0
Light Poster
thanks
0
Newbie Member
Below is some code I wrote w/ pradeep's help. It was an assignment in a book I'm going through.

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Code:
#include<stdio.h>
main()
{
int month, year, n, leap, notLeap;

printf("\n\nEnter a month and year:   ");
scanf("%d %d", &month, &year);

if(year % 400 == 0 || (year % 100 != 0 && year % 4 == 0))

year = leap;

else
year = notLeap;

if (year == leap && month == 2)
n = 29;

if (year == notLeap && month == 2)
n = 28;

if (month == 1) n = 31;
if (month == 3) n = 31;
if (month == 4) n = 30;
if (month == 5) n = 31;
if (month == 6) n = 30;
if (month == 7) n = 31;
if (month == 8) n = 31;
if (month == 9) n = 30;
if (month == 10) n = 31;
if (month == 11) n = 30;
if (month == 12) n = 31;

printf("\n\nThere are %d days in that month.", n);

getch();
return;
}

Last edited by shabbir; 14Feb2011 at 11:59.. Reason: Code blocks
0
Newbie Member
Quote:
The logic is that the year is either divisible by both 100 and 4 , OR its only divisible by 4 not by hundred
Your code is good but your logic isn't. Specifically, it's not just that the number is divisible by 100 and 4, it's that the number is divisible by 400. 200 is divisible by 100 and 4, but isn't a leap year.

To say that a number is divisible by 100 and 4, or 100 or 4, is that same as saying a number is divisible by 4.

That is: (A ^ B) v (A ^ !B) = A ^ (B v !B) = A.

This is the same logic that makes all those ads so hilarious: "You can make up to \$50,000 a year, or more!" So I guess the only amount you can't make is \$50,000!

Sorry, anal logician here

-Plast