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.examplesdialect "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);endrule "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);endrule "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。
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