多线程求素数

package test;
成都创新互联公司是一家企业级云计算解决方案提供商,超15年IDC数据中心运营经验。主营GPU显卡服务器，站群服务器，西云机房，海外高防服务器，机柜大带宽、租用·托管，动态拨号VPS，海外云手机，海外云服务器，海外服务器租用托管等。
 
import java.util.concurrent.*;
 
public class test {
	/*
	 * sum : the total of prime number. 
	 * n : the range. 
	 * nPart,eachPart : divide n into nPart,eachPart is n/nPart.
	 */
	public static void main(String[] args) {
		int i, sum = 0, n = 10000000, nPart = 16, eachPart = n / nPart, LRange = 1, RRange = eachPart;
		long begin, end;
		Future[] future = new Future[nPart];
		ExecutorService threadPool = Executors.newCachedThreadPool();
		begin = System.nanoTime();
		for (i = 0; i < nPart; i++)
			future[i] = threadPool.submit(new MyThread(LRange + i * eachPart,
					RRange + i * eachPart));
		threadPool.shutdown();
		while (!threadPool.isTerminated())
			;
		try {
			for (i = 0; i < 16; i++)
				sum += (Integer) future[i].get();
		} catch (Exception e) {
			// TODO: handle exception
			e.printStackTrace();
		}
		end = System.nanoTime();
		System.out.println((double) (end - begin) / 1000000000);
		System.out.println(sum);
	}
}
 
class MyThread implements Callable {
	int sum = 0, LRange, RRange;// LRange: range left ; RRange : range right.
 
	public MyThread(int lRange, int rRange) {
		LRange = lRange;
		RRange = rRange;
	}
 
	public Integer call() throws Exception {
		int i, j;
		for (i = LRange; i <= RRange; i += 2) {
			for (j = 2; j * j <= i; j++)
				if (i % j == 0)
					break;
			if (j * j > i)
				sum++;
		}
		return sum;
	}
}
            
            
                        

            分享题目：多线程求素数            

            转载注明：http://cqcxhl.cn/article/jgiodo.html