Find out the day of the week from a given date.
Day of week from a given date
Write a program that finds out the day of the week from a given date. The formula for calculating the day is-
day = (y + j + f - h + fh) % 7;
j = julian day of the date
y = year of given date (in 4 digit)
f = int part of (y-1)/4
h = int part of (y-1)/100
fh = int part of (y-1)/400
The value of variable day tells us the day of week
|
Value of the variable date |
Name of day of week |
|
0 |
Saturday |
|
1 |
Sunday |
|
2 |
Monday |
|
3 |
Tuesday |
|
4 |
Wednesday |
|
5 |
Thursday |
|
6 |
Friday |
In the above formula Julian day of a date represent the day of year on which the date fall. Julian day of the 1st Jan is 1 , of 2nd Feb is 33, of 31st Dec is 365 (366 if leap year).
|
Month |
days |
Not leap year |
Leap year |
|
January |
31 |
31 |
31 |
|
February |
28/29 |
32-59 |
32-60 |
|
March |
31 |
60-90 |
61-91 |
|
April |
30 |
91-120 |
92-121 |
|
May |
31 |
121-151 |
122-152 |
|
June |
30 |
152-181 |
153-182 |
|
July |
31 |
182-212 |
183-213 |
|
August |
31 |
213-243 |
214-244 |
|
September |
30 |
244-273 |
245-274 |
|
October |
31 |
274-304 |
275-305 |
|
November |
30 |
305-334 |
306-335 |
|
December |
31 |
335-365 |
336-366 |
Now let's decide the structure of our program. Once we get the value of variable day, we can use switch to print the name of day of the week from the value of day.
To calculate the value of day we will have to find out the value of Julian day. Let's month and year of entered date are stored in the variable d, m, y respectively. We take another variable for value of Julian day.
Initially the value of j is taken equal to d, and then days of previous month are added to j to get the value of Julian day.
suppose date is 29/05/2022
d = 29
m = 05
y = 2022
j = 29 + 31 + 28 + 31 + 30 = 149
if the year is leap then add 1 mean j = j+1;
The above logic applied in switch statement was simple but whole switch statement is confusing read and understand.
now it is easy how to find the day by using the above formula is -
day = (y + j + f - h + fh)%7
we have only y and j not f ,h,and fh
now find f = (y-1)/4
= (2022-1)/4
= 2021/4
= 505.25
but f is int so f = 505
similarly
h = (y-1)/100
h = (2022-1)/100
h = 20
fh = (2022-1)/400
fh = 2021/400
fh = 5
day = (2022 + 149 + 505 - 20 +5)%7
day = 2661%7
day = 1 mean (Sunday)
see in program :-
/* write a program that finds out the day of week from a given date*/
#include <stdio.h>
#include <conio.h>
int main()
{
int d, m, y, j, f, h, fh, day;
/* where
d=date, m = month,y= year,j = julian day of the date,
f= int part of (y-1)/4,h = int part of (y-1)/100, fh=int part of (y-1)/400
julian day = 1,2,3,4,5,6,7,8,9,.............31,32,33,34........365/366*/
printf("Enter the date (dd/mm/yyyy):\n");
scanf("%d/%d/%d", &d, &m, &y);
if (d < 0 && d >= 31 || m < 0 && m > 12)
{
printf("Invalid Date!");
}
else
{
j = d;
if (y % 4 == 0 && y % 100 != 0 || y % 400 == 0)
// for 366 days in the leap year.
{
if (m != 1 && m != 2) // not the month Jan and Feb.
{
j = j + 1;
}
}
switch (m - 1)
{
case 11:
j += 30; // ---> November month
do not use break statement b/c they can add the month also.
case 10:
j += 31; //---> October month
case 9:
j += 30; //---> September month
case 8:
j += 31; //---> August month
case 7:
j += 31; //---> July month
case 6:
j += 30; //---> June month
case 5:
j += 31; //---> May month
case 4:
j += 30; //---> April month
case 3:
j += 31; //---> March month
case 2:
j += 28; //---> February month
case 1:
j += 31; //---> January month
}
}
f = (y - 1) / 4;
h = (y - 1) / 100;
fh = (y - 1) / 400;
// formula to calculate the day
day = (y + j + f - h + fh) % 7;
switch (day)
{
case 0:
printf("Saturday");
break;
case 1:
printf("Sunday");
break;
case 2:
printf("Monday");
break;
case 3:
printf("Tuesday");
break;
case 4:
printf("Wednesday");
break;
case 5:
printf("Thursday");
break;
case 6:
printf("Friday");
break;
}
return 0;
}
Output :
Enter the date (dd/mm/yyyy):
29/05/2022
Sunday
***** Thank You ******
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