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
2022-03-03 10:43:41
###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