C programming example 4

Question: enter a day of a year to determine the day of the year?

1. program Analysis: Taking June 1, March 5 as an example, we should add up the previous two months, and then add 5 days, that is, the day of the current year. In special cases, an additional day is required when the input month is greater than 3 in a leap year.

2. program source code:

# Include <stdio. h>

Int main (void)
{
Int day, month, year, countday, leap;
Printf ("\ nplease input year, month and day \ n ");
Scanf ("% d", & year, & month, & Day );
If (Year % 4 = 0 & amp; Year % 100! = 0) | (Year % 400 = 0 ))
Leap = 1;
Else
Leap = 0;
Switch (month)
{
Case 1:
Countday = 0;
If (day <= 0 | day> 31)
Return 1;
Break;
Case 2:
Countday = 31;
If (leap! = 1)
{
If (day <= 0 | day> 28)
Return 1;
}
Else
{
If (day <= 0 | day> 29)
Return 1;
}
Break;
Case 3:
Countday = 59;
If (day <= 0 | day> 31)
Return 1;
Break;
Case 4:
Countday = 90;
If (day <= 0 | day> 30)
Return 1;
Break;
Case 5:
Count day = 120;
If (day <= 0 | day> 31)
Return 1;
Break;
Case 6:
Count day = 151;
If (day <= 0 | day> 30)
Return 1;
Break;
Case 7:
Count day = 181;
If (day <= 0 | day> 31)
Return 1;
Break;
Case 8:
Count day = 212;
If (day <= 0 | day> 31)
Return 1;
Break;
Case 9:
Count day = 243;
If (day <= 0 | day> 30)
Return 1;
Break;
Case 10:
Count day = 273;
If (day <= 0 | day> 31)
Return 1;
Break;
Case 11:
Count day = 304;
If (day <= 0 | day> 30)
Return 1;
Break;
Case 12:
Count day = 334;
If (day <= 0 | day> 31)
Return 1;
Break;
Default:
Printf ("input month error! \ N ");
Return 1;
Break;
}
Countday = countday + day + 1;
Printf ("this is the % DTH of this years! \ N ", countday );
Return 0;
}