In other words, you are counting by one in binary. When you count to
2^N - 1, it is on.

On Sun, May 9, 2010 at 1:00 PM, Abizern <[email protected]> wrote:
> Here's a simple analysis:
> Imagine 3 lights plugged into a wall at the right and the state is 1 if it
> is on, and 0 if it is off.
> so for each 'click' the states will be:
> 0        0  0  0
> 1        0  0  1
> 2        0  1  0
> 3        0  1  1
> 4        1  0  0
> 5        1  0  1
> 6        1  0  1
> 7        1  1  1
> 8        0  0  0
> 9        0  0  1
> 10      0  1  1
> And so on. You can see a pattern here: that for the light to be on, all the
> switches need to be set to 1. And as a binary number -> so:
> for 1 snappers the number is 1 which is 1 in decimal
> for 2 snappers the number is 1 1 which is 3 in decimal
> for 3 snappers the number is 1 1 1 which is 7 in decimal.
> So can you see the pattern. for N snappers, all the digits in binary are 1
> for the number (2^N)-1. As this wraps around we bring in modulus which is
> where we see that k % 2^N must be (2^N)-1.
>
>
>
> On 9 May 2010 20:06, Douglas Drumond <[email protected]> wrote:
>>
>> > Is anyone interested in discussing the algorithm for Snapper Chain.
>> > What was the right approach to it?
>> >
>> > PS: I have seen some solved source code. But, would like to know about
>> > the
>> > approach.
>>
>>
>> You should check contest analysis:
>> http://code.google.com/codejam/contest/dashboard?c=433101#s=a
>>
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>
>
>
> --
> Abizer
>
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