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From "01/01/1900" to "monday". How???

Given a date, how can I work out the day of the week?
Since I'm in England and we do dates as dd/mm/yy, I want to enter a date, for example 15/04/1976 and the function to return a value for Thursday (which was the day of the week for 15/04/1976 if I remember). The way I'm thinking would probably work but it would be so long that it would just be dumb to use. No grace or finesse. Come on, astound me.
I think this could be hard so there's 200 points riding on it!
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wierd_dave
Asked:
wierd_dave
1 Solution
 
ahTriCommented:
I'm not really sure to attemp to take the 200 points u offer but i'm thinking of setting a fix date for example today (Saturday, 16/1/1999) calculate the number of days from today to the date u want to query, devide that number to 7, the remain of the division can be calculate to make the Date.

If u think this way work I will send u the set of objects to calculate the number of days from one day to another. Sorry for your 200 points if u not satisfy with this answer
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ozoCommented:
int dd=1;
int mm=1;
int yy=1900;
int jdn;
char *weekday[]={"Tuesday","Wednesday","THursday","Friday","Saturday","Sunday","Monday"};
if( mm < 3 ){ mm += 12; yy -= 1; }
jdn = dd+1720996+(mm+1)*306/10+yy*365 + yy/4 - yy/100 + yy/400;
printf("%s\n",weeday[jdn%7]);
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wierd_daveAuthor Commented:
Sorry ahTri, that was fairly much the way I was thinking. I was going to work out the number of days from 01/01/1900 and use the remainder as you said to get the date.
Rejection seems so negative, but i've tried ozos' method and it is far batter than I would have some up with.
If ozo wants to submit an answer, they can have the 200 points.

Thanks to both of you.
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The_BrainCommented:
I have written a program to work out what day of the week it is on any specific date, in that there is a function that tells you the day number from 1900...

Well?  Interested?
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ozoCommented:
char *weekday[]={"Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday","Monday"};
 dd=15;
 mm=4;
 yy=1976;
if( mm < 3 ){ mm += 12; yy -= 1; }
JulianDayNumber = 1721088 + dd + (mm-2)*367/12 + yy*365 + yy/4 - yy/100 + yy/400;
printf("%s\n",weeday[JulianDayNumber%7])
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