Answer:
To check if the year comes under each 100th year, lets check if the remainder when dividing with 100 is 0 or not.
Similarly check for 400th year and multiple 0f 4. The following C program describes the function.
#include<stdio.h>
#include<stdbool.h>
bool is_leap_year(int year);
void main()
{
int y;
bool b;
printf("Enter the year in yyyy format: e.g. 1999 \n");
scanf("%d", &y); // taking the input year in yyyy format.
b= is_leap_year(y); //calling the function and returning the output to b
if(b==true)
{
printf("Thae given year is a leap year \n");
}
else
{
printf("The given year is not a leap year \n");
}
}
bool is_leap_year(int year)
{
if(year%100==0) //every 100th year
{
if(year%400==0) //every 400th year
{
return true;
}
else
{
return false;
}
}
if(year%4==0) //is a multiple of 4
{
return true;
}
else
{
return false;
}
}
Explanation:
Output is given as image
// This command takes input from the user.
Scanner input = new Scanner(System.in);
int response= input.nextInt();
// Making the variables required
int noCount =0 ;
int yesCount =0;
// Checking the response if it is 1 or 2 and reporting accordingly
if(response == 1 || response ==2){
yesCount+=1;
System.out.println("YES WAS RECORDED");
}
// Checking if the input is 3 or 4 then printing the required lines
else if (response == 3 || response ==4){
noCount+=1;
System.out.println("NO WAS RECORDED");
}
// if the input is not valid than printing INVALID
else{
System.out.println("INVALID");
}
I am guessing. My guess is code.