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基于面向对象实现:猫抓老鼠小游戏 有一只猫和一只老鼠,当猫的体重是小于老鼠体重的5倍时,猫可以抓到老鼠。

程序员文章站 2022-06-10 14:51:49
###有一只猫和一只老鼠,当猫的体重是小老鼠体重的5倍时,猫可以抓到老鼠,####猫每吃掉一只老鼠,体重增加老鼠的体重的0.5倍,####但是当猫的体重超过了老鼠体重的5倍,猫无法抓到老鼠,老鼠逃脱并对猫发出嘲讽。在这里插入代码片......

有一只猫和一只老鼠,当猫的体重是小于老鼠体重的5倍时,猫可以抓到老鼠,

猫每吃掉一只老鼠,体重增加老鼠的体重的0.5倍,

但是当猫的体重超过了老鼠体重的5倍,猫无法抓到老鼠,老鼠逃脱并对猫发出嘲讽。

创建一个Cat类
public class Cat {
 private String name;
 private double weight;
 
 Cat() { 
 }
public Cat(String name, double weight) {
  super();
  this.name = name;
  this.weight = weight;
 }
 public String getName() {
  return name;
 }
public void setName(String name) {
  this.name = name;
 }
 public double getWeight() {
  return weight;
 }
public void setWeight(double weight) {
  this.weight = weight;
 }
 public void catchMouse(Mouse jerry) {
  System.out.println("抓到啦");
  
 }      
创建一个Mouse类
public class Mouse {
 private String name;
 private double weight;
 
 Mouse() {
  
 }
 public Mouse(String name, double weight) {
  super();
  this.name = name;
  this.weight = weight;
 }
 public String getName() {
  return name;
 }
 public void setName(String name) {
  this.name = name;
 }
 public double getWeight() {
  return weight;
 }
public void setWeight(double weight) {
  this.weight = weight;
 }
 public void haha() {
  System.out.println("jerry说 :hahaa,你这个笨蛋");
  
 }
 }
创建一个Test_Mouse_Cat类
public class Test_Mouse_Cat {
public static void main(String[] args) {
 Mouse jerry = new Mouse("jerry",2);
 Cat Tom = new Cat("Tom",6);
 int count =0;
 
 double j =Tom.getWeight();
 while (j/2<5) {
 Tom.catchMouse(jerry);
   j+=1;
   count++;
   System.out.println("抓到"+count+"次");
   }jerry.haha();
 }
 }

本文地址:https://blog.csdn.net/weixin_45985600/article/details/109551597

相关标签: java