猴子吃桃

猴子第一天摘下若干桃子，当即吃了一半，觉得好不过瘾，于是又多吃了一个第二天又吃了前天剩余桃子数量的一半，觉得好不过瘾，于是又多吃了一个以后每天都是吃前天剩余桃子数量的一半，觉得好不过瘾，又多吃了一个等到第 10 天的时候发现桃子只有 1 个了。

请问猴子第一天摘了多少个桃子？1534 个

递归问题 - 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JSRUN 在线运行

  1. public static int f(int n){
  2.     if(n == 10){
  3.         return 1;
  4.     }else {
  5.         return 2 * f(n + 1) + 2;
  6.     }
  7. }
  9. @Test
  10. public void test(){
  11.     System.out.println(f(1));
  12. }

啤酒问题

啤酒 2 元 1 瓶，4 个盖子可以换一瓶，2 个空瓶可以换一瓶，请问 10 元钱可以喝多少瓶酒，剩余多少空瓶和盖子。（15 瓶 3 盖子 1 瓶子）
JSRUN 在线运行

  1. // 定义一个静态的成员变量用于存储可以买的酒数量
  2. public static int totalNumber; // 总数量
  3. public static int lastBottleNumber; // 记录每次剩余的瓶子个数
  4. public static int lastCoverNumber; // 记录每次剩余的盖子个数
  7. public static void main(String[] args) {
  8.     // 1、拿钱买酒
  9.     buy(10);
  10.     System.out.println("总数：" + totalNumber);
  11.     System.out.println("剩余盖子数：" + lastCoverNumber);
  12.     System.out.println("剩余瓶子数：" + lastBottleNumber);
  13. }
  15. public static void buy(int money){
  16.     // 2、看可以立马买多少瓶
  17.     int buyNumber = money / 2; // 5
  18.     totalNumber += buyNumber;
  20.     // 3、把盖子和瓶子换算成钱
  21.     // 统计本轮总的盖子数和瓶子数
  22.     int coverNumber = lastCoverNumber + buyNumber;
  23.     int bottleNumber = lastBottleNumber + buyNumber;
  25.     // 统计可以换算的钱
  26.     int allMoney = 0;
  27.     if(coverNumber >= 4){
  28.         allMoney += (coverNumber / 4) * 2;
  29.     }
  30.     lastCoverNumber = coverNumber % 4;
  32.     if(bottleNumber >= 2){
  33.         allMoney += (bottleNumber / 2) * 2;
  34.     }
  35.     lastBottleNumber = bottleNumber % 2;
  37.     if(allMoney >= 2){
  38.         buy(allMoney);
  39.     }
  41.     Integer[] arr2 = new Integer[]{11, 22, 33};
  42.     Arrays.sort(arr2);
  43. }