Problem 1732. GJam: 2013 Rd1a Bullseye Painting - Large Numbers

Created by Richard Zapor

Solve Later

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<a href = "http://code.google.com/codejam/contests.html">Google Code Jam</a> 2013 Round 1a Bullseye challenge is to determine how many full rings can be painted given an initial radius and an amount of paint.Given a radius r, central white zone, create black rings of width 1 cm separated by 1 cm given P ml of paint that covers pi sq-cm per ml of paint provided.The test cases are a few from the 6000 in the Large number set. In contest mode the player has 8 minutes to submit a response after downloading the data set.<img src = "http://code.google.com/codejam/contest/images/?image=bullseye.png&p=2464487&c=2418487">Input: [r, p] uint64, 1&lt;=r&lt;=1E18, 1&lt;=P&lt;=2E18. Always enough P for one ringOutput: RingsExamples:<pre class="language-matlab">[1 9] 1; % Normal number examples to understand concept
[1 10] 2;
[3 40] 3;
</pre><pre class="language-matlab">[1 1000000000000000000] 707106780 for Bullseye Large Number Challenge
[1 2000000000000000000] 999999999
[760961177192651897 1521922354385303795] 1
</pre>Google Code Jam:The next competition starts in April 2014. See details from above link.Solutions to the various past Challenges using Matlab can be found via <a href = "http://www.go-hero.net/jam/13/solutions">GJam Solutions</a>.Large number solutions require more elegant methods to avoid time-outs.Related Challenges:1) <a href = "http://www.mathworks.com/matlabcentral/cody/problems/1731-text-file-read-64-bit-data-set">Reading 64 bit input file</a>2) <a href = "http://www.mathworks.com/matlabcentral/cody/problems/1730-gjam-2013-rd1a-bullseye-painting">Bullseye Regular Numbers</a>Usage of regexp is verboten

<http://code.google.com/codejam/contests.html Google Code Jam> 2013 Round 1a Bullseye challenge is to determine how many full rings can be painted given an initial radius and an amount of paint.

Given a radius r, central white zone, create black rings of width 1 cm separated by 1 cm given P ml of paint that covers pi sq-cm per ml of paint provided.

The test cases are a few from the 6000 in the Large number set. In contest mode the player has 8 minutes to submit a response after downloading the data set.

<<http://code.google.com/codejam/contest/images/?image=bullseye.png&p=2464487&c=2418487>>

*Input:* [r, p] uint64, 1<=r<=1E18, 1<=P<=2E18. Always enough P for one ring

*Output:* Rings

*Examples:*

[1 9] 1; % Normal number examples to understand concept
  [1 10] 2;
  [3 40] 3;
  
  [1 1000000000000000000] 707106780 for Bullseye Large Number Challenge
  [1 2000000000000000000] 999999999
  [760961177192651897 1521922354385303795] 1

*Google Code Jam:*

The next competition starts in April 2014. See details from above link.

Solutions to the various past Challenges using Matlab can be found via <http://www.go-hero.net/jam/13/solutions GJam Solutions>.

Large number solutions require more elegant methods to avoid time-outs.

*Related Challenges:*

1) <http://www.mathworks.com/matlabcentral/cody/problems/1731-text-file-read-64-bit-data-set Reading 64 bit input file>

2) <http://www.mathworks.com/matlabcentral/cody/problems/1730-gjam-2013-rd1a-bullseye-painting Bullseye Regular Numbers>

*Usage of regexp is verboten*

Solve

Solution Stats

25 Solutions
8 Solvers

Last Solution submitted on May 28, 2021

Last 200 Solutions

Problem Comments

2 Comments

2 Comments

Tim on 22 Jul 2013

Does r=uint64(308436464205151562) make sense? It seems like the argument to uint64 is a double precision number, so you have already lost the precision you are trying to gain by using uint64.

Richard Zapor on 22 Jul 2013

r=uint64(308436464205151562) outputs for uint64(r) 308436464205151562. However, r=308436464205151562 outputs for uint64(r) 308436464205151552. A very small loss in precision, but still a loss. The Challenge set has some numbers that use all the paint if full precision math was used. If residual paint was required as an output then full precision is needed. For solutions using p=p-2*r-1 the errors in r and p may cause issues.

Solution Comments

Solution 286810

2 Comments

2 Comments

Richard Zapor on 22 Jul 2013

Very novel input parameter adjustment to allow direct solution.

Rafael S.T. Vieira on 21 Sep 2020

That's Muller's method which is more computationally realiable than Bhaskaracharya's.

Solution 286604

1 Comment

1 Comment

Richard Zapor on 22 Jul 2013

Nice bisecting search method and anonymous function for clean appearance.

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