A nice explaination with screenshots is given at
http://www.rawkam.com/?p=1156
On Tue, Aug 24, 2010 at 8:12 PM, Ajay Mohan wrote:
> @ Dave
>
> Tat s really a cool solution Dave. So i thot of generalizing the solution.
> If there are 'N' pirates then
>
> * The eldest should give 1 gold coin to N
techinterview.org
On Tue, Aug 24, 2010 at 8:12 PM, Ajay Mohan wrote:
> @ Dave
>
> Tat s really a cool solution Dave. So i thot of generalizing the solution.
> If there are 'N' pirates then
>
> * The eldest should give 1 gold coin to N/2 -1 ppl and he takes 100 -N/2
> +1
>
> Am i rite?
>
> --
> Y
@ Dave
Tat s really a cool solution Dave. So i thot of generalizing the solution.
If there are 'N' pirates then
* The eldest should give 1 gold coin to N/2 -1 ppl and he takes 100 -N/2 +1
Am i rite?
--
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"Algorithm Geeks"
This is not a computer problem. Think about it backwards.
If there are only 2 pirates left, then the elder can claim all 100
coins, at worst the vote will be tied, and so he survives and takes
100 coins.
If there are 3 pirates left, then the oldest can claim 99 coins and
give the youngest 1 coin,
