Java多线程求π

利用泰勒展开
$\dfrac{\pi}{4}=1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\cdots$

import java.util.*;

class Calc extends Thread{
    private double a1;//首项
    private double dif;//分母公差
    private double res;//线程结果
    private static double eps=1e-9;//控制精度
    private double sign;//正负号

    public Calc(double a1,double dif,int id){
        this.a1=a1;
        this.dif=dif;
        res=0;
        if(id%2>0) sign=-1;
        else sign=1;
    }

    public double getRes(){
        return res;
    }
    public void run(){
        for(double i=a1;;i+=dif){
            if(Math.abs(1.0/i)<eps) break;
            res+=1.0/i;
        }
        res*=sign;
    }
}

public class Main{
    public static void main(String[] args){
        int threadnum;//线程个数
        double sum=0;//结果
        double op=1;//用于串行算法正负号变换
        long start,end;//开始与结束时间
        System.out.println("请输入并行算法线程个数：");
        Scanner in = new Scanner(System.in);
        while(in.hasNext()){
            threadnum=in.nextInt();
            System.out.println("------串行算法------");
            start=System.currentTimeMillis();
            for(double i=1;;i+=2){
                if(Math.abs(1.0/i)<1e-9) break;
                sum+=op*1.0/i;
                op=-op;
            }
            end=System.currentTimeMillis();
            System.out.println("结果："+sum*4.0);
            double time1=(double)(end-start)/1000.0;
            System.out.println("用时："+time1+"s");

            sum=0;
            Calc[] c = new Calc[threadnum];
            int cnt=0;
            for(int i=1;i<=1+2*(threadnum-1);i+=2){
                c[cnt] = new Calc(i, 2*threadnum,cnt);
                cnt=cnt+1;
            }
            System.out.println("------并行算法------");
            start=System.currentTimeMillis();
            for(int i=0;i<threadnum;i++){
                c[i].start();
            }
            try {
                for(int i=0;i<threadnum;i++){
                    c[i].join();
                    //System.out.println(i+": "+c[i].getRes());
                    sum+=c[i].getRes();
                }
            } catch (Exception e) {
                e.printStackTrace();
            }
            end=System.currentTimeMillis();
            System.out.println("结果："+sum*4.0);
            double time2=(double)(end-start)/1000.0;
            System.out.println("用时："+time2+"s");
            System.out.println("串行与并行时间之比："+time1/time2);
        }
        in.close();
    }
}