leap year - by only using arithmetical operators in C

Question

I have the weirdest exercise I've ever seen: I have to find a leap year by scanning a year in the console and to control if that is a leap year.

I can only use + - / * % as arithmetical operators; I am not allowed to use any other operators or functions.

Here is what I have so far:

   int year = 0;
   bool b = false;
    
   printf ("Type in a year: ");
    
   int helpVar = 1000;
   for (int i = 0; i < 4; i++) {
       year += (getchar() - '0') * helpVar;
       helpVar = helpVar / 10;
   }
    
   b = (((year % 4) + (year % 100) + (year % 400)) + 1) % 2;

So I don't understand what I am doing wrong here. It works so far, the only case that's freaking me out is for year "1900". It shouldn't be a leap year, but appears to, by my code.

What am I missing here?

Answer 1

Here is one possibility (perhaps not the shortest -- obviously only works for the Gregorian calendar):

b = (((year-1)%4)+1)/4 - (((year-1)%100)+1)/100 + (((year-1)%400)+1)/400;

The idea is that ((year-1) % n) + 1 equals n only if year is a multiple of n (for positive year), and is smaller than n otherwise. Thus, if you divide that by n, you get 1 if and only if (year % n == 0).

Since year%100==0 cannot be true if year%4==0 is not, you can subtract that from each other, but add the year%400==0 term at the end.