一个小学题目的解: 采用规则引擎Drools实现

同学帮他侄儿问我一个问题：
资源：
1、小明的钱可以购买50瓶汽水。
2、老板搞促销，两个空汽水瓶子可以换一瓶汽水。

问：小明最多能喝多少瓶汽水？

开始还楞了一下， 现在的小学题目还真不简单。不过马上给出答案：
先是50瓶，然后用50个空瓶换来25瓶，喝完后用25个空瓶换12瓶再多个空瓶，
喝完后用13个空瓶换6瓶，然后是7个空瓶换3瓶，然后是4个空瓶换2瓶，
然后是2个空瓶换1瓶，最后问老板借1个空瓶，再用2个空瓶换1瓶，
剩下来那个空瓶还给老板。所以喝到50+25+12+6+3+2+1+1=100

不过最近在学习规则引擎， 于是就想到了怎么用规则引擎来实现。
我的想法是：
1: 假设汽水一元一瓶， 则小明最初有50元。
2: 如果兜里有超过一元钱， 则买一瓶汽水喝， 此时：
钱减少一元，同时拥有的空瓶增加一个。
3: 如果拥有至少两个空瓶， 则两个空瓶卖给老板， 兜里的钱加一。
根据如上想法， 有了如下规则引擎的实现的实现(Drool)：

java文件SodaWater.java：

package org.drools.examples;

import java.io.InputStreamReader;
import java.io.Reader;

import org.drools.FactHandle;
import org.drools.RuleBase;
import org.drools.RuleBaseFactory;
import org.drools.StatefulSession;
import org.drools.WorkingMemory;
import org.drools.compiler.PackageBuilder;
import org.drools.rule.Package;

public class SodaWater
{
    public static final void main(String[] args) throws Exception {
        final PackageBuilder builder = new PackageBuilder();
        builder.addPackageFromDrl( new InputStreamReader( SodaWater.class.getResourceAsStream( "SodaWater.drl" ) ) );

        final RuleBase ruleBase = RuleBaseFactory.newRuleBase();
        ruleBase.addPackage( builder.getPackage() );

        final StatefulSession session = ruleBase.newStatefulSession();

        Customer customer = new Customer( "XiaoMing", 50);
        session.insert(customer);
        session.fireAllRules();
    }
   
    public static class Customer
    {
        private String name;
       
        private int money;
       
        private int drinkSum;
       
        private int blankCup;
       
        public Customer(String name, int money)
        {
            this.name = name;
            this.money = money;
            this.drinkSum = 0;
            this.blankCup = 0;
        }       

        public int getMoney()
        {
            return money;
        }
       
        public void setMoney(int money)
        {
            this.money = money;
        }
       
        public int getBlankCup()
        {
            return blankCup;
        }

        public void setBlankCup(int blankCup)
        {
            this.blankCup = blankCup;
        }

        public int getDrinkSum()
        {
            return drinkSum;
        }

        public void setDrinkSum(int drinkSum)
        {
            this.drinkSum = drinkSum;
        }

        public String getName()
        {
            return name;
        }

        public void setName(String name)
        {
            this.name = name;
        }      
       
    }
}

 

rule文件：

package org.drools.examples

dialect "mvel"

import org.drools.examples.SodaWater.Customer
 

rule "buy a soda water and drink"
    when
        $c : Customer(money > 0, $m:money, $b:blankCup, $d:drinkSum)
    then
        $c.money = $m - 1;
        $c.blankCup = $b + 1;
        $c.drinkSum = $d + 1;
        System.out.println( "Customer " + $c.name + " now buy a soda water and drink: money=" + $c.money + " and blankCup=" + $c.blankCup);
        update($c);
end

rule "sale blank cup and get money"
    when
        $c : Customer(blankCup > 1, $b:blankCup, $m:money )
    then
        $c.blankCup = $b - 2;
        $c.money = $m + 1;
        System.out.println("Customer " + $c.name + " now sale 2 cups and get money: money=" + $c.money + ", blankCup=" + $c.blankCup);
        update($c);
end

rule "finish drink"           
    no-loop true   
    dialect "java"
    when
        $c : Customer(blankCup < 2, money == 0)
    then
        System.out.println( "Customer " + $c.getName() + " finished drink, and drink number is " + $c.getDrinkSum() + " blankCup=" + $c.getBlankCup());
end

 

运行结果如下：
Customer XiaoMing now buy a soda water and drink: money=49 and blankCup=1
Customer XiaoMing now buy a soda water and drink: money=48 and blankCup=2
Customer XiaoMing now sale 2 cups and get money: money=49, blankCup=0
Customer XiaoMing now buy a soda water and drink: money=48 and blankCup=1
Customer XiaoMing now buy a soda water and drink: money=47 and blankCup=2

...

Customer XiaoMing now buy a soda water and drink: money=1 and blankCup=1
Customer XiaoMing now buy a soda water and drink: money=0 and blankCup=2
Customer XiaoMing now sale 2 cups and get money: money=1, blankCup=0
Customer XiaoMing now buy a soda water and drink: money=0 and blankCup=1
Customer XiaoMing finished drink, and drink number is 99 blankCup=1

不过小明最后手上还有最后一个瓶子， 计算机只能到此了。
向老板借一个瓶子换一瓶汽水，喝完在还一个空瓶子，估计就是人类智慧来。

 

此程序需要drools的jar包，具体可见：

http://lcllcl987.iteye.com/blog/254381

看看大家还有什么好的算法。