Write a function that calculates the day of the week for any particular date in the past or future. A typical application is to calculate the day of the week on which someone was born or some other special event occurred.
Following is a simple function suggested by Sakamoto, Lachman, Keith and Craver to calculate day. The following function returns 0 for Sunday, 1 for Monday, etc.
Understanding the Maths:
14/09/1998
dd=14
mm=09
yyyy=1998 //non-leap year
Step 1: Informations to be remembered.
Magic Number Month array.
For Year: {0,3,3,6,1,4,6,2,5,0,3,5}
DAY array starting from 0-6: {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
Century Year Value: 1600-1699 = 6
1700-1799 = 4
1800-1899 = 2
1900-1999 = 0
2000-2099 = 6..
Step 2: Calculations as per the steps
Last 2 digits of the year: 98
Divide the above by 4: 24
Take the date(dd): 14
Take month value from array: 5 (for September month number 9)
Take century year value: 0 ( 1998 is in the range 1900-1999 thus 0)
-----
Sum: 141
Divide the Sum by 7 and get the remainder: 141 % 7 = 1
Check the Day array starting from index 0: Day[1] = Monday
**If leap year it will be the remainder-1
Implementation
/* A program to find day of a given date */
#include <bits/stdc++.h>
using namespace std;
int dayofweek(int d, int m, int y)
{
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;
}
// Driver Code
int main()
{
int day = dayofweek(30, 8, 2010);
cout << day;
return 0;
}
Output :
1 (Monday)