空瓶子换汽水问题

       用空瓶子换汽水经常遇到 今天在论坛上有人提出来，给的算法用递归实现的 如果数量在百万那么时间复杂度我认为不合理

      其实我认为这个问题很简单 如果几十个人需要喝汽水 完全可以根据数学模型口算出来 也就是说时间复杂度为n

      数学模型：

      N个瓶子换一瓶汽水。（假设瓶子的价钱为1）

     瓶子价钱x 一瓶汽水除去瓶子本身的价钱实际价钱为y那么

     N * x = x + y；

    y = （N - 1 ) * x;

   那么至少需要的money就很容易计算出来就可以计算出开始需要买几瓶汽水了

  java代码如下：

  package cn.xy.csdn;public class Bottletosoda {public static final int bottoleprice = 1;// 一个瓶子的价钱private int percount = 3;// 每瓶汽水需要几个空瓶子换private int sodacount = -1;// 需要的汽水的数量public int sodaprice = 2;public boolean flag = false;//是否允许使用最后的两个瓶子换一瓶汽水在将空瓶抵给卖方public int getPercount() {return percount;}public void setPercount(int percount) {this.percount = percount;}public int getSodacount() {return sodacount;}public void setSodacount(int sodacount) {this.sodacount = sodacount;}public void compsodaprice() {this.setSodaprice(bottoleprice * (this.getPercount() - 1));}public int getSodaprice() {return sodaprice;}public void setSodaprice(int sodaprice) {this.sodaprice = sodaprice;}public int buysodanumber() {compsodaprice();int needminmoney = this.getSodaprice() * this.getSodacount();int lucysocle = needminmoney % (this.getSodaprice() + bottoleprice);int minnumber = needminmoney/ ((this.getSodaprice() + bottoleprice));float realnum = (float) needminmoney / (float) (this.getSodaprice() + bottoleprice);if ((!this.flag && lucysocle == 0) || realnum > (float)minnumber) {minnumber += 1;}return minnumber;}public static void main(String[] s) {Bottletosoda bt = new Bottletosoda();bt.setSodacount(27);System.out.println(bt.getSodacount() + "个人都喝到汽水至少应买的汽水瓶数为："+ bt.buysodanumber());}public boolean isFlag() {return flag;}public void setFlag(boolean flag) {this.flag = flag;}}