Here is Sakamoto’s algorithm for calculating the day of the week, taken from the comment that introduces the code:

Jan 1st 1 AD is a Monday in Gregorian calendar.
So Jan 0th 1 AD is a Sunday [It does not exist technically].

Every 4 years we have a leap year. But xy00 cannot be a leap unless xy divides 4 with reminder 0.

y/4 – y/100 + y/400 : this gives the number of leap years from 1AD to the given year. As each year has 365 days (divdes 7 with reminder 1), unless it is a leap year or the date is in Jan or Feb, the day of a given date changes by 1 each year. In other case it increases by 2.

y -= m So y + y/4 – y/100 + y/400 gives the day of Jan 0th (Dec 31st of prev year) of the year. (This gives the reminder with 7 of the number of days passed before the given year began.)

Array t: Number of days passed before the month ‘m+1’ begins.

So t[m-1]+d is the number of days passed in year ‘y’ upto the given date.

(y + y/4 – y/100 + y/400 + t[m-1] + d) % 7 is reminder of the number of days from Jan 0 1AD to the given date which will be the day (0=Sunday,6=Saturday).

int dow(int y, int m, int d) {
static int t[] = {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4};
y -= m < 3;
return (y + y/4 - y/100 + y/400 + t[m-1] + d) % 7;

Another description is given here.

Your task is to write a program that implements the day-of-week algorithm shown above. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.


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