Question: d. A common Gregorian year has 364 days and 6 hours. A leap year is a year that is either: 1 multiple of 400

   d. A common Gregorian year has 364 days and 6 hours. A  

d. A common Gregorian year has 364 days and 6 hours. A leap year is a year that is either: 1 multiple of 400 2 or it is a multiple of 4 but not a multiple of 100. A leap Gregorian year has 365 days; one day is added to the month of February which will have 29 days. Write a program to check whether a year entered by a user is leap or not. Example of leap years: 2000, 2004, 2008, 2012, 2016, 2020. Example of common years (not leap): 1900 (divisible by 4 and by 100).

Step by Step Solution

3.42 Rating (161 Votes )

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock

Answer include using namespace std int main int year cout year year entered by user ifyear40 y... View full answer

blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Electrical Engineering Questions!