Write a program in C that reads a month and a day in that month
and computes how many days it is since
December 26th of previous year. For example, if input is 1 10
(January 10th), the number of days since
December 26th is 15. The program computes the number of days for a
nonleap year (February has always
28 days). The input is two integers, one between 1 and 12
(inclusive) and one between 1 and 31 (inclusive)
and the output will be a single line “X days”, where X is the
number of days.
Hello, here is the answer to this question.
EXPLANATION:
First we will declare few variables & initialize them.
We will take a variable fixed and initialize it with 26 as we have
to calculate value from 26th december.
We will take a variable monthend and initialize it with 31 as there
are 31 days in a month according to the question.
We will initialize the result variable with 0 value, which will
later hold the solution.
Coming on to the algorithm:
We will make 2 cases:
1) If the month is January or Feb:
We will calculate by this formula : result =
(monthend * (month-1)) + (monthend - fixed) + day;
monthend - fixed will give the value of
the remaining days of the december month
monthend * month-1 will give the days in the new
year.
2) If the month is any other than January or February:
We will calculate using this formula : result =
(monthend * (month-1)) + (monthend - fixed) + day - 3;
We are subtractiing by 3 because Feb month has 3
days less than the other months.
CODE:
#include <stdio.h>
int main()
{
// VARIABLE DECLARATION
int month,day;
int fixed = 26;
int monthend = 31;
int result = 0;
printf("Please Enter the Month and Day :
");
scanf("%d %d",&month,&day);
// if the month is Jan or Feb
if( month == 1 || month == 2 ) {
result = (monthend *
(month-1)) + (monthend - fixed) + day;
}
// Any month except jan and feb.
if( month > 2) {
result = (monthend *
(month-1)) + (monthend - fixed) + day - 3;
}
printf(" There are : %d Days!",result);
return 0;
}
OutPut:

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