温馨提示×

温馨提示×

您好,登录后才能下订单哦!

密码登录×
登录注册×
其他方式登录
点击 登录注册 即表示同意《亿速云用户服务条款》

多线程求素数

发布时间:2020-06-14 22:54:07 来源:网络 阅读:320 作者:沙漏半杯 栏目:编程语言


<pre name="code" class="java">package test;

 

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;

}

}


向AI问一下细节

免责声明:本站发布的内容(图片、视频和文字)以原创、转载和分享为主,文章观点不代表本网站立场,如果涉及侵权请联系站长邮箱:is@yisu.com进行举报,并提供相关证据,一经查实,将立刻删除涉嫌侵权内容。

AI