Kieran Grayshon|Mayan Calendar

The idea of this program was to challenge understanding of C, the objective of this program is to convert a Mayan Date to a Gregorian date (given the date is after Jan 2000). The way the code achieves this is by implementing a sort of division by repeated subtraction. We cannot do a division due to the subtraction not being constant due to edge cases such as leap years and non-constant month lengths. We also take advantage of 2 structs to use as data structures as it makes the most sense Problem from: Olympiad

To view more of the code on Github click here

        
            
/*
  Mayan Calendar
  Description: Conversion between the Mayan Calendar and Gregorian Calendar
  Author: Kieran Grayshon
*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

// https://www.mayaarchaeologist.co.uk/public-resources/maya-calendar-converter-long-count/
struct Mayan{
  int baktun;
  int katun;
  int tun;
  int uinal;
  int kin;
};

struct Date{
  int year;
  int month;
  int day;
  char* day_name;
};

/**
 * mayan_to_gregorian:Converts Mayan date to Gregorian date
 *
 * @param: date as a pointer to a Mayan struct defined above
 *
*/
struct Date mayan_to_gregorian(struct Mayan* date){
  //Day names in this order due to offset as 1st Jan 2000 is a Saturday
  const char day_names[7][3]={"sat", "sun", "mon", "tue", "wed", "thu", "fri"};
  const char month_names[12][3]={"jan", "feb", "mar", "apr", "may", "jun", "jul", "aug", "sep", "oct", "nov", "dec"};
  int month_lengths[12]={31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  const int jan_2000=1728000+136800+2160+300+2; //Number of kins in 1st Jan 2000

  //Convert the Mayan date into days
  int days=(date->baktun)*144000+(date->katun)*7200+(date->tun)*360+(date->uinal)*20+date->kin;
  int difference=days-jan_2000; //Find number of days between the date and "anchor" date

  //Get setup to store the result
  struct Date result;

  //Calculate the day by dividing my days in a week
  result.day_name=malloc(3);
  strncpy(result.day_name, day_names[difference%7], 3);

  int year=0;
  int month=0;
  int day=0;

  /*
  Algorithm works by:
  Take away from the number of days until a subtraction
  makes the difference go negative
  This is now the remainder and we move down to the next level
  and repeat
  */

  int days_per_year=365;
  while(1){
    days_per_year=(year%4==0)?366:365;
    if(difference-days_per_year<0){
      break;
    }
    difference-=days_per_year;
    year++;
  }

  int days_per_month;
  if(year%4==0){month_lengths[1]=29;}
  while(1){
    days_per_month=month_lengths[month];
    if(difference-days_per_month<0){
      break;
    }
    difference-=days_per_month;
    month++;
  }
  month++;
  day=difference+1;

  result.year=year+2000;
  result.month=month;
  result.day=day;

  return result;
}

int main(){
  //Setup the test date and run the program
  //Print out the result
  struct Mayan* test=malloc(sizeof(struct Mayan));
  //Wednesday 12th Oct 2022
  test->baktun=13;
  test->katun=0;
  test->tun=9;
  test->uinal=17;
  test->kin=2;

  struct Date result=mayan_to_gregorian(test);
  free(test);
  printf("%s %i %i %i\n", result.day_name, result.day, result.month, result.year);
  return 0;
}