Problem 42774. GJam March 2016 IOW: Clock Dance

Created by Richard Zapor

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This Challenge is derived from <a href = "http://code.google.com/codejam/contest/8274486/dashboard#s=p1">GJam March 2016 Annual I/O for Women Dance Around the Clock</a>. This is a mix of the small and large data sets.The GJam story goes that D even dancers are arranged clockwise 1:D. Who is adjacent to dancer K after N pair swaps. The odd move swaps positions 1/2,3/4,..D-1/D. The even move swaps positions D/1, 2/3,...D-2/D-1. D=8 [1:8] becomes [2 1 4 3 6 5 8 7] after the first swap. After second swap we see [7 4 1 6 3 8 5 2].<img src = "http://code.google.com/codejam/contest/images/?image=dance_around_the_clock.png&p=5733344260128768&c=8274486">Input: [D K N] , 4&lt;=D&lt;=1E8, 1&lt;=K&lt;=D, 1&lt;=N&lt;=1E8Output: [KCW KCCW] , KCW is dancer to left of Kth dancer after N movesExamples: [m] [v]<pre class="language-matlab">[8 3 1] creates v=[6 4]
[8 4 2] creates v=[1 7]
[6 6 1] creates v=[5 3]
</pre><a href = "http://code.google.com/codejam">Google Code Jam 2016 Open Qualifier: April 8, 2016</a>Contest Theory: The small case was D&lt;10 and N&lt;10. This could be done brute force in the 5 minutes of entry submission allowed. However, this is clearly a case of modulo math so might as well solve the unbounded case of D/N &lt;1E8. The number being assigned as 1:D leads to confusion as mod maps to 0:D-1. Suggest offset the K dancer number by 1 for processing and then correct for the final answer. Mod is fun as mod(-1,8) yields a useful 7. Quick observation is for offsetted K vector [0:D-1] the Evens move CW and the Odds move CCW. One method used 5 mod calls to solve.

This Challenge is derived from <http://code.google.com/codejam/contest/8274486/dashboard#s=p1 GJam March 2016 Annual I/O for Women Dance Around the Clock>. This is a mix of the small and large data sets.

The GJam story goes that D even dancers are arranged clockwise 1:D. Who is adjacent to dancer K after N pair swaps. The odd move swaps positions 1/2,3/4,..D-1/D. The even move swaps positions D/1, 2/3,...D-2/D-1. D=8 [1:8] becomes [2 1 4 3 6 5 8 7] after the first swap. After second swap we see [7 4 1 6 3 8 5 2].

<<http://code.google.com/codejam/contest/images/?image=dance_around_the_clock.png&p=5733344260128768&c=8274486>>

*Input:* [D K N] , 4<=D<=1E8, 1<=K<=D, 1<=N<=1E8

*Output:* [KCW KCCW] , KCW is dancer to left of Kth dancer after N moves

*Examples:* [m] [v]

[8 3 1] creates v=[6 4]
  [8 4 2] creates v=[1 7]
  [6 6 1] creates v=[5 3]

*<http://code.google.com/codejam Google Code Jam 2016 Open Qualifier: April 8, 2016>*

*Contest Theory:* The small case was D<10 and N<10. This could be done brute force in the 5 minutes of entry submission allowed. However, this is clearly a case of modulo math so might as well solve the unbounded case of D/N <1E8. The number being assigned as 1:D leads to confusion as mod maps to 0:D-1. Suggest offset the K dancer number by 1 for processing and then correct for the final answer. Mod is fun as mod(-1,8) yields a useful 7. Quick observation is for offsetted K vector [0:D-1] the Evens move CW and the Odds move CCW. One method used 5 mod calls to solve.

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Problem 42774. GJam March 2016 IOW: Clock Dance

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Problem 42774. GJam March 2016 IOW: Clock Dance

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