CareerCup

//One more approach
//======================================
#include<stdio.h>
int main()
{
    int months[]={31,28,31,30,31,30,31,31,30,31,30,31};
    int i,j,year;
    printf("Enter year ");
    scanf("%d",&year);
    if(year%4==0)
        months[1]=29;
    for(i=0;i<12;i++)
        for(j=0;j<months[i];j++)
            printf("%d/%d/%d\n",i+1,j+1,year);
}

//================================================

import calendar

def printCal():
    year = raw_input(int("Which year?"))
    print calendar.calendar(year)

@manoj
ain't we supposed to write full code for printing the calendar and not use any existing functionality to do the same?

Have a look here
http://coverage.livinglogic.de/Lib/calendar.py.html

Do we need to print out the weekday corresponding to a date in the calendar? for example, Jan. 1st is Friday.

How do u know the day of the week for 1st Jan of the given year??

we can make the calender by using the concept of odd days
like for example......
suppose the give year is 1980 now calculate the odd days for the 1 Jan 1980.
Odd days in a year 365mod7 gives 1
Odd days in a year 366mod7 gives 2 leap year.
so by this we can calculate the odd days in
100 year = 5 odd days
200 year = 3 odd days
300 year = 1 odd days
and 400,800,1600 have 0 odd days.
now claculate the odd days for 1 Jan 1980
No. of Year odd days
1600 0
300 1
79 0 ((79/4)*2+(79-(79/4)))mod7=0
-----------------
1979 1
n for 1 Jan 1
-----------------
total 2 odd days

and 0 is Sunday 1 is Monday 2 is Tuesday and so on
so the year 1980 start with Tuesday now we can easily make the calender for the year.

i hope this will help............

Sorry this is print all dates for a year. Not calendar

it will print, u just have to set according to the months.
means u have day of year's very first date(1st January). if it is a leap year then set all months according to feb of 29 othervise set it normaly.

any good answer? c'mon guys. put some solution/ideas!

if only printing date, it'll be much easier

I am telling you guys. When I said calendar, I meant just dates not the real calendar.

//The following program should print the calender given a year

// to preserve whitespace
//============================================================================
// Name        : CalendarCPP.cpp
// Author      : 
// Version     :
// Copyright   :
//============================================================================

#include <iostream>
#include <string>
using namespace std;

int main() {
    int year;
    bool leap = false;
    char *months[] = {"January","February", "March", "April", "May", "June",
            "July", "August", "September", "October", "November", "December"};
    cout << "Enter the year: " << endl;
    cin>>year;
    // check for leap year
    /*
    **    The logic is that the year is either divisible by both
    **    100 and 4 , OR its only divisible by 4 not by hundred
    */
    if(year%400 ==0 || (year%100 != 0 && year%4 == 0)) {
        cout<<"Year %d is a leap year\n"<<year;
        leap = true;
    }
    else {
        cout<<"Year %d is not a leap year\n"<<year;
    }

    //January to June
    for(int i=0; i<=6; i++) { // month
        for(int j=1; j<=31; j++) { // days
            if(i!=1) { //Not February
                if(i%2==0){ //31 days
                    cout<<months[i]<<" "<<j<<","<<year<<endl;
                } else { //30 days
                    if(j<31) {
                        cout<<months[i]<<" "<<j<<","<<year<<endl;
                    }
                }
            } else { //February
                if(j<30) {
                    if(leap) {
                        cout<<months[i]<<" "<<j<<","<<year<<endl;
                    } else {
                        if(j<29) {
                            cout<<months[i]<<" "<<j<<","<<year<<endl;
                        }
                    }
                }
            }
        }
    }
    //August to December
    for(int i=7;i<31;i++) { //month
        for(int j=1;j<=31;j++){ //days
            if(i%2!=0){ //31 days
                cout<<months[i]<<" "<<j<<","<<year<<endl;
            } else { //30 days
                if(j<31)
                    cout<<months[i]<<" "<<j<<","<<year<<endl;
            }
        }
    }
    return 0;
}
//================================================================

// to preserve whitespace

Take the current date and day of the week.
Find the difference in days between the current date and 1 Jan <given year>
(diff number) mod 7 should return you the correct day of the week for given year
Start printing the calendar (in which ever way you want!)

