CODEWITHSUNDEEP
cdate-arithmetic
Optimus Prime
Optimus Prime

Reputation: 6907

simple program to calculate day not working

I am trying to write a simple program, that outputs the day given a date, using 1/1/0001 as Saturday. I have written the program, but have passed about 4 hours just to find some mistake which is making my answer wrong. The correct answer can be found here. http://www.timeanddate.com/date/durationresult.html?d1=1&m1=1&y1=1&d2=1&m2=1&y2=400&ti=on

The following is my program:

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

long long int leap_days, days;

long long int *no_leap_years(int day, int month, long long int year){
    leap_days = (year-1)/4;
    leap_days -= (year-1)/100;
    leap_days += (year-1)/400;

    int is_leap;

    if(!(year%400)){
        is_leap = 1;
    } else if(!(year%100)){
        is_leap = 0;
    } else if(!(year%4)){
        is_leap = 1;
    } else {
        is_leap = 0;
    }

    printf("Leap year - %d\n", is_leap);
    printf("Leap days - %lld\n", leap_days);
    if(is_leap){
        if(month>2) leap_days++;
        else if(month==2){
            if(day>28) leap_days++;
        }
    }
    printf("Leap days - %lld\n", leap_days);
    return &leap_days;
}

long long int *no_of_days(int day, int month, long long int year){
    long long int leap_days = *no_leap_years(day, month, year);
    days = leap_days + (365*(year-1));
    if(month>1) days+=31;
    if(month>2) days+=28;
    if(month>3) days+=31;
    if(month>4) days+=30;
    if(month>5) days+=31;
    if(month>6) days+=30;
    if(month>7) days+=31;
    if(month>8) days+=31;
    if(month>9) days+=30;
    if(month>10) days+=31;
    if(month>11) days+=30;
    days+=day;
    printf("total - %lld\n", days);
    return &days;
}

char *day_computer(int day, int month, long long int year){
    long long int days = *no_of_days(day, month, year);
    printf("total - %lld\n", days);
    days %= 7;
    printf("remain - %lld\n", days);
    if(days==1) return "Saturday";
    else if(days==2) return "Sunday";
    else if(days==3) return "Monday";
    else if(days==4) return "Tuesday";
    else if(days==5) return "Wednesday";
    else if(days==6) return "Thursday";
    else return "Friday";
}

int main()
{
   int  dd = 1;
   int mm = 1;
   long long int yy = 400;

   printf("%s\n", day_computer(dd, mm, yy));

   return 0;
}

According to the website, total no.of days from 1/1/1 to 1/1/400 should be 145,735 days, but according to my code, it is 145732. I cannot find where I am missing 3 days. Please help me.

You may use this online editor to compile my program. http://www.compileonline.com/compile_c_online.php

Upvotes: 2

Views: 136

Answers (1)

Jongware
Jongware

Reputation: 22457

The online calendar program is correct; then again, so is your code. The problem lies in the specific time period for which you are calculating.

Your code adjusts for leap years: every 4 years, except for centuries (which are not) and quad-centuries (which again are). However, both adjustments "once a century" were not observed in the Julian Calendar:

The Julian reform lengthened seven months and replaced the intercalary month with an intercalary day to be added every four years to February.

For the period 46 BC - 4 October 1582, you should use the Julian way of calculating leap years. For dates after that "closing date" (which is 15 October 1582), you can use the Gregorian (current) calculation.

Without the 2 adjustment for leap years in whole centuries, you get the same result as the online calendar.

Upvotes: 3

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