find_day(int dd, int mm, int yy)
{
// dd = day in question, used in odddayday function
// mm = month in question, used in odddaymonth function
// yy = year in question, used in odddayyear and isleapyear functions
int odddays, leapyear;
// odddays is the value returned
// leapyear = 1 if yy is leap year, 0 if not
leapyear=isleapyear(yy);
// Adds oddyear + oddmonth + oddday
odddays=odddayyear(yy)+odddaymonth(mm,leapyear)+odddayday(dd);
// Get modulus of return day divived by 7
odddays %= 7;
return odddays;
}
/*Every 4th year is a leap year, however centuries are not leap years
except those divisible by 400
for example:
1990 is NOT a leap year
2000 is a leap year
2004 is a leap year
*/
isleapyear(int yy)
{
if(yy%4==0 && (yy%100!=0 || yy%400==0))
return 1; // leap year
else
return 0; // not leap year
}
//Calculates odd days upto first day of year
odddayyear(int yy)
{
// odddays is return value
int odddays=0;
// subtract 1 from year
yy-=1;
// if year > than 400
if(yy>=400)
yy%=400; // Get the modulus of year / 400
// if year (or modulus) is greater or equal to 100
if(yy>=100)
{
odddays=yy/100; // return value is year / 100
odddays *= 5; // return value is multipled by 5
yy%=100; // get modulus of year or year modulus
}
odddays += yy; // add return value to year or year modulus
odddays += yy/4; // add return value to year / 4
return odddays;
}
odddaymonth(int mm, char leapyear)
{
// mm is month
// leapyear = 0 or 1, see above. Note C++ treats char as int
switch(mm)
{
// January returns 0
case 1:
return 0;
// February returns 3
case 2:
return 3;
// March and November returns 4 if leap year or 3 if not
case 3:
case 11:
return leapyear?4:3;
// April and July return 0 if leap year or 6 if not
case 4:
case 7:
return leapyear?0:6;
// May returns 2 if leap year or 1 is not
case 5:
return leapyear?2:1;
// June returns 5 if leap year of 4 if not
case 6:
return leapyear?5:4;
// August returns 3 if leap year or 2 is not
case 8:
return leapyear?3:2;
// September and December returns 6 if leap year or 5 if not
case 9:
case 12:
return leapyear?6:5;
// October returns 1 if leap year or 0 if not
case 10:
return leapyear?1:0;
}
}
// return remainder of day divided by 7
odddayday(int dd)
{
// dd is day
return dd%7;
}
/*******************H I N T S *********************
Every 4th year is a leap year,
however centuries are not leap years except those divisible by 400
for example:
1990 is NOT a leap year
2000 is a leap year
2004 is a leap year
for example:
-------------
1 ordinary year = 365 days = (52 weeks + 1 day) = 1 odd day
that's why if this year 23th August is on Tuesday,
it will be on Wednesday next year
1 leap year = 366 days = (52 weeks + 2 days) = 2 odd days
100 years = 76 ordinary years + 24 leap years
= [(76*52) weeks+76 days] + [(24*52) weeks+48 days]
= 5200 weeks + 124 days
= 5217 weeks + 5 days
= 5 odd days
200 years = 10 odd days = 3 odd days
300 years = 15 odd days = 1 odd days
400 years = 20 + 1 odd days = 0 days
800 years = 0 odd days
Sunday is the 0th odd day, Monday the 1st odd day and so on*/int first_day = find_day(3, 8, 2006);int odddays, leapyear;
leapyear=isleapyear(yy);
odddays=odddayyear(yy)+odddaymonth(mm,leapyear)+odddayday(dd);
odddays %= 7;
return odddays;odddayyear(int yy)
{
// yy will be 2006
int odddays=0;
// subtract 1 from year
yy-=1; // yy will now be 2005
// This will be true since 2006 > 400
if(yy>=400)
yy%=400; // yy now equals 5
// This condition is false since yy < 100
if(yy>=100)
{
odddays=yy/100; // return value is year / 100
odddays *= 5; // return value is multipled by 5
yy%=100; // get modulus of year or year modulus
}
odddays += yy; // odddays equals 5
odddays += yy/4; // oddays now equals 6 as 5 / 4 = 1.25 and since odddays is an integer it drops the .25 so we have 1 + 5 = 6
return odddays; // This function returns 6;
}