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これは、ニュートン法を使用して、ユーザーが入力した数値の平方根を推定するための宿題であり、< .0001 の結果を返す必要があります。コードを実行して数値を入力しても、その後何も起こりません。デバッグモードでは、「値」が増加します。これは、私がやりたいこととは逆です。前もって感謝します。

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;  
    }
}
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2 に答える 2

0

あなたの条件が

while (Number >= 0.0)

条件が満たされたときに実際に関数を終了した場合は問題ありません。

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

そのため、最後の行を変更すると機能します。

デモ: 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 に答える