Java写在控制台运行的英国著名猜数字游戏

女爷i 2022-05-24 07:24 123阅读 0赞

关于猜数字规则请参考[百度][Link 1]，效果图

![20180523103750617][]

代码如下（个人思路有不足请指出）

import java.util.*;  
    class bullsCows  
    {  
        public static void main(String[]args)  
        {  
            System.out.println("猜数字");  
            int[]sj = new int[4];  
            for (int i = 0; i < 4; i++)  //随机生成4个数字  
            {  
                sj[i] = (int)(1+Math.random() * 9);  
                //System.out.print(sj[i]); //用于测试显示随机数结果  
            }  
            while (sj[0] == sj[1] || sj[0] == sj[2] || sj[0] == sj[3] || sj[1] ==  
                sj[2] || sj[1] == sj[3] || sj[2] == sj[3])  
            {  
                System.out.print(" ");  
                for (int i = 0; i < 4; i++) //如果四个数字有相同的数字再重新生成  
                {  
                    sj[i] = (int)(1+Math.random() * 9);  
                  //System.out.print(sj[i]); //用于测试显示随机数结果  
                }  
            }  
            int a = 0;  
            int b = 0;  
            int ci = 0;  
            while (a != 4)  
            {  
                Scanner input = new Scanner(System.in); //用户输入四位数  
                System.out.print("\n请输入四位数(每位数字不相同):");  
                double num = input.nextDouble();  
                int num1 = (int)num;  
                int[]sr = new int[4];  
                sr[0] = num1 / 1000 % 10;  
                sr[1] = num1 / 100 % 10;  
                sr[2] = num1 / 10 % 10;  
                sr[3] = num1 % 10;  
                a = 0;  
                b = 0;  
                for (int i = 0; i < 4; i++)  //判断几A几B  
                {  
                    if (sr[i] == sj[i])  
                    {  
                        a++;  
                    }  
                    for (int j = 0; j < 4; j++)  
                    {  
                        if (i != j && sr[i] == sj[j])  
                        {  
                            b++;  
                        }  
                    }  
                }  
                if (a != 4){  System.out.println(a + "A" + b + "B ,不正确请重新输入 ,消耗" + (++ci) + "次");  } //输出结果  
            }  
            System.out.println(a + "A" + b + "B, 完全正确 ! 消耗" + (++ci) + "次");  
        }  
    }

[Link 1]: https://baike.baidu.com/item/%E7%8C%9C%E6%95%B0%E5%AD%97/83200
[20180523103750617]: /images/20220524/a85d381c566f4e5b87d0bb5130e06651.png