在Java中使用循环而不是类来获取要舍入的数字

Question

我试图得到一个数字的平方根，在该数字的0.000001范围内四舍五入。例如，10的sqrt=3.1622766。。。。。用双人床。我有，但四舍五入到3.162267是我的问题。我必须使用循环，不能使用类。谢谢你，杰瑞德

import java.util.Scanner;

public class squareRoot {
    public static void main(String[] args) {
        Scanner kb = new Scanner(System.in);
        System.out.println("Please enter a non-negative integer.");
        int myInt = kb.nextInt();
        {
            double testNum;
            double squareRoot = myInt / 2;
            do {
                testNum = squareRoot;
                squareRoot = (testNum + (myInt / testNum)) / 2;
            }
            while (squareRoot - (testNum * testNum) > .000001);
            System.out.println("\nThe square root of " + myInt + " is " + squareRoot);
        }
    }
}

Answer 1

您可以编写自己的round()方法，并使用它而不是在java类中构建：

private static double round(double number, int places){
    double result;
    if(number == (int) number){
        return number;
    }
    result = (int) number;
    int multiplier = 1;
    for(int i = 0 ; i < places ; i++){
        multiplier*= 10;
    }
    double fraction = number - (int)number;
    result = result + ((int)(multiplier * fraction))*1.0/multiplier;
    return result;
}

public static void main(String[] args) {
    double d = 1.6546213;
    System.out.println(round(d, 2));
}

Answer 2

问题是，使用此算法时，最后的数字有一个错误元素，因此可以增加有效数字的数量，然后四舍五入到所需的数字：

double myDouble = kb.nextDouble();
{
    double testNum;
    double squareRoot = myInt/2;
    do
    {
        testNum=squareRoot;
        squareRoot = (testNum + (myInt/testNum))/2;
    }
    while(Math.abs(myDouble - (testNum * testNum)) > .00000001);//←decrease the error here

    //round squareRoot down to the num of decimals you want        

    System.out.println("\nThe square root of " + myDouble + " is " + squareRoot);
    }
}