BlogJava-A life of continued exploration is a life worth living.-随笔分类-Data Structure && Algorithm

一个处理逻辑表达式的java程序

曲强 Nicky — Tue, 26 Feb 2008 11:05:00 GMT

摘要: 前两天写了一个处理逻辑表达式的小程序，可以用来处理专家系统实事等逻辑关系表达式。阅读全文

曲强 Nicky 2008-02-26 19:05 发表评论

classcial dialogue about Rule Engine_Rete...

曲强 Nicky — Thu, 29 Nov 2007 02:42:00 GMT

摘要: Have you read Jess In Action by Earnest J. Freedman-Hill? It's one of the better books out there on rule engines, though it does require some experience to get the most out of the book.
A variation on the bookstore could be recommendation engine. That would provide an opportunity to use a rule engine for direct matching on category/subcategory and aggregations. 阅读全文

曲强 Nicky 2007-11-29 10:42 发表评论

Trace Iamge

曲强 Nicky — Tue, 30 Oct 2007 13:36:00 GMT

摘要: Given a picture composed entirely of horizontal and vertical line segments, calculate the minimum number of times you must lift your pen to trace every line segment in the picture exactly n times.
Each line segment will be of the form " " (quotes for clarity), representing a segment from (x1,y1) to (x2,y2). Segments may cross each other. Segments may also overlap, in which case you should count the overlapping region as appearing in the drawing only once. For example, say t 阅读全文

曲强 Nicky 2007-10-30 21:36 发表评论

Lottery Again

曲强 Nicky — Wed, 24 Oct 2007 13:20:00 GMT

摘要: In most states, gamblers can choose from a wide variety of different lottery games. The rules of a lottery are defined by two integers (choices and blanks) and two boolean variables (sorted and unique). choices represents the highest valid number that you may use on your lottery ticket. (All integers between 1 and choices, inclusive, are valid and can appear on your ticket.) blanks represents the number of spots on your ticket where numbers can be written. 阅读全文

曲强 Nicky 2007-10-24 21:20 发表评论

A Q of Encrypting String

曲强 Nicky — Tue, 23 Oct 2007 05:34:00 GMT

摘要:
Let's say you have a binary string such as the following:
011100011
One way to encrypt this string is to add to each digit the sum of its adjacent digits. For example, the above string would become:
123210122 阅读全文

曲强 Nicky 2007-10-23 13:34 发表评论

Maze Problem

曲强 Nicky — Sun, 21 Oct 2007 13:28:00 GMT

摘要: People enjoy mazes, but they also get them dirty. Arrows, graffiti, and chewing gum are just a few of the souvenirs people leave on the walls. You, the maze keeper, are assigned to whiten the maze walls. Each face of the wall requires one liter of paint, but you are only required to paint visible faces. You are given a map of the maze, and you must determine the amount of paint needed for the job. 阅读全文

曲强 Nicky 2007-10-21 21:28 发表评论

TopCoder_2

曲强 Nicky — Sun, 21 Oct 2007 12:14:00 GMT

摘要: In written languages, some symbols may appear more often than others. Expected frequency tables have been defined for many languages. For each symbol in a language, a frequency table will contain its expected percentage in a typical passage written in that language. For example, if the symbol "a" has an expected percentage of 5, then 5% of the letters in a typical passage will be "a". If a passage contains 350 letters, then 'a' has an expected count of 17.5 for that passage (17.5 = 350 * 5%). Pl 阅读全文

曲强 Nicky 2007-10-21 20:14 发表评论

TopCoder_1

曲强 Nicky — Sat, 20 Oct 2007 05:25:00 GMT

摘要: A speed radar is installed in a highway zone where the maximum speed limit is maxLimit mph, and the minimum speed limit is minLimit mph. 阅读全文

曲强 Nicky 2007-10-20 13:25 发表评论

拐点提取法在曲线拟合中的简单应用

曲强 Nicky — Tue, 14 Aug 2007 09:27:00 GMT

摘要: 本程序为优化，比较好的方法是，进行曲线积分，算误差比进行方程约束，由于时间比较少，稍候会给出优化程序。阅读全文

曲强 Nicky 2007-08-14 17:27 发表评论

Stack Simple implement

曲强 Nicky — Fri, 03 Aug 2007 10:14:00 GMT

摘要: A simple implement of own stack 阅读全文

曲强 Nicky 2007-08-03 18:14 发表评论

To find a max segment in a array which includes negative and positive no.

