char days[] = { "S", "M", "T", "W", "T", "F", "S" }; // day abbrs
 
cout << "your month here" << "\n\n";
 
for(int i = 0; i < 7; i++)
    cout << days[i] << " ";
cout << "\n";
 
for(int i = 0; i < 14; i++)
    cout << "-";
cout << "\n";

#include <stdio.h>
#include <string.h>
 
 
#define MIN_YEAR 1582
#define MAX_YEAR 9999
#define DAYS_PER_WEEK 7
#define MAX_FILE_NAME_SIZE 25
#define MAX_MONTH_NAME_SIZE 10
#define PREVIOUSDAYSPACE 3
 
//function declarations
int count_leap_years(int);
int find_jan_first(int);
int print_month(int, int, int);
 
void print_calendar(int);
void print_dayNames();
 
 
int main()
{
 
	int year;
 
	char fileName[MAX_FILE_NAME_SIZE];
    FILE * inFile;                         //file pointer
	printf("Please enter a date file: "); // user enters text file
	scanf("%s",fileName);
	inFile = fopen (fileName,"r");
	while(!inFile)                   //error check if not valid file
	{
		printf("Try Again ");
		printf("Please enter a date file: ");
		scanf("%s",fileName);
		inFile = fopen (fileName,"r");
	}
	while ( fscanf ( inFile, "%d", &year )!= EOF ) //error if date not in Gregorian calendar
	{
		if (year < MIN_YEAR || year > MAX_YEAR)
		{
			printf("\nCalendar may be incorrect,\n"); 
			printf("the year may not exist in\n");
			printf("the Gregorian Calendar range.\n");
		}
		print_calendar (year);
	}
 
	fclose (inFile);
 
}
 
void print_calendar(int year)  //calling first day and year
{
	int firstday; //first day of month	
    int month;    // month of the year
 
	printf("\n       %d\n\n", year);
	firstday = find_jan_first(year);
 
	for(month= 1; month <= 12; month++)
	{
		firstday= print_month (month, firstday, year);
	}
 
}
 
int print_month (int month, int day, int year) //length of month and name of month
{
 
	int days, numspacefirst, spaces, datetoprint;
 
	char monthName[MAX_MONTH_NAME_SIZE];
	switch(month)
	{
	case 1: 
        days = 31;
		strcpy(monthName,"January");
        break;
    case 3: 
        days = 31;
		strcpy(monthName,"March");
        break;
    case 5: 
        days = 31;
		strcpy(monthName,"May");
        break;
    case 7: 
        days = 31;
		strcpy(monthName,"July");
        break;
    case 8: 
        days = 31;
		strcpy(monthName,"August");
        break;
    case 10: 
        days = 31;
		strcpy(monthName,"October");
        break;
    case 12: 
        days = 31;
		strcpy(monthName,"December");
        break;
    case 4: 
        days = 30;
		strcpy(monthName,"April");
        break;
    case 6: 
        days = 30;
		strcpy(monthName,"June");
        break;
    case 9: 
        days = 30;
		strcpy(monthName,"September");
        break;
    case 11:
        days = 30;
		strcpy(monthName,"November");
        break;
    case 2: 
        days = 28 +
			   1 * (count_leap_years(year+1) -count_leap_years(year));
        strcpy(monthName,"February");
	}
 
	printf("     %s\n", monthName);
	print_dayNames();
	numspacefirst = PREVIOUSDAYSPACE * day;   //spacing of date numbers
 
	for (spaces=0;spaces<numspacefirst;spaces++)
	{
		printf("%c",' ');
	}
	for (datetoprint=1;datetoprint<=days;datetoprint++)
	{
		printf("%2d",datetoprint);
		printf("%c", ' ');
		day++;
		if (day == DAYS_PER_WEEK)
		{
			day = 0;
			printf("\n");
		}
	}
	printf("\n\n");
 
	return day;
 
}// find the day
 
void print_dayNames() // print day identations
    {
    printf("S  M  T  W  T  F  S \n");
	printf("--------------------\n");
    }
 
int find_jan_first
(int year)              // the input year, must be >= MIN_YEAR 
 
{  // find_jan_first 
 
   int firstDayInJan; // 0 is Sunday, 1 is Monday, ..., 6 is Saturday 
 
   // starting on January 1, 1582, add days to find the first day 
   // in January of the year chosen 
   firstDayInJan = ((5 + (year - MIN_YEAR) + 
   count_leap_years (year)) % DAYS_PER_WEEK);
 
   return (firstDayInJan);
 
}  // find_jan_first 
 
int count_leap_years
(int year)               // user's choice of year, must be >= MIN_YEAR 
 
{  // count_leap_years 
 
   int leapYears;     // number of leap years to return 
   int hundreds;      // number of years multiple of a hundred 
   int fourHundreds;  // number of years multiple of four hundred 
 
   // determine number of years in interval that are a multiple of 4 
   leapYears = (year - (MIN_YEAR - 1)) / 4;
 
   // determine number of years in interval that are a multiple of 100 
   // and subtract, because they are not leap years 
   hundreds = (year - 1501) / 100;
   leapYears -= hundreds;
 
   // determine number of years in interval that are a multiple of 400 
   // and add these back in, since they are leap years 
   fourHundreds = (year - 1201) / 400;
   leapYears += fourHundreds;
 
   return (leapYears);
 
}  // count_leap_years