Farmer raising cattle

Here I saw a math in http://polaris.blog.51cto.com/1146394/399738:

A cow can have a calf every year when it is 3-10 years old. The ratio of a bull to a cow is 50%. When it is 12 years old, it is sent to a slaughterhouse for purchase.
Now there is a farmer who has a one-year-old cow. When the cow is 3 years old, it is sent to a nearby farm for breeding. How many cows may this farmer have in 40 years,
It is best to write the relevant code or answer ideas, and use object-oriented.

The following code is provided by myself. If there is any error, please correct it ~

Cow. Java:

Package test. question. niu; public class cow {int age; char sex; // F M malchar childsex; // Gender of the recently generated child public cow (INT age, char sex) {This. age = age; this. sex = sex;} public char getchildsex () {return childsex;} public void setchildsex (char childsex) {This. childsex = childsex;} public int getage () {return age;} public void setage (INT age) {This. age = age;} public int getsex () {return sex;} public void setsex (char sex) {This. sex = sex ;}}

Farmer class, farmer. Java

package test.question.niu;import java.util.ArrayList;import java.util.List;public class Farmer {List<Cow> cowList = new ArrayList<Cow>();public Farmer(){Cow cow = new Cow(1,'F');this.cowList.add(cow);}public void addCow(Cow cow){this.cowList.add(cow);}}

Test. Java

Package test. question. niu; import Java. util. arraylist; import Java. util. list;/***** a cow can have a calf every year at the age of 3 to 10. The ratio of a bull to a cow is 50%, when Niu was 12 years old, he sent it to the slaughterhouse to buy it. * Now there is a farmer who has a one-year-old cow. When the cow is 3 years old, it is sent to a nearby farm for breeding. How many cows may this farmer have in 40 years, * It is best to write related code or answer ideas using object-oriented programming. **/Public class test {public static void main (string ARGs []) {int year = 40; farmer = new farmer (); For (INT I = 1; I <= year; I ++) {// every year, every ox increases by one year. For (Int J = 0; I! = 1 & J <farmer. cowlist. size (); j ++) {farmer. cowlist. get (j ). setage (farmer. cowlist. get (j ). getage () + 1);} int size = farmer. cowlist. size (); // number of cattle (Int J = 0; j <size; j ++) {cow = farmer. cowlist. get (j); If (cow. getage ()> = 3 & cow. getage () <= 10 & cow. getsex () = 'F') {// a calf is generated only when the conditions are met. // here, each cow alternate to generate a bull or a cow, obtain the value with a probability of 50%. // you can also use (INT) math. round (math. random () to simulate probability events, char childsex = cow. getchildsex () = 'F '? 'M': 'F'; cow. setchildsex (childsex); // record the Gender cow newcow = New Cow (0, childsex) of the recently generated cattle; // The farmer, 0, just born. cowlist. add (newcow);} else if (cow. getage ()> 11) {// sell farmer if he is later than 11 years old. cowlist. remove (j) ;}} system. out. println ("owned by farmer after 40 years" + farmer. cowlist. size () + "head Ox ");}}