Our year must be greater than 1752 since our calendar changed. So 1752 is an anonmly and before that we would need a different algorithm. For the curious here is what the calendar for 1752 looked like:
1752
January February March
Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa
1 2 3 4 1 1 2 3 4 5 6 7
5 6 7 8 9 10 11 2 3 4 5 6 7 8 8 9 10 11 12 13 14
12 13 14 15 16 17 18 9 10 11 12 13 14 15 15 16 17 18 19 20 21
19 20 21 22 23 24 25 16 17 18 19 20 21 22 22 23 24 25 26 27 28
26 27 28 29 30 31 23 24 25 26 27 28 29 29 30 31
April May June
Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa
1 2 3 4 1 2 1 2 3 4 5 6
5 6 7 8 9 10 11 3 4 5 6 7 8 9 7 8 9 10 11 12 13
12 13 14 15 16 17 18 10 11 12 13 14 15 16 14 15 16 17 18 19 20
19 20 21 22 23 24 25 17 18 19 20 21 22 23 21 22 23 24 25 26 27
26 27 28 29 30 24 25 26 27 28 29 30 28 29 30
31
July August September
Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa
1 2 3 4 1 1 2 14 15 16
5 6 7 8 9 10 11 2 3 4 5 6 7 8 17 18 19 20 21 22 23
12 13 14 15 16 17 18 9 10 11 12 13 14 15 24 25 26 27 28 29 30
19 20 21 22 23 24 25 16 17 18 19 20 21 22
26 27 28 29 30 31 23 24 25 26 27 28 29
30 31
October November December
Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa
1 2 3 4 5 6 7 1 2 3 4 1 2
8 9 10 11 12 13 14 5 6 7 8 9 10 11 3 4 5 6 7 8 9
15 16 17 18 19 20 21 12 13 14 15 16 17 18 10 11 12 13 14 15 16
22 23 24 25 26 27 28 19 20 21 22 23 24 25 17 18 19 20 21 22 23
29 30 31 26 27 28 29 30 24 25 26 27 28 29 30
31
Between 1 and 12.
Between 1 and the maximum for that month.
So 1..31 for January, March, May, July, August, October and December.
1..30 for April, June, September and November
February is either 1 .. 28 or 29 depending if it is a leap year.
A year that is divisible by 4. But it it is divisible by 100, it must be divisible by 400. So
| year | divisible by 4 |
divisible by 100 |
divisible by 400 |
Leap |
|---|---|---|---|---|
| 2006 | No | - | - | No |
| 2004 | Yes | No | - | Yes |
| 1900 | Yes | Yes | No | No |
| 2000 | Yes | Yes | Yes | Yes |
If we invert the year divisble by 100 test it looks like a logical or condition. so our test looks like:
| year | Y%4 ==0 | Y%100 > 0 | Y%400 == 0 | Leap |
|---|---|---|---|---|
| 2006 | F | - | - | F |
| 2004 | T | T | - | T |
| 1900 | T | F | F | F |
| 2000 | T | F | T | T |
To calculate the day of the week for a given date, Zeller's congruence can be applied. A date is expressed as follows:
- "m" month number, with January and February taken as months 11 and 12 of the preceding year. Then March is 1, April is 2.. December is 10.
- "k" day of the month
- "C" The century
- "D" The year in the century.
- "f" The day of the week where 0 = Sunday, 1 = Monday ... 6 = Saturday
f={[2.6m-0.2]+k+D+[D/4]+[C/4]-2C} mod 7
[] denotes the greatest integer in.
//
// what day of the week is it?
//
#include <iostream>
using namespace std;
int main()
{
int iMonth, iDay, iYear; // input
bool bValid=true; // valid input flag
int iMaxdays; // Days in month
//
// variables for Zellers
//
int m,k,C,D,f; // as in the write up
//
// get input
//
cout << "Please enter the Month :";
cin >> iMonth;
cout << "Enter the day of the month :";
cin >> iDay;
cout << "Finally enter the year : ";
cin >> iYear;
//
// validate the input
//
if(iMonth <= 1 || iMonth > 12) {
cout << "Invalid Month " <<endl;
cout << " Must be 1..12" <<endl;
bValid=false;
}
if(iYear < 1753 ) {
cout << "Invalid year " <<endl;
cout << "must be greater than 1753 " << endl;
bValid=false;
}
switch(iMonth) {
case 1:
case 3:
case 5:
case 7:
case 8:
case 10:
case 12: iMaxdays=31;
break;
case 4:
case 6:
case 9:
case 11: iMaxdays=30;
break;
case 2: if(iYear%4 == 0 &&
(iYear%100 >0 || iYear%400 == 0))
iMaxdays=29;
else iMaxdays=28;
break;
default: iMaxdays=1;
}
if(iDay < 1 || iDay > iMaxdays) {
cout << "Invalid day of Month " << endl;
cout << "For month " << iMonth
<< " must be 1 .. " << iMaxdays << endl;
bValid=false;
}
//
// Zeller's congruence
//
if(bValid) {
m=iMonth-2;
k=iDay;
D=iYear;
if(m <=0) {
m+=12;
D--;
}
C=D/100;
D%=100;
f=((int)(2.6*m-0.2) + k + D + D/4 + C/4 - 2*C) % 7;
if(f < 0 ) f+=7; // % to mod fix
switch(f) {
case 0 : cout << "Sunday" << endl; break;
case 1 : cout << "Monday" << endl; break;
case 2 : cout << "Tuesday" << endl; break;
case 3 : cout << "Wednesday" << endl; break;
case 4 : cout << "Thursday" << endl; break;
case 5 : cout << "Friday" << endl; break;
case 6 : cout << "Saturday" << endl; break;
}
}
system("pause");
return(0);
}
Please enter the Month :9 Enter the day of the month :27 Finally enter the year : 2006 Wednesday Press any key to continue . . .
Please enter the Month :12 Enter the day of the month :7 Finally enter the year : 1941 Sunday Press any key to continue . . .
Please enter the Month :9 Enter the day of the month :11 Finally enter the year : 2001 Tuesday Press any key to continue . . .
Please enter the Month :2 Enter the day of the month :29 Finally enter the year : 2006 Invalid day of Month For month 2 must be 1 .. 28 Press any key to continue . . .
Please enter the Month :13 Enter the day of the month :23 Finally enter the year : 1624 Invalid Month Must be 1..12 Invalid year must be greater than 1753 Invalid day of Month For month 13 must be 1 .. 1 Press any key to continue . . .