Question

我要做的是创建一个闰年计算器，它不仅可计算一年，而且可计算用户可输入的多年。现在，我只计算一年。

这是输出应该是什么样的：

该程序测试给定年份是否为闰年。

输入起始年份：2008

输入结束年份：2013

2008年是闰年

2012年是闰年

非常感谢您的回复！帮助我学到的东西远远超过我的讲师实际教授的内容:)

import java.util.Scanner;
public class LeapYears
{
public static void main (String [] args)
{
Scanner keyboard = new Scanner (System.in); 

System.out.println ("This program tests whether a given " + "year is a leap year.");

    System.out.print ("Type the starting year: ");
    int year = keyboard.nextInt ();

    System.out.print ("Type the ending year: ");
    int EndYear = keyboard.nextInt ();

if (((year%4 == 0) && (year%100 != 0)) || (year%400 == 0)) 

System.out.println (year + " is a leap year.");
    else
        System.out.println (year + " is not a leap year.");

} 
}

Answer 1

您可以简单地将闰年评估代码包装在方法中，并在循环中调用它，例如：

public boolean isLeapYear(int year){
   if (((year%4 == 0) && (year%100 != 0)) || (year%400 == 0)) 
       return true;
   else 
       return false;
}

并在您的主要方法中：

System.out.print("Type the starting year: ");
int startYear = keyboard.nextInt();

System.out.print("Type the ending year: ");
int endYear = keyboard.nextInt();

for(int currentYear=startYear;currentYear++;currentYear<=endYear){
    if(isLeapYear(currentYear))
       System.out.println(currentYear + " is a leap year.");
    else
       System.out.println(currentYear + " is not a leap year.");
}

除此之外，最好为案例添加一些防御性代码，例如用户输入不是整数，startYear大于endYear。

Answer 2

试试这个

    GregorianCalendar gc = new GregorianCalendar();
    for(int y = year; y <= endYear; y++) {
        if (gc.isLeapYear(y)) {
            System.out.println(y + " is leap year");
        } else {
            System.out.println(y + " is not leap year");
        }
    }

Answer 3

使用while循环进行多年检查，但是一旦获得第一个闰年，下一次迭代将增加4年（即年=年+4; ）。这样就可以避免不必要的检查。

    Scanner keyboard = new Scanner(System.in);

    System.out.println("This program tests whether a given "
            + "year is a leap year.");

    System.out.print("Type the starting year: ");
    int year = keyboard.nextInt();

    System.out.print("Type the ending year: ");
    int EndYear = keyboard.nextInt();

    while (year <= EndYear) {
        if (((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0)) {

            System.out.println(year + " is a leap year.");
            year = year + 4;
        } else {
            //System.out.println(year + " is not a leap year.");
            year++;

        }
    }

Answer 4

是的，这是一个循环问题。尝试while循环：

import java.util.Scanner;
public class LeapYear {
    public static void main (String [] args)
    {
        Scanner keyboard = new Scanner (System.in); 

        System.out.println ("This program tests whether a given " + "year is a leap year.");

        System.out.print ("Type the starting year: ");
        int year = keyboard.nextInt ();

        System.out.print ("Type the ending year: ");
        int EndYear = keyboard.nextInt ();
        int temp=year;
        while(temp<EndYear)
        {   if (((temp%4 == 0) && (temp%100 != 0)) || (temp%400 == 0)) {

                System.out.println (temp + " is a leap year.");
            }
            else
            {
                System.out.println (temp + " is not a leap year.");
            }
            temp++;

        }


    } 
}

计算多年，这是Java的闰年

4 个答案: