大神们~~~~~~~~~~~~·

就是三个线程类，操作一个static，一个每隔10秒（时间随意）加15，一个减8，第三个线程检测static大于10000了吗，大于10000就结束。基本思路就这样

大神求程序啊

4个线程，执行每一次循环加或减，并等待所有线程都执行完，然后继续下一次加减

package DomeTB;
public class Dome implements Runnable{
int water=0;
int time=0 ;
public void run() {
while(true) {
if(water>=10000){
System.out.println(Thread.currentThread().getName()+"用了"+time+"小时");
return;
}else xsc();
}
}
public synchronized void xsc(){
water+=15;
water-=(8/3);
time++;
}
}

package DomeTB;
public class DomeZ {
public static void main(String[] args) {
Dome dome=new Dome();
Thread do1=new Thread(dome,"进水口1");
Thread do2=new Thread(dome,"进水口2");
Thread do3=new Thread(dome,"进水口3");
do1.start();
do2.start();
do3.start();
}
}

我也刚学到这里，手打了，你试试，我试了没问题，就共享资源嘛，接口实现

public class WaterFlooding implements Runnable{
private static Double WATERCOUNT=0.0;
private static Integer TIME=0;
@Override
public void run() {
while(true){
if(WATERCOUNT>=10000){
System.out.println(Thread.currentThread().getName()+"用了"+TIME+"个小时将水池注满");
break;
}else {
exe();
}
}
}
private synchronized void exe(){
WATERCOUNT+=15;
WATERCOUNT-=8.0/3;
TIME++;
}
public static void main(String[] args) {
WaterFlooding waterFlooding=new WaterFlooding();
Thread t1=new Thread(waterFlooding,"进水口A");
Thread t2=new Thread(waterFlooding,"进水口B");
Thread t3=new Thread(waterFlooding,"进水口C");
t1.start();
t2.start();
t3.start();
}
}