Question

为什么我的代码报告说这个过程最多需要5秒才能完成，即使它甚至不能实时拍摄四分之一秒？

我将尝试使用与秒表特别相关的代码加入，以防止您不必仔细查看。请善意，因为这是我的第一篇文章，所以如果它笨拙，我很抱歉。如果代码没有加粗，看起来就会有问题。

*背景：这是一篇数学论文。它应该是一个程序，它可以找到素数因子并告诉它找到它们需要多长时间。它正在努力寻找主要因素，但秒表在几秒钟内报告了一个荒谬的数字。此外，此代码受

影响最大

http://www.geeksforgeeks.org/print-all-prime-factors-of-a-given-number/

使用秒表，用户输入功能，并通过我自己的想法或在其他人的帮助下添加重复*

    // Program to print all prime factors
import java.io.*;
import java.lang.Math;
import java.util.Scanner;
import java.text.DecimalFormat;

class primeFactorer4
{
    **static long startTime = System.nanoTime();**
    // A function to print all prime factors
    // of a given number n
    public static void primeFactors(long n)
    {
        // Print the number of 2s that divide n
        while (n%2==0)
        {
            System.out.print(2 + " ");
            n /= 2;
        }

        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i+= 2)
        {
            // While i divides n, print i and divide n
            while (n%i == 0)
            {
                System.out.print(i + " ");
                n /= i;
            }
        }

        // This condition is to handle the case whien
        // n is a prime number greater than 2
        if (n > 2)
            System.out.print(n);
    }

    public static void main (String[] args)
    {
        Console console = System.console();
        String input = console.readLine("Enter input:");
        long n = Long.valueOf(input);
        for (int k=1; k<=10; k++)
    {
        primeFactors(n);
        System.out.println(" Try " + k);
    }
        **double endTime = System.nanoTime();
        double totalTime = endTime - startTime;
        DecimalFormat totalTimeFormat = new DecimalFormat("##.###");
        System.out.println("    Time taken in seconds:" + totalTimeFormat.format(totalTime/10/1000000000));**
        primeFactorer4.main(args);
    //reason for the weird division is for clarity. "totalTime" is the time surpassed 
    //to repeat all the methods, the "10" in the middle is to get the mean total time
    //of all the primeFactors cycles, and the "1000000000" at the end is to convert nanoseconds into seconds
    }
}

我之所以打电话给primeFactors是因为我希望我的电脑为我做结果的平均值，因为任何学校会告诉你，在试验时，你需要重复你的IV等级3（或更多）时间来获得更准确的结果

Answer 1

好的，没关系，我解决了我的问题。我在startTime和endTime变量下面放了一个println命令，我发现startTime变量是在程序初始化时启动的，而不是当用户输入他们想要的数字时。它现在给了我适当的结果，不觉得与我个人输入数字的速度有关。

对于那些对该程序感兴趣的人，这个问题的解决方案适用于您，或者您只是想看到解决方案与问题之间的对比，这里是新代码。

    // Program to print all prime factors
import java.io.*;
import java.lang.Math;
import java.util.Scanner;
import java.text.DecimalFormat;

class primeFactorer4
{

    // A function to print all prime factors
    // of a given number n
    public static void primeFactors(long n)
        {
            // Print the number of 2s that divide n
            while (n%2==0)
            {
                System.out.print(2 + " ");
                n /= 2;
            }

            // n must be odd at this point.  So we can
            // skip one element (Note i = i +2)
            for (int i = 3; i <= Math.sqrt(n); i+= 2)
            {
                // While i divides n, print i and divide n
                while (n%i == 0)
                {
                    System.out.print(i + " ");
                    n /= i;
                }
            }

            // This condition is to handle the case whien
            // n is a prime number greater than 2
            if (n > 2)
                System.out.print(n);
        }



    public static void main (String[] args)
    {
        Console console = System.console();
        String input = console.readLine("Enter input:");
        long n = Long.valueOf(input);
        long startTime = System.nanoTime();
        System.out.println(startTime);
        for (int k=1; k<=10; k++)
        {
            primeFactors(n);
            System.out.println(" Try " + k);
        }
        double endTime = System.nanoTime();
            System.out.println(endTime);
            double totalTime = endTime - startTime;
            DecimalFormat totalTimeFormat = new DecimalFormat("##.##########");
            System.out.println("    Time taken in seconds:" + totalTimeFormat.format(totalTime/10/1000000000));
        primeFactorer4.main(args);
    //reason for the weird division is for clarity. "totalTime" is the time surpassed 
    //to repeat all the methods, the "10" in the middle is to get the mean total time
    //of all the primeFactors cycles, and the "1e9" at the end is to convert nanoseconds into seconds
    }
}

Answer 2

我为你写了一些东西，我认为这可能更快，因此更准确但请记住，执行任何陈述需要时间，因此它不能完全准确。

long start= System.currentTimeMillis(); //start time
//Insert code here
long difference = System.currentTimeMillis(); //finish time
difference -= start;
System.out.println("Time took to run code was " + difference);; //Print the amount of time that it took

秒表没有报告正确的时间

2 个答案: