Answer:
import java.util.Scanner;
public class LabProgram {
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int inputYear;
boolean isLeapYear;
isLeapYear = false;
inputYear = scnr.nextInt();
// If a year is divisible by 400, then it is a leap year
if (inputYear % 400 == 0)
isLeapYear = true;
// If a year is divisible by 100, then it is not a leap year
if (inputYear % 100 == 0)
isLeapYear = false;
// If a year is divisible by 4, then it is a leap year
if (inputYear % 4 == 0)
isLeapYear = true;
if(isLeapYear)
System.out.println(inputYear + " is a leap year.");
else
System.out.println(inputYear + " is not a leap year.");
}
}
Explanation:
- Take the year as an input from user and store it to inputYear variable.
- If the year is a century year, check if the year is divisible by 400 ( the year must be evenly divisible by 400 ), then set the boolean isLeapYear to true. If a year is divisible by 100, then set the boolean isLeapYear to false. If a year is divisible by 4, then set the boolean isLeapYear to true.
- Check if isLeapYear is true, then print that it is a leap year. Otherwise, print that it is not a leap year.
Output:
1712
1712 is a leap year.
Answer:
i dont know spanish but i am gussing that you want to find differnce in virus information?
Explanation:
192.168 suggests class C networks which have a 24 bit netmask (255.255.255.0) but you haven't provided enough info, like the netmask of a working machine, to be definite.
B is better than any other one's I think.
