我正在尝试计算F的值,其中F = 1 /(1 ^(1/3))+ 1 /(2 ^(1/3))+ ...... + 1 /(600000000 ^(1/3))。我正在使用以下类,但是每个F1,F2,F3和F4的输出返回150000000。仔细检查后,我发现每次将值加到总和时,总和就加1。我如何才能获得非常小的小数位数的准确值,而不是将其变为1?
import java.util.Date;
public class Problem1 {
public static void main(String[] args) throws InterruptedException{
//start measuring time
long begTime = new Date().getTime();
//Measure execution time of following statements
MyThread t1 = new MyThread();
MyThread t2 = new MyThread();
MyThread t3 = new MyThread();
MyThread t4 = new MyThread();
t1.n1 = 1;
t1.n2 = 150000000;
t2.n1 = 150000001;
t2.n2 = 300000000;
t3.n1 = 300000001;
t3.n2 = 450000000;
t4.n1 = 450000001;
t4.n2 = 600000000;
t1.start();
t2.start();
t3.start();
t4.start();
t1.join();
t2.join();
t3.join();
t4.join();
System.out.printf("Value of F1 is %f\n", t1.sum);
System.out.printf("Value of F2 is %f\n", t2.sum);
System.out.printf("Value of F3 is %f\n", t3.sum);
System.out.printf("Value of F4 is %f\n", t4.sum);
System.out.println();
System.out.printf("Value of F is %f\n", t1.sum+t2.sum+t3.sum+t4.sum);
//Compute and print elapsed time
double timeElapsed = (new Date().getTime() - begTime) * 0.001;
System.out.printf("Time elapsed = %f secs\n", timeElapsed);
}
}
public class MyThread extends Thread {
int n1, n2;
double sum = 0.0;
public void run(){
for (int i=n1; i<= n2;i++){
sum = sum+ (1/(Math.pow((double) i, (double) (1/3))));
}
}
}
答案 0 :(得分:3)
您可以使用BigDecimal而不是double。 https://docs.oracle.com/javase/7/docs/api/java/math/BigDecimal.html
这可能也有帮助。 Double vs. BigDecimal?
答案 1 :(得分:2)
如果数字类型没有问题,则应使用BigDecimal。
它比Doubles更强大,并且可以放置更多的小数,以及其他诸如本机取舍的操作
https://docs.oracle.com/javase/7/docs/api/java/math/BigDecimal.html
该线程的解释要好得多。
答案 2 :(得分:0)
这是 float 和 double 的正常行为,如here所示。
考虑使用 BigDecimal 对象,因为它们对于精确的算术运算更强大。