Question

我实际上正在制作累积奖金游戏，它会生成随机数并要求用户猜测随机数，如果它符合打印的条件胜利，如果不符合则转移到其他条件语句。

我差不多完成了我的代码编写，只是我对这一步感到困惑。我的疑问是：

我在randInt(max)上设置了一个限制a，即随机数，但我想知道如果它超出了限制，它必须显示一条我无法自拔的消息。

如果有人可以，请分享您的想法。

public class Jackpot {

static Scanner input = new Scanner(System.in);
public static void main(String[] args) {
    int mode; 
    System.out.println("Choose the difficulty level mode");
    System.out.println("1: Easy (0-15)");
    System.out.println("2: Medium (0-30)");
    System.out.println("3: Difficult (0-50)");
    System.out.println("or type another number to quit");
    mode = input.nextInt();
    if(mode == 1){
        Start_Game(randInt(15));
    }else if(mode == 2){
        Start_Game(randInt(30));
    }else if(mode == 3){
        Start_Game(randInt(50));
    }else if(mode <= 4 || mode >= 0){
        System.out.println("Quit");
        Restart_Game();
    }    
}
public static int randInt(int max){
    int randomNum = (int) (Math.random() * ((max) + 1));

    return randomNum;
}
 public static int Start_Game(int max){  
    System.out.println("Please enter your guessed number");
    int GuessNum , life = 5;; 

    do{
        GuessNum = input.nextInt();

         if(GuessNum == max){
            System.out.println("You WIN");
            break;
        }else if(GuessNum > max){
            System.out.println("BIG");
            life--;
            System.out.println("life: " + life);
        }else if(GuessNum < max){
            System.out.println("SMALL");            
            life--;
            System.out.println("life: " + life);
        }

    }while(life != 0);
    Restart_Game();
    return 0;
}
public static void Restart_Game(){
    System.out.println("if you want to restart the game \nPress Y to continue or N to exit");
    char restart = input.next().charAt(0);
    if(restart == 'Y' || restart == 'y'){
        //Start_Game(randInt(10));
        int mode; 
        System.out.println("Choose the difficulty level mode");
        System.out.println("1: Easy (0-15)");
        System.out.println("2: Medium (0-30)");
        System.out.println("3: Difficult (0-50)");
        System.out.println("or type another number to quit");
        mode = input.nextInt();
        if(mode == 1){
            Start_Game(randInt(15));

        }else if(mode == 2){
            Start_Game(randInt(30));
        }else if(mode == 3){
            Start_Game(randInt(50));
        }else if(mode <= 4 || mode >= 0){
            System.out.println("Quit");
            Restart_Game();
        }
    } else if(restart == 'n'|| restart == 'N'){
        System.out.println("Thank you for playing");
        System.exit(0);
    }
}

private static int randIt(int max) {
   throw new UnsupportedOperationException("Not supported yet."); //To change body of generated methods, choose Tools | Templates.
}
}

Answer 1

你的“游戏”没有实际范围的线索，即使你有一个参数调用“max”，它只是找到的值，你可以重命名它或你真的可以发送接受的最大值：

Start_Game(15);

让游戏基于此生成随机值。您已找到max值和random value。现在很容易检查你的情况。

public static int Start_Game(int max){  
    int toFind = randInt(max);

    do{
        GuessNum = input.nextInt();

        if(GuessNum > max){
            System.out.println("Out of range");
        } else if(GuessNum == toFind ){
            System.out.println("You WIN");
            break;
        }else if(GuessNum > toFind ){
            System.out.println("BIG");
            life--;
            System.out.println("life: " + life);
        }else if(GuessNum < toFind ){
            System.out.println("SMALL");            
            life--;
            System.out.println("life: " + life);
        }

    }while(life != 0);
}

注意：你应该重命名你的方法和变量，JAVA有一个约定（就像其他语言一样），不要命名一个以大写字母开头的变量。

public static int startGame(int max){

或

int guessNum;

超出java中随机数的限制？

1 个答案: