再犯尼姆游戏中的错误

Question

我不确定我是否应该为this...but创建一个新的问题，现在出现了一个新的问题。当计算机在所谓的“智能模式”中运行时，程序运行得很好，但是当程序以“愚蠢模式”运行时会出现错误

每当我接近赢得这场游戏(Snickerdoodles，我刚刚输掉了游戏)时，当计算机处于“愚蠢的mode...the错误”时，我总是会收到一个错误：

java.lang.IllegalArgumentException: n必须在java.util.Random.nextInt(Random.java:294)在GameOfNim.main(GameOfNim.java:62)呈阳性

我理解在这个实例中，n必须是肯定的，但是我甚至有一个if语句，它告诉代码如果实际发生这种情况应该做什么.

你们有什么解决办法吗？

哦，这里又是代码(我不包括智能模式方法，因为这与它无关)：

import java.util.*;
public class GameOfNim
{
    public static void main (String [] args)
    {
        Scanner in = new Scanner (System.in);
        Random num = new Random ();
        int numberLeft = num.nextInt(101-10) + 10;
        int computerMode = num.nextInt(2);
        int subtraction = numberLeft;
        boolean turn = num.nextBoolean();
        String computer = "";

        System.out.println ("The number you start out with is: " + numberLeft);
        System.out.println ("-------------------");

        if (computerMode == 0)
        {
            System.out.println ("The computer is in smart mode");
            System.out.println ("-------------------");
            computer = "Smart";
        }
        if (computerMode == 1)
        {
            System.out.println ("The computer is in dumb mode");
            System.out.println ("-------------------");
            computer = "Dumb";
        }

        while (numberLeft > 1)
        {
            if (turn == true)
            {
                System.out.println ("-------------------");
                System.out.println ("It is your turn...");
                System.out.printf ("Please enter the number you wish to take from the pile (Remember it has to be less than " + ((numberLeft/2) +1) + "): ");
                subtraction = in.nextInt();
                numberLeft -=subtraction;
                System.out.println ("The number left is " + numberLeft);
                turn = false;

            }
            if (turn ==false)
            {
                System.out.println ("-------------------");
                System.out.println ("It is the computer's turn...");
                if (computer.equals("Smart"))
                {
                    numberLeft = smartComputer(numberLeft);
                    System.out.println ("The number left is " + numberLeft);

                }

                if (computer.equals("Dumb"))
                {
                    if (numberLeft - num.nextInt(numberLeft/2) <= 1)
                    {
                        return;
                    }
                    else
                    {
                        numberLeft -= (num.nextInt(numberLeft/2 - 1) + 1);
                    }
                    System.out.println ("The number left is " + numberLeft);
                }
                turn = true;
            }
        }

        if (numberLeft <= 1)
        {
            if (turn = false)
            {
                System.out.println ("~~~~~~~~~~~~~~~~~~~~~~~~");
                System.out.println ("You Win!");
            }
            else
            {
                System.out.println ("~~~~~~~~~~~~~~~~~~~~~~~~");
                System.out.println ("You're horrible...you lost to a computer.");
            }
        }
    }

Answer 1

if (numberLeft <= 1)
    {
        if (turn = false)
        {
            System.out.println ("~~~~~~~~~~~~~~~~~~~~~~~~");
            System.out.println ("You Win!");
        }
    }

您在这里有一个任务，而不是一个等式检查。虽然我不认为这是整个问题的原因

Answer 2

这部分代码看上去很粗略。

if (turn == true)
            {
                System.out.println ("-------------------");
                System.out.println ("It is your turn...");
                System.out.printf ("Please enter the number you wish to take from the pile (Remember it has to be less than " + ((numberLeft/2) +1) + "): ");
                subtraction = in.nextInt();
                numberLeft -=subtraction;
                System.out.println ("The number left is " + numberLeft);
                turn = false;

            }
            if (turn ==false)
            {
               ...

在当轮到播放器时处理的代码块中，您将更改numberLeft的值，并设置turn=false。这意味着测试turn==false将为真，并将输入计算机程序生成随机整数的代码块。

您可以尝试这样做：

else if (turn == false)
{
    ...