1

这是一个家庭作业，使用牛顿法估计用户输入的数字的平方根，它应该返回 < .0001 的结果。当我运行代码并输入一个数字时，之后什么也没有发生。在调试模式下，“值”会增加，这与我想要它做的相反。提前致谢。

import java.text.DecimalFormat;
import java.util.Scanner;

public class Newton {

    public static void main(String[] args) 
      {
        // declare a Scanner class object
        Scanner sc = new Scanner(System.in);
        // declare a DecimalFormat class object
        DecimalFormat fourDecimal =  new DecimalFormat("0.0000");

        float Number = 0;

        System.out.println("Program: find square roots by Newton's Method");
        System.out.println("Please enter a number: ");

        Number = sc.nextFloat();

        System.out.println("The square root of " + Number + " is " + fourDecimal.format(Compute(Number)));
        }

    public static float Compute(float Number)
    {
    // define variable sqrRoot to hold the approximate square root
    float sqrRoot = 0;
    // define temporary variable temp to hold prior value of iteration
    float temp = 0;
    // divide variable num by 2 to start the iterative process
    // and assign the quotient to temp
    temp = Number/2;
    // open a while() loop that continues as long as num >= 0.0
    while (Number >= 0.0)
    {
    // construct the main iterative statement
        sqrRoot = temp - (temp * temp - Number) / (2 * temp);
    // open an if block to check if the absolute value of the difference of
    // variables temp and sqrRoot is below a small sentinel value such as 0.0001
    // if this condition is true then break the loop
        float value;
        value = Math.abs(temp - sqrRoot);
        if (value < .0001)
            // return sqrRoot as the answer
            Number = sqrRoot;
            // if this condition is not true then assign sqrRoot to temp
            else temp = sqrRoot;

    // close the while() loop
    }
    return Number;  
    }
}
4

2 回答 2

0

您的循环不会终止，因为您的条件是

while (Number >= 0.0)

如果您在满足条件时实际退出了该功能，那将是可以的：

   if (value < .0001)
        // return sqrRoot as the answer
        return sqrRoot;

所以 - 改变最后一行，它会起作用。

演示：http: //ideone.com/XzJXLv

public static float Compute(float Number)
{
    // define variable sqrRoot to hold the approximate square root
    float sqrRoot = 0;
    // define temporary variable temp to hold prior value of iteration
    float temp = 0;
    // divide variable num by 2 to start the iterative process
    // and assign the quotient to temp
    temp = Number/2;

    // open a while() loop that continues as long as num >= 0.0
    while (Number >= 0.0)    // <<<< you might reconsider this condition: iteration count?
    {
        // construct the main iterative statement
        sqrRoot = temp - (temp * temp - Number) / (2 * temp);

        // open an if block to check if the absolute value of the difference of
        // variables temp and sqrRoot is below a small sentinel value such as 0.0001
        // if this condition is true then break the loop
        float value;
        value = Math.abs(temp - sqrRoot);

        if (value < .0001)
            // return sqrRoot as the answer
            return sqrRoot;  // <<<<< this is the line you needed to change
        // if this condition is not true then assign sqrRoot to temp
        else temp = sqrRoot;
    } // close the while() loop

    return Number;  // <<<<< you will never reach this line
}
于 2014-02-09T22:35:32.953 回答
0

1970 年代之前工程师和学童使用的巨大的平方根表现在已被牛顿方法的五行程序所取代。

public static Number squareRoot(Number value) {
    double temp = value.doubleValue() / 2;
    for(double sqrRoot; ; temp = sqrRoot) {
        sqrRoot = temp - (temp * temp - value.doubleValue()) / (2 * temp);
        if (Math.abs(temp - sqrRoot) < 1e-10) return sqrRoot;
    }
}
于 2019-02-04T04:42:01.490 回答