A title . . .

Farm heifer ,
rope heifers each year ,
old heifers cows ,
two decades later how much cattle ?


java app out there . . .

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you have to consider the bull breeding problems . . .
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import java.util.ArrayList;
  
  public class TestCow {
  
  	/**
  	 * @param args
  	 */
  	public static void main(String[] args) {
  		ArrayList<Cow> farm = new ArrayList<Cow>();
  		Cow source = new Cow();
  		farm.add(source);
  		// 每年都让农场的所有母牛过一次生日
  		for (int year = 1; year <= 20; year++) {
  			// 遍历所有的母牛，让它过生日
  			for (int i = 0; i < farm.size(); i++) {
  				Cow mother = farm.get(i);
  				Cow daughter = mother.birthday();
  				// 如果生了母牛则加入农场
  				if (daughter != null) {
  					farm.add(daughter);
  				}
  			}
  			System.out.println("第" + year + "年，农场规模：" + farm.size());
  		}
  	}
  
  }
  
  /**
   * 母牛类
   * 
   * @author mfanw
   * 
   */
  class Cow {
  	/**
  	 * 刚出生的年龄是0岁
  	 */
  	private int age = 0;
  
  	public int getAge() {
  		return age;
  	}
  
  	public void setAge(int age) {
  		this.age = age;
  	}
  
  	/**
  	 * 母牛过生日，如果超过5岁则生一个母牛
  	 * 
  	 * @return
  	 */
  	public Cow birthday() {
  		age++;
  		if (age >= 5) {
  			System.out.println("我超过5岁了，生一个母牛！");
  			return new Cow();
  		}
  		return null;
  	}
  
  }

------ Solution ------------------------------ --------------
recursive

public class BornOx {
  	public static void main(String[] args) {
  		System.out.println(born(1, 6, 20));
  	}
  
  	public static int born(int num, int age, int year) {
  		if (year <= 0) {
  			return num;
  		} else {
  			if (age >= 5) {
  				return born(num, age + 1, year - 1) + born(1, 1, year - 1);
  
  			} else {
  				return born(num, age + 1, year - 1);
  			}
  		}
  
  	}
  }

------ Solution ------------------------------------- -------

//农场类
  public class Farm {
  	// 所有的母牛
  	private List<Cattle> cattles = new ArrayList<Cattle>();
  
  	// 初始化农场，最初只有一只5岁的母牛
  	public Farm() {
  		Cattle cattle = new Cattle();
  		cattle.setAge(5);
  		cattles.add(cattle);
  	}
  	// 每过一年，计算出母牛数
  	public int cattleCountAfterOntYear() {
  		//遍历所有母牛，判断的没有超过5岁的母牛
  		for (int i = 0; i < cattles.size(); i++) {
  			Cattle cattle = cattles.get(i);
  			//由于过了一年，所以所有的母牛都加了一岁
  			cattle.increaseAge();
  			//判断母牛是不是5岁了
  			if (cattle.getAge() >= 5) {
  				//母牛生小母牛，初始化小母牛的年龄为0
  				Cattle newCattle = new Cattle();
  				cattles.add(newCattle);
  			}
  		}
  		// 返回总母牛数
  		return cattles.size();
  	}
  	public static void main(String[] args) {
  		Farm cattleCount = new Farm();
  		//20年后，的母牛数，有点多啊！！！900多。。。。。。
  		for (int i = 0; i < 20; i++) {
  			int thisYearNewCattleCount = cattleCount.cattleCountAfterOntYear();
  			System.out.println("第 " + (i + 1) + " 年的母牛数： "
  					+ thisYearNewCattleCount + " 头");
  		}
  	}
  }
  // 定义母牛类
  class Cattle {
  	int age;
  	public int getAge() {
  		return age;
  	}
  	public void setAge(int age) {
  		this.age = age;
  	}
  	// 每过一年，母牛会长一岁
  	public void increaseAge() {
  		age++;
  	}
  }

------ For reference only ------------------------- --------------
you have to consider the bull breeding problems . . .  

you can really pull . . . . .
------ For reference only -------------------------------------- -
you have to consider the bull breeding problems . . .          
  
you can really pull . . . . .   Hey hey hey , is being written. . .
------ For reference only -------------------------------------- -
you have to consider the bull breeding problems . . .                
    
you can really pull . . . . .        Hey hey hey , is being written. . .  
my math is not very good ah. . . . They did not come to this theme . . . Oh / pull the nose /
------ For reference only --------------------------------- ------
are ox bar, which can also be side-tracked procedure