To calculate day of the week

When I was reading through some placement question papers, I found this problem asked quite a few times.
Q: Given a date, we have to find out the day of the week.

Solution:
1) C = 2(3-(cent%4))
where cent = first 2 digits of the century
2) y = year + floor(year/4)
where year = last two digits of the year given
3)m = choose from the table

Month	Value
January	0(leap year-3)
Febuary	3(leap year-6)
March	3
April	6
May	1
June	4
July	6
August	2
September	5
October	0
November	3
December	5

4)sum = c + y + m + date given
5)val = sum%7
6) sun - 0,mon-1,tue-2,... sat - 6, use value and find the day

Sample program:
#include

int leapyearMonValue[] = {0,3,3,6,1,4,6,2,5,0,3,5};
int nonleapyearMonValue[] = {3,6,3,6,1,4,6,2,5,0,3,5};

void calculateDay(int date,int mon,int year)
{
int val,cent,monval,yr,i=3,temp,leapyear=0;
int splitcent[4],wrongInput=0;
if(year%4 == 0)
leapyear = 1;
if(mon>12 || mon <1)
wrongInput = 1;
if(date==30 && !(mon%2) || mon==7 || mon==1)
wrongInput = 1;
if(mon==2 && date>28 && !leapyear)
wrongInput = 1;
if(!wrongInput)
{
while(year)
{
splitcent[i--] = year%10;
year = year/10;
}
temp = splitcent[0]*10 + splitcent[1];
cent = 2*(3-(temp%4));
temp = splitcent[2]*10 + splitcent[3];
yr = temp + floor(temp/4);
if(leapyear)
monval = leapyearMonValue[mon-1];
else
monval = nonleapyearMonValue[mon-1];
val = cent + yr + monval + date;
val = val%7;
switch(val)
{
case 0:
{
printf("day is Sunday\n");
break;
}
case 1:
{
printf("day is Monday\n");
break;
}
case 2:
{
printf("day is Tuesday\n");
break;
}
case 3:
{
printf("day is Wednesday\n");
break;
}
case 4:
{
printf("day is Thursday\n");
break;
}
case 5:
{
printf("day is Friday\n");
break;
}
case 6:
{
printf("day is Saturday\n");
break;
}
default:
printf("wrong calculation\n");
}
}
else
printf("wrong Input... check the date, mon \n");
}

int main()
{
int date,mon,year;
printf("Enter the date in format DD/MM/YYYY\t eg: 14/11/2007\n");
scanf("%d/%d/%d",&date,&mon,&year);
calculateDay(date,mon,year);
getch();
return 0;
}


Courtesy:
Wikipedia