曲强 Nicky — Tue, 31 Jul 2007 04:45:00 GMT

摘要: To find a max segment in a array which includes negative and positive no.There r several methods to solve this question. 阅读全文

曲强 Nicky 2007-07-31 12:45 发表评论

大家来试试运气啊...欢迎给出缩减版本

曲强 Nicky — Thu, 12 Oct 2006 10:45:00 GMT

/**
    *@ the titlt about a Random example about choose 7 from 33
    *@ the auther Nicky (EN) QuQiang(CH)
    *@ the date   2006.9.1
**/

/** the rules

//一等奖：选中６个正选号及特别号；

//二等奖：选中５个正选号及特别号；

//三等奖：选中５个正选号；

//四等奖：选中４个正选号及特别号；

//五等奖：选中４个正选号或选中３个正选号及特别号；

//六等奖：选中３个正选号。

**/

import java.util.*;

public class NotSameRandoml{
   private static String transform;
   private static String match="00";
   private static int special;

//产生彩票主逻辑函数
   private static void Nicky(int[] guess){
      Random r = new Random();     //构造伪随机生成器
      //某些映射实现可明确保证其顺序，如 TreeMap 类；某些映射实现则不保证顺序，如 HashMap 类
      Map map = new TreeMap();   //Map 接口的实现
      int n = 0;
      int nt = 1;
      String[] temps=new String[7];

      while(true){
          n = r.nextInt(33)+1; //产生1~33的随机数
          //if( map.get(new Integer(n))!=null){
          //   nt = ((Integer)map.get(new Integer(n))).intValue();
          //}
          //避免了产生的随机数字重复
          if(map.containsValue(n)){
            continue;
          }
          map.put(new Integer(nt),new Integer(n));//将指定的值与此映射中的指定键相关联
          if(map.size()==7){
             break;
          }
          nt++;
      }

      Iterator it = map.keySet().iterator(); //返回此映射中包含的键的 set 视图
      for(int i=0;it.hasNext();i++){
       Object o = it.next();
        // 为了更符合现实中33选7，数字为01。。。2位
       int temp=((Integer)map.get(o)).intValue();
       if(temp>=1&&temp<10){
           transform="0"+Integer.toString(temp);
           match=match+" "+transform;
           temps[i]=transform;
           if(((Integer)o).intValue()==7){
                   special=temp;
                   System.out.println(""+transform+"为产生的特别中奖中奖号码");
              }else
                 System.out.println(""+transform+"为产生的第"+((Integer)o).intValue()+"个中奖号码");
        }else{
        temps[i]=Integer.toString(temp);
        match=match+" "+temps[i];
        if(((Integer)o).intValue()==7){
           System.out.println(""+transform+"为产生的特别中奖中奖号码");
        }else
        System.out.println(""+temp+"为产生的第"+((Integer)o).intValue()+"个中奖号码");
       }
      }
      String creat=match.substring(3);
      System.out.println("所产生的中奖号码串为:"+creat);
      //System.out.println("对产生的中奖号码顺序排序为:"+creats);
      Sort(temps);
      check(map,guess);
   }

   //实现排序，也可以调用方法，但是却必须要解决Void问题
    private static void Sort(String[] temps) {
         for(int i=0;i         for(int j=i+1;j         if(Integer.parseInt(temps[i])>Integer.parseInt(temps[j])){
            String k;
            k=temps[i];temps[i]=temps[j];temps[j]=k;
         }
        }
      }
      System.out.println("对产生的中奖号码顺序排序为:");
      for(int i=0;i          System.out.print(temps[i]+" ");
      }
      System.out.println("\n");
    }

