題目：古典問題：有一對兔子，從出生後第3個月起每個月都生一對兔子，小兔子長到第三個月後每個月又生一對兔子，假如兔子都不死，問每個月的兔子對數為多少？
程式分析：哇是斐波那契數列1,1,2,3,5,8,13,21… 用遞迴fn = fn-1 +fn-2

吐槽：題幹別讀成小兔子長到第三個月後（的第四個月）又生了一對兔子，這樣遞迴會變成fn = fn-2+fn-3+fn-4…，小兔子是2周月就成熟！

遞迴解法：從第三月開始，總數為前2個月的和：

	public static int countPairsOfRabbits(int month) throws Exception{
 

		if(month == 1 || month == 2){
			return 1;
		}else if(month <=0){
			throw new Exception("invalid argument of month");
		}else{
			return countPairsOfRabbits(month-1)+countPairsOfRabbits(month-2);
		}
	}

如果要問每個月小兔子和大兔子的對數怎麼辦？

對數： 小兔+大兔=兔子
初始1對小兔，花2個月變成大兔子，然後大兔子生同樣數量的小兔。

可以把小兔放進一個長度2的陣列，模擬佇列，出列合併大兔生仔~

	public static void countRabbitAndBunny(int month) throws Exception{

		if(month == 1 || month == 2){
			System.out.println("Pairs: Sum: 1 || Rabbit: 0 || Bunny: 1");
		}else if(month <=0){
			throw new Exception("invalid argument of month");
		}else{
			//where we found our bunnies
			int[] bunny =
 
 new int[2];
			
			/*
			 * the second month status
			 */
			bunny[1] = 1; //bunny of month 2 in the array  
			int pairOfBunny = 1;
			int pairOfRabbit = 0;	
			int sum = pairOfBunny + pairOfRabbit;
			
			//count the "i"th month status
			for(int i=3;i<=month;i++){
				//bunnies became rabbits
				pairOfRabbit = pairOfRabbit + bunny[1];
				pairOfBunny = pairOfBunny - bunny[1];
				
				//some bunnies grew up
				bunny[1] = bunny[0];
				//some bunnies were just born
				bunny[0] = pairOfRabbit;
				pairOfBunny = pairOfBunny + bunny[0];
				
				//count all rabbits alive
				sum = pairOfBunny + pairOfRabbit;
			}
			
			//we now have the result
			System.out.println("Pairs: Sum: " + sum + "|| Rabbit:  "
			+ pairOfRabbit + " || Bunny: "+pairOfBunny);
		}
	}