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

Solution 286604

Submitted on 22 Jul 2013 by Tobias Schäfer

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Richard Zapor on 22 Jul 2013

Nice bisecting search method and anonymous function for clean appearance.

Test	Status	Code Input and Output
1	Pass	%% tic r=uint64(308436464205151562); pmax=uint64(1850618785230909388); rings=solve_rings(r,pmax); assert(isequal(rings,3),sprintf('rings=%i, Expected 3',rings)) x = @(r0,K)(2K+2r0-1)*K
2	Pass	%% r=uint64(760961177192651897); pmax=uint64(1521922354385303795); rings=solve_rings(r,pmax); assert(isequal(rings,1),sprintf('rings=%i, Expected 1',rings)) x = @(r0,K)(2K+2r0-1)*K
3	Pass	%% r=uint64(1); pmax=uint64(2000000000000000000); rings=solve_rings(r,pmax); assert(isequal(rings,999999999),sprintf('rings=%i, Expected 999999999',rings)) x = @(r0,K)(2K+2r0-1)*K
4	Pass	%% r=uint64(1000); pmax=uint64(2000000000000000000); rings=solve_rings(r,pmax); assert(isequal(rings,999999500),sprintf('rings=%i, Expected 999999500',rings)) x = @(r0,K)(2K+2r0-1)*K
5	Pass	%% r=uint64(3); pmax=uint64(1243678947483597075); rings=solve_rings(r,pmax); assert(isequal(rings,788567988),sprintf('rings=%i, Expected 788567988',rings)) x = @(r0,K)(2K+2r0-1)*K
6	Pass	%% r=uint64(432); pmax=uint64(1800692162794074077); rings=solve_rings(r,pmax); assert(isequal(rings,948865465),sprintf('rings=%i, Expected 948865465',rings)) toc x = @(r0,K)(2K+2r0-1)*K Elapsed time is 0.039589 seconds.