//输出结果类别
   private static void check(Map map ,int[] guess){
    int flag=0;
       for(int i=0;i           if(map.containsValue(guess[i])){
                  flag++;
             }
       }
       if(guess[guess.length-1]==special){
          flag=flag+10;
       }
       switch(flag){
            case 16: System.out.println("恭喜您中一等奖");break;
            case 15: System.out.println("恭喜您中二等奖");break;
            case 5: System.out.println("恭喜您中三等奖");break;
            case 14: System.out.println("恭喜您中四等奖");break;
            case 13: System.out.println("恭喜您中五等奖");break;
            case 4: System.out.println("恭喜您中五等奖");break;
            case 3: System.out.println("恭喜您中六等奖");break;
            default: System.out.println("谢谢参与，祝您下次中奖");
       }
   }

//说明

   private static void usage(){
      System.out.println("Usage:java Randomol program [the number you guess for the lucky nums.]");
      System.out.println("\t And the nums. you must typed 7,else you will be cancel by the game rules");
      System.out.println("\t The first 6 nums is your basic nums.,the last one is your special num.");
      System.exit(0);
   }

//主函数
   public static void main(String []args){
    if(args.length==0||args.length>7){
          usage();
       }//带入参数
       int[] guess=new int[7];
       for(int i=0;i            guess[i]=Integer.parseInt(args[i]);
       }
       //判断所输入的号码是否相同
    List ls= new ArrayList();
    for(int i=0;i          if(ls.contains(guess[i])){
            System.out.println("您所买的号码不可以相同");
            System.exit(0);
         }else ls.add(guess[i]);
    }
    Nicky(guess);
    System.exit(0);
   }
}

曲强 Nicky 2006-10-12 18:45 发表评论

今天突然想写了个折半查找

曲强 Nicky — Sun, 27 Aug 2006 09:02:00 GMT

呵呵不知道写的如何。但是总感觉，在UltraEdit下写，还是很爽的，或许是因为学校办公室的电脑内存太小的缘故吧。

1 /**
2  * @the author:Nicky(EN) QuQiang(CH)
3  * @the data :2006.8.27
4  **/
5 // 当然本算法最好编制成两到3个文件
6
7 public class HalfSearch{
8     private static int lengthValue;
9     private static byte  flag=0;
10     private static int SearchValue=0;
11
12     //定义返回数组类型的方法  此方法给出的public权限
13     public static int[] Sort(int[] b){
14         for(int i=0;i<b.length;i++){
15             for(int j=i+1;j<b.length;j++){
16                 int temp;
17                 if(b[i]>b[j]){
18                     temp=b[i];b[i]=b[j];b[j]=temp;
19                 }
20             }
21         }
22         System.out.println("\u6309\u5e8f\u6392\u5217\u4e3a:");  //对应汉字为     按序排列为:
23         for(int i=0;i<b.length;i++){
24           System.out.println(b[i]);
25         }                            //冒泡排序法排序  或直接使用ＪＤＫ提供的sort()函数进行排列
26         return b;
27     }
28
29     //折半查找
30     private static int Half(int[] a){
31         int m=a.length,k,n=1;
32         System.out.println("\u5f53\u524d\u5171\u8f93\u5165"+m+"\u4e2a\u6570\u7b26"); //对应汉字为  "当前共输入" 个数符
33         int index=0;
34         if(m==1){
35            if(a[0]==SearchValue){
36              flag=1;
37              index=1;
38            }
39            else{
40               flag=0;
41               index=1;
42            }
43         }             //考虑到用户除输入查找数符外，只输入一个数符，即n=m=1
44         for(int i=1;;i++){
45           if(n<m){
46              index+=1;
47              k=(n+m)/2;
48              if(a[k-1]>SearchValue){
49                  m=k-1;      //此时定义最大值为当前中间值的前一个值  //第次查找未成功，当前值大于SearchValue
50                  System.out.println("\u7b2c"+index+"\u6b21\u67e5\u627e\u672a\u6210\u529f\uff0c \u5f53\u524d\u503c\u5927\u4e8e"+SearchValue);//否则将9改为变量
51              }else if(a[k-1]<SearchValue){
52                  n=k+1;          //未找到，则在后半区间进行查找  //小于
53                  System.out.println("\u7b2c"+index+"\u6b21\u67e5\u627e\u672a\u6210\u529f\uff0c \u5f53\u524d\u503c\u5c0f\u4e8e"+SearchValue);//否则将9改为变量
54              }else{
55                  System.out.println("\u67e5\u627e\u6210\u529f");  //查找成功
56                  flag=1;
57                  break;
58              }
59            }else break;
60          }
61         return index;
62     }                      //也可以定义一个用户需输入的查找值
63
64     //使用说明
65    private static void usage(){
66       System.out.println("Usage:java HalfSearch program [the number you search] [the array list you give]");
67       System.out.println("\t [the number you search] the only number you want to use it to test this program");
68       System.out.println("\t [the Array list you give] A array list may be it contains it may be not");
69       System.exit(0);
70    }
71
72     //主函数的功能  传入查找数组
73     public static void main(String[] args){
74         if(args.length==0||args.length==1){
75             usage();
76             System.exit(0);
77         }
78         SearchValue=Integer.parseInt(args[0]);
79         lengthValue=args.length;
80         int index=0;
81         int[] TransArray=new int[args.length-1];
82         for(int i=1;i<args.length;i++){
83             TransArray[i-1]=Integer.parseInt(args[i]);
84             System.out.println(TransArray[i-1]);
85         }
86         index=HalfSearch.Half(HalfSearch.Sort(TransArray));
87         //输出查询结果
88         switch(flag){
89              //在本次折半查找中共经过  次查找,未成功   成功
90             case 0: System.out.println("\u5728\u672c\u6b21\u6298\u534a\u67e5\u627e\u4e2d\u5171\u7ecf\u8fc7"+index+"\u6b21\u67e5\u627e,\u672a\u6210\u529f");break;
91             case 1: System.out.println("\u5728\u672c\u6b21\u6298\u534a\u67e5\u627e\u4e2d\u5171\u7ecf\u8fc7"+index+"\u6b21\u67e5\u627e,\u6210\u529f");break;
92         }
93         System.exit(0);
94     }
95 }

曲强 Nicky 2006-08-27 17:02 发表评论

求任意正整数的阶乘

曲强 Nicky — Sat, 22 Jul 2006 09:21:00 GMT

在Java中，我们只要利用BigInteger类，可以完成同样功能；这里也测试了异常以及Dialog的产生。

开发环境：Windows Server 2003 Standard Edition SP1, J2SDK 1.5.0_06, Eclipse 3.1.2
源代码如下：

//Factorial.java
import java.math.BigInteger;
import javax.swing.*;


/**
 * 计算任意正整数的阶乘
 *  
 * 
 */
public class Factorial {
 public static void main(String[] args) {
  BigInteger x = BigInteger.valueOf(1); //存储结果
  int num = 1; //待计算的整数
  String s = null;
  boolean correct = false;
  do {
   try {
    s = JOptionPane.showInputDialog(null, "请输入要计算的数(正整数)：");
    if (s == null)
     break;
    else {
     num = Integer.parseInt(s);
     if (num < 0) 
      throw new IllegalArgumentException();
     else correct = true; 
    }  
   } catch (NumberFormatException e) {
    JOptionPane.showMessageDialog(null, "数据格式错误！");
    continue;
   } catch (IllegalArgumentException e) {
    JOptionPane.showMessageDialog(null,"请输入一个正整数！");
    continue;
   }
   break;
  } while (true);
  if (correct == true) {
   for (int i = 1; i <= num; i++)
    x = x.multiply(BigInteger.valueOf(i));
   JTextArea textArea = new JTextArea(x.toString(), 5, 30);
   textArea.setEditable(false);
   textArea.setLineWrap(true);
   Object[] object = {num + " ! : ",new JScrollPane(textArea)};
   JDialog dialog = new JOptionPane(object).createDialog(null,"阶乘的结果");
   dialog.setVisible(true);
  }
  System.exit(0);
 }
}

运行结果如下：

曲强 Nicky 2006-07-22 17:21 发表评论