(C syntax) print the number of days in a month, syntax 17

(C syntax) print the number of days in a month, syntax 17

Knowledge point:

Flexible use of logical expressions

Method 1:

Switch () usage, pay attention to the use of case 'A ':......; Break; case 6 :......; Break;

Do not forget break;

Method 2:

If ...... Else usage

Check whether it is a leap year algorithm.

 

Content: print the number of days in a month.

Input description:

Two integers in one row, followed by the year and followed by the month

Output description:

An integer

 

Input example:

2009 6

Output example:

30

 

Method 1:

#include <stdio.h>
int main()
{
 int y,m;
 scanf("%d%d",&y,&m);
 if((y%4==0&&y%100!=0)||y%400==0)
 {
  switch (m)
  {
  case 1: printf("31\n");break;
  case 2: printf("29\n");break;
  case 3: printf("31\n");break;
  case 4: printf("30\n");break;
  case 5: printf("31\n");break;
  case 6: printf("30\n");break;
  case 7: printf("31\n");break;
  case 8: printf("31\n");break;
  case 9: printf("30\n");break;
  case 10: printf("31\n");break;
  case 11: printf("30\n");break;
  case 12: printf("31\n");break;
  }
 }
 else
  switch (m)
 {
  case 1: printf("31\n");break;
  case 2: printf("28\n");break;
  case 3: printf("31\n");break;
  case 4: printf("30\n");break;
  case 5: printf("31\n");break;
  case 6: printf("30\n");break;
  case 7: printf("31\n");break;
  case 8: printf("31\n");break;
  case 9: printf("30\n");break;
  case 10: printf("31\n");break;
  case 11: printf("30\n");break;
  case 12: printf("31\n");break;
 }
 return 0;
}

 

Method 2:

#include <stdio.h>
int main()
{
 int y,m;
 scanf("%d%d",&y,&m);
 if (m==1||m==3||m==5||m==7||m==8||m==10||m==12)
 {
  printf("31\n");
 }
 else 
 {
  if (m==2)
  {
   if((y%4==0&&y%100!=0)||y%400==0)
   {
    printf("29\n");
   }
   else
   {
    printf("28\n");
   }
  }
  else 
  {
   printf("30\n");
  }
 }
 return 0;
}

 

How many days is a program written in java language? It must be a java language and not too complex syntax. I just learned

Calendar c = Calendar. getInstance ();
C. set (Calendar. YEAR, year );
C. set (Calendar. MONTH, month-1); // This value must be reduced by 1, that is, January is 0.
C. set (Calendar. DAY_OF_MOUTH, day );

Int result = c. get (Calendar. DAY_OF_YEAR );
 
I have a problem with the C language. This is caused by many Syntax problems.

1. Pay attention to scanf statements.
2 getDate (y, days, & m, & d );
3 pay attention to the calculation of the number of days swith ()
4 getDate () function
# Include <stdio. h>
Int isLeap (int year );
Void getDate (int y, int days, int * pm, int * pd );
Int getDays (int y, int m, int d );
Int main (void)
{
Int y, m, d, days;
Printf ("Enter _ year_month_day: \ n ");
Scanf ("% d", & y, & m, & d); // quotes of the scanf statement
Days = getDays (y, m, d); // Add d to the number of days
Printf ("% d day of % d: \ n", y, days );

GetDate (y, days, & m, & d );
Printf ("Enter % d: \ n", y, m, d );
Return 0;
}
// Write a subfunction to calculate the day of the year.
Void getDate (int y, int days, int * pm, int * pd)/* y indicates the Year, and days indicates the day of the year, pm points to the monthly value variable of the day in the main function, and * // * pd points to the Daily Value variable of the day in the main function */
{
Int m, d, I;
Int day;
For (I = 1; days> getDays (y, I, 0); I ++)
{
Days = days-getDays (y, I, 0 );
}
M = I;/* m: the monthly value calculated after one day */
D = days;/* d: The daily value calculated after one day */
* Pm = m;
* Pd = d;
}
// Write a sub-function to find that the day of a month is the day of the year.
Int getDays (int y, int m, int d)
{
Int n = 0;
Switch (m)
{
Case 12:
N + = 30;
Case 11:
N + = 31;
Case 10:
N + = 30;
Case 9:
N + = 31;
Case 8:
N + = 31;
Case 7:
N + = 30;
Case 6:
N + = 31;
Case 5:
N + = 30;
Case 4:
N + = 31;
Case 3:
If (isLeap (y) = 1)
N + = 29;
Else
N + = 28;
Case 2:
N + = 31;
Case 1:
N + = d;
Break;
Default:
N + = d;
}
Return n;
}
// Write a subfunction to determine the leap year.
Int isLeap (int year)
{
If (year % 4 = 0) & (year % 100! = 0) | (year % 400 = 0 ))
Return 1;
Else
Return 0; ...... the remaining full text>