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多线程求素数

作者:互联网


<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;

}

}


标签:eachPart,sum,nPart,int,素数,LRange,RRange,多线程
来源: https://blog.51cto.com/14028890/2409019