If the year divides evenly by 4 it's a leap year. If it's a century year it has to divide evenly by 400. 2000 was a leap year. 2100 will not be a leap year.
leap yr = ((y%100!=0 && yr %400==0)?1:(yr%4==0)?1:0; leap yr =1; printf ("leap yr"); else printf ("not leap yr");
It depends on when you start - whether in a leap year or not, whether your time span includes a xy00 year or not. So there is no simple answer.
dono
#include using namespace std;int main(){short numberOfDays = 0;cout > numberOfDays;if (numberOfDays == 365){cout
The Impossible Leap in One Hundred Simple Steps was created in 2003.
You can use a number of nested "if" commands. The rules are the following: If the year is a multiple of 400, it IS a leap year. Else, if the year is a multiple of 100, it's NOT a leap year. Else, if the year is a multiple of 4, it IS a leap year. Else it's NOT a leap year. In Java, to check for divisibility, you use the "%" operator, which gives you the remainder of a division. For example: if (year % 400 = 0) ... To show results on screen, you use: System.out.println(...)
Simple answer - no.
If you want to determine whether a year is a leap year. Use the cal command for the month of February. So to check 2012, for example, do: cal 2 2012 and see if it prints 28 or 29 days.
Yes. It is divisible by 4. Whether or not leap years were recognized that long ago, I am not sure.
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Simple past tense of the verb to leap.