import java.io.IOException;

 import java.util.concurrent.Semaphore;

 public class PrintABC {

     public static int MAX_TIME = 10;

     public static class PrintThread extends Thread {

         private String printChar;

         private Semaphore curSemaphore, nextSemaphore;

         public PrintThread(String printChar, Semaphore curSemaphore,

                 Semaphore nextSemaphore) {

             this.printChar = printChar;

             this.curSemaphore = curSemaphore;

             this.nextSemaphore = nextSemaphore;

         }

         public void run() {

             for (int i = 0; i < MAX_TIME; ++i) {

                 try {

                     curSemaphore.acquire();  /* 请求打印需要的信号量，资源数-1，即down操作  */

                     System.out.println(printChar);

                     nextSemaphore.release(); /* 释放下一个线程打印需要的信号量，资源数+1，即up操作  */

                 } catch (InterruptedException e) {

                     // TODO Auto-generated catch block

                     e.printStackTrace();

                 }

             }

         }

     }

     public static void main(String[] args) throws IOException {

         Semaphore semaphoreA = new Semaphore(1); /* 只有A信号量的初始资源数为1，保证从A开始打印  */

         Semaphore semaphoreB = new Semaphore(0);

         Semaphore semaphoreC = new Semaphore(0);

         new PrintThread("A", semaphoreA, semaphoreB).start();

         new PrintThread("B", semaphoreB, semaphoreC).start();

         new PrintThread("C", semaphoreC, semaphoreA).start();

     }

 }