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public class Solution { public static void main(String[] args) throws Exception { BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in)); String x =bufferedReader.readLine(); int a = Integer.parseInt(x); int leapyearby400 = a%400; int leapyearby100 = a%100; int leapyearby4 = a%4; if (leapyearby400 != 0) { int z = 366; System.out.println("Number of days in the year: "+z); return; } if (leapyearby100 != 0) { int z = 365; System.out.println("Number of days in the year: "+z); return; } if (leapyearby4 != 0) { int z = 366; System.out.println("Number of days in the year: "+z); return; } if (leapyearby400 != 0 && leapyearby4 != 0 && leapyearby100 != 0) { int z = 365; System.out.println("Number of days in the year: "+z); return; } } }

Use the keyboard to enter a year, and then determine the number of days in the year. Display the result as follows:
Number of days in the year: x
, where
x is 366 for a leap year, and
x is 365 for an ordinary year.

Hint:
A leap year is 366 days, but an ordinary year is 365 days.
1) if the year is evenly divisible by 400, then it is a leap year;
2) for other years, if the year is evenly divisible by 100, then it is a regular year;
3) for other years, if the year is evenly divisible by 4, then it is a leap year;
4) all remaining years are not leap years.
Thus, the years 1700, 1800, and 1900 are not leap years, since they are multiples of 100 but not 400.
The years 1600 and 2000 are leap years, since they are multiples of 100 and multiples of 400.
The years 2100, 2200 and 2300 are not leap years.

Requirements:

The program should read a number from the keyboard.
The program should display text on the screen according to the task conditions.
If the entered year is a leap year, you should display: "Number of days in the year: 366".
If the entered year is not a leap year, you should display: "Number of days in the year: 365".

package com.codegym.task.task04.task0414; /* Number of days in the year */ import java.io.*; import java.io.BufferedReader; public class Solution { public static void main(String[] args) throws Exception { BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in)); String x =bufferedReader.readLine(); int a = Integer.parseInt(x); int leapyearby400 = a%400; int leapyearby100 = a%100; int leapyearby4 = a%4; if (leapyearby400 != 0) { int z = 366; System.out.println("Number of days in the year: "+z); return; } if (leapyearby100 != 0) { int z = 365; System.out.println("Number of days in the year: "+z); return; } if (leapyearby4 != 0) { int z = 366; System.out.println("Number of days in the year: "+z); return; } if (leapyearby400 != 0 && leapyearby4 != 0 && leapyearby100 != 0) { int z = 365; System.out.println("Number of days in the year: "+z); return; } } }

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Agent Smith

Level 38

17 August 2020, 12:01

solution

Проверь условия ещё раз, например если год делится без остатка на 400, то 366 дней. У тебя наоборот, ибо вместо (==) в if стоит (!=). Ну и пара советов по коду: 17-19: Зачем 3 дополнительных переменных? Получается, что программа выполняет лишние действия, например если год делится без остатка на 400 - сразу понятно, что остальные сравнения делать уже не нужно. Вместо if + return, if + return, if + return, if + return лучше использовать if, else-if, else-if, else. Это тоже самое, только без лишнего кода и проще для восприятия. Вот этот повторяющийся кусок текста "Number of days in the year: " выгоднее сохранить в переменную и просто ссылаться на неё, например:

String msg = "Number of days in the year: ";
System.out.println(msg + "366");

Переменную z можно вообще убрать, а просто добавлять нужную цифру в println(). В итоге кода станет меньше, читабельность улучшится и проще будет искать баги.

Maxim Raitin

Level 20 , Tomsk, Russian Federation

17 August 2020, 17:41

Спасибо!