
October 19th, 2012, 11:22 AM
#1
Party Algorithm
I have got problem with this task from my university algorithm course (translated from german):
> John have got a party. He prepared $x$ glasses of vodka and $k*x$
> glasses of rum.
>
> Number of glasses is n. Each guests drink one
> round of alcohol (one round is $k$ glasses of rum and $one$ of vodka) but
> guest must chose glass that every full glass must have got one full
> neighbor.
>
> Write the program which simulate John's party. The party ends when all
> glasses are empty.
>
> In input we have got $n$ and $k$ which means number of glasses and number
> of glasses of rum drunk in single turn).
Simple Input : (assume that r is rum and v  vodka ).
> 16 3 vvrrrvrvrrrrrrrr
Simple Output : (we numbered glass 1...n)
> 1 14 15 16
> 2 11 12 13
> 3 4 5 6
> 7 8 9 10
Explain :
Guest 1 drink 1, 14, 15 and 16 glass.
Guest 2 drink 2, 11, 12, 13 glass.
etc.
We write it in ascending order. Notice that we have got $\frac{n}{(k+1)}$ guests, $n$ glasses, $k$ glasses of rum in every round. And in all inputs $k+1$ divides $n$. And $n$ can be very big  about 2 000 000. And we have about 12s to answer the question.
Find the fastest as possible algorithm to solve this problem.
Thanks for every help. If you have got any questions, i will answer it.
Edger.

October 19th, 2012, 11:40 AM
#2

October 19th, 2012, 05:02 PM
#3
Re: Party Algorithm
Originally Posted by Edger521
Thanks for every help. If you have got any questions, i will answer it.
I'm slightly Lost in Translation. Could you please post the German original.
Is this an algorithmic problem or is it a question of writing a simulation program to study an algorithm. If it's the latter I wouldn't expect anyone to supply such a program.
Have you tried something yourself? Some effort on your part is expected.
Posting Permissions
 You may not post new threads
 You may not post replies
 You may not post attachments
 You may not edit your posts

Forum Rules

Click Here to Expand Forum to Full Width
This a Codeguru.com survey!
OnDemand Webinars (sponsored)
