Java并行程序设计——求素数个数（Runnable接口实现）

【项目说明】： 使用Runnable接口实现求10000以内素数个数的Java多线程并行程序

求10000以内素数个数，输出结果、并行、串行时间，并计算加速比。

【项目代码】：

public class And {
    public static void main(String[] args) throws InterruptedException {
        work work1=new work(1, 10000);
        work work2=new work(2, 10000);
        Thread thread1=new Thread(work1);
        Thread thread2=new Thread(work2);  
        long startTime=System.currentTimeMillis();
        thread1.start();
        thread2.start();
        thread1.join();
        thread2.join();
        long endTime=System.currentTimeMillis();
        System.out.println("并行结果="+(work1.getSum()+work2.getSum()));
        System.out.println("并行时间="+(endTime-startTime));
        long t1=endTime-startTime;
        startTime=System.currentTimeMillis(); 	
        work work=new work(1, 10000);
        long sum=work.sum();
        endTime=System.currentTimeMillis();
        System.out.println("串行结果="+sum);
        System.out.println("串行时间="+(endTime-startTime));
        long t2=endTime-startTime;
        System.out.println("加速比="+t2*1.0/t1);
    }
}
class work implements Runnable{
    private  long start;
    private  long end;
    private  long sum=0;  
    
    public work(long start, long end) {
        super();
        this.start = start;
        this.end = end;
    }

    public void run() {
    	long t;
        for(long i=start;i<=end;i+=2)
        {  	t=0;
		for(long j=2;j<=i;j++)
			if(i%j!=0)
				t++;
		if(i!=0&&i!=1)
			if(t>=i-2)
				sum++;
		}
    }
	
    public long sum(){
    	long t;
        for(long i=start;i<=end;i++)
        {  	t=0;
		for(long j=2;j<=i;j++)
			if(i%j!=0)
				t++;
		if(i!=0&&i!=1)
			if(t>=i-2)
				sum++;
		}
        return sum;
    }

    public long getSum() {
        return sum;
    }
}

【运行截图】：