there are -1 stick(s) left.

How many do you want to take (1-3): 1
That is an invalid number of sticks!!! Try again.


Player 3 there are -1 stick(s) left.

How many do you want to take (1-3): 3
That is an invalid number of sticks!!! Try again.

Answer 1

...而不是赔率或平均值确定它的1号或2号球员只是递增玩家所以它变为0,1,2,3,4,5 ...

当然可以。每次转弯后你都会++currentPlayer将currentPlayer设置为下一个玩家，那么为什么会做但从0到1，到2，......等等。即便如此，你对代表球员的赔率/均匀的想法仍然有效。唯一的问题是你需要测试currentPlayer奇数/偶数到处你想知道哪个玩家被代表，而不仅仅是在宣布获胜者的时候。

最好选择特定的值currentPlayer来代表两个玩家中的每一个，然后确保currentPlayer只能设置currentPlayer这些价值中的一个或另一个。然后你可以随时查看currentPlayer并立即知道哪个玩家的意思。这里有一个方便的技巧：如果让0和1代表两个玩家，你可以方便地在两个玩家之间切换currentPlayer = 1 - currentPlayer;：

numOfSticks < 0

我不知道把声明放在哪里numOfSticks＆lt; 0

不要把它放在任何地方！它太少，太晚了。如果if (someInt < 1 || // can't take less than one stick... someInt > 3 || // ... nor move then three ... someInt > numOfSticks) // ... nor more than are actually in the pile throw new InvalidOptionException();为真，则意味着之前的播放器被错误地允许使用比实际堆叠更多的棒，并且我们现在才发现它是目前的球员正试图移动。好的，很好，你知道发生了一件坏事，但是你无法做任何事情。太少，太晚了。

你必须首先阻止误入歧途的玩家拿太多的棍子，如下：

paramsFor()

对决定当前播放器的类方法感到困惑

1 个答案: