多线程之模拟售票

请你编写一个程序用于模拟多窗口售票。要求：1、窗口数目由键盘输入。2、总的票数也由键盘输入。3、每一张票都有一个编号，从1号开始。4、每个窗口售票都是先售出号码最小的票。5、输出每个窗口售票的情况：窗口号+所售票号。6、最后输出每个窗口售出的票的总数。7、当票售完后，窗口输出“票已售完”，并结束线程。

import java.util.Scanner;
public class Main {
public static void main(String[] args) {
  // TODO 自动生成的方法存根
  Scanner cin = new Scanner(System.in);
  while(cin.hasNext()){
   int ticket = cin.nextInt();
   int window = cin.nextInt();
   Thread thread[] = new Thread[window+1];
   Myth my = new Myth(ticket,window);
   for(int i = 1;i<=window;i++)
   {
    thread[i] = new Thread(my,"售票窗口"+i);
    thread[i].start();
   }
 }
 }
}
class Myth implements Runnable{
 private int tickets;
 private int windows;
 private int num=1;
 private int a[];
 public Myth(int tickets, int windows) {
  super();
  this.tickets = tickets;
  this.windows = windows;
  a = new int[windows+1];
 }
public int getTickets() {
  return tickets;
 }
 public void setTickets(int tickets) {
  this.tickets = tickets;
 }
 public int getWindows() {
  return windows;
 }
 public void setWindows(int windows) {
  this.windows = windows;
 }
 public void run()
 {
  synchronized (this) {
   for(int i=1;i<=tickets;i++) {
    if(this.tickets>0) {
     try {
      Thread.sleep(1000);
     }catch(InterruptedException e)
     {
      e.printStackTrace();
     }
     System.out.println(Thread.currentThread().getName()+"正在售票。当前余票是"+this.tickets--+".售出票号是:"+num);
     num++;
     a[Thread.currentThread().getName().charAt(4)-'0']++;
    }
    if(this.tickets==0) {
     for( i=1;i<=windows;i++) {
      System.out.println("票已售完.售票窗口"+i+"共售票"+a[i]+"张");
     }
    }
    }
  }
 }
}