Yes, in order to find the current date/day of the week, you would use inbuilt functions -- which are not allowed. I believe Rock's method looks simple and sweet. Anyway, in the assessment test they just ask you the psudo code and the logical approach.

@ravi

how you calculated odd days in 100 years or so ?
Did you find how many leap and non-leap years in a hundred years, then added all the days in these 100 years. Finally did mod 7 ?

Is there a shorter way ?

@ravi

I got what you are saying ! Thanks.

Unix
cal year

#include <iostream>

using namespace std;

void printMonthTitle(int year, int month);
int getTotalNumOfDays(int year, int month);
int getNumOfDaysInMonth(int year, int month);
int getStartDay(int year, int month);
bool isLeapYear(int year);
void printMonthBody(int startDay, int numOfDaysInMonth);
const char* getMonthName(int month);

  /** Print the calendar for a month in a year */
   void printMonth(int year, int month) {
    // Get start day of the week for the first date in the month
    int startDay = getStartDay(year, month);

    // Get number of days in the month
    int numOfDaysInMonth = getNumOfDaysInMonth(year, month);

    // Print headings
    printMonthTitle(year, month);

    // Print body
    printMonthBody(startDay, numOfDaysInMonth);
  }

  /** Get the start day of the first day in a month */
   int getStartDay(int year, int month) {
    // Get total number of days since 1/1/1800
    int startDay1800 = 3;
    int totalNumOfDays = getTotalNumOfDays(year, month);

    // Return the start day
    return (int)((totalNumOfDays + startDay1800) % 7);
  }

  /** Get the total number of days since Jan 1, 1800 */
   int getTotalNumOfDays(int year, int month) {
     static int tillYear=0;
    
    if (!tillYear)
    {
        // Get the total days from 1800 to year -1
        for (int i = 1800; i < year; i++)
        {
            if (isLeapYear(i))
              tillYear = tillYear + 366;
            else
              tillYear = tillYear + 365;
        }
    }
    int total = tillYear;
    // Add days from Jan to the month prior to the calendar month
    for (int i = 1; i < month; i++)
      total = total + getNumOfDaysInMonth(year, i);

    return total;
  }

  /** Get the number of days in a month */
   int getNumOfDaysInMonth(int year, int month) {
    if (month == 1 || month==3 || month == 5 || month == 7 ||
      month == 8 || month == 10 || month == 12)
      return 31;

    if (month == 4 || month == 6 || month == 9 || month == 11)
      return 30;

    if (month == 2)
      if (isLeapYear(year))
        return 29;
      else
        return 28;

    return 0; // If month is incorrect.
  }

  /** Determine if it is a leap year */
   bool isLeapYear(int year) {
    if ((year % 400 == 0) || ((year % 4 == 0) && (year % 100 != 0)))
      return true;

    return false;
  }

  /** Print month body */
   void printMonthBody(int startDay, int numOfDaysInMonth) {
    // Pad space before the first day of the month
    int i = 0;
    for (i = 0; i < startDay; i++)
      cout <<"    ";

    for (i = 1; i <= numOfDaysInMonth; i++) {
      if (i < 10)
        cout <<"   "<<i;
      else
        cout <<"  "<< i;

      if ((i + startDay) % 7 == 0)
        cout<<endl;
    }

    cout <<"\n";
  }

  /** Print the month title, i.e. May, 1999 */
   void printMonthTitle(int year, int month) {
    
    cout <<"-----------------------------\n";
    cout <<" Sun Mon Tue Wed Thu Fri Sat\n";
    cout <<"         "<< getMonthName(month) <<", "<< year<<endl;
  }

  /** Get the English name for the month */
   const char* getMonthName(int month) {
    switch (month) {
      case 1: return "January"; break;
      case 2: return "February"; break;
      case 3: return "March"; break;
      case 4: return "April"; break;
      case 5: return "May"; break;
      case 6:  return "June"; break;
      case 7:  return "July"; break;
      case 8:  return "August"; break;
      case 9:  return "September"; break;
      case 10:  return "October"; break;
      case 11:  return "November"; break;
      case 12:  return "December";
    }

    return "";
  }


  int main() 
  {
    // Convert string into integer
    int month = 1;
    int year = 2015;

    // Print calendar for the month of the year
    printMonth(year, month);
    return 0;

}

raj:
use the base 1970 epoc and then get moving