Problem 44077. GJam 2017 Kickstart: Vote (Large)

Created by Richard Zapor

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This Challenge is derived from <a href = "http://code.google.com/codejam/contest/6304486/dashboard#s=p1">GJam 2017 Kickstart Vote</a>. This is the first 100 large cases with 0&lt;=M&lt;N&lt;=2000.<a href = "http://code.google.com/codejam">Google Code Jam 2017 Qualifier</a> is March 7, 2017. Typically four challenges with small and large aspects with 27 hours to complete.The GJam story is given N and M votes, where N&gt;M, determine the number of voting sequences for which N is always in the lead divided by the total number of sequences(N+M)!.Input: [N,M], the quantity of votes to N and MOutput: [V], the ratio of N always leading sequences to total sequencesExamples: [N,M] [V]; [2,1] [0.33333333]For the case [2,1] there are 6 sequences [N1N2M1,N2N1M1,N1M1N2,N2M1N1,M1N1N2,M1N2N1] with only the first two always having N in the lead.Theory: Brute force permutations and counting will not succeed in a timely manner for 0&lt;=M&lt;N&lt;=2000 as 2000! may be a bit large. Determining the inherent mathematical pattern is usually the best way to succeed in GJam. <a href = "http://code.google.com/codejam/contest/6304486/scoreboard#vf=1">GJam Kickstart solutions(C++,Python)</a>.

This Challenge is derived from <http://code.google.com/codejam/contest/6304486/dashboard#s=p1 GJam 2017 Kickstart Vote>. This is the first 100 large cases with 0<=M<N<=2000.

<http://code.google.com/codejam Google Code Jam 2017 Qualifier> is March 7, 2017. Typically four challenges with small and large aspects with 27 hours to complete.

The GJam story is given N and M votes, where N>M, determine the number of voting sequences for which N is always in the lead divided by the total number of sequences(N+M)!.

*Input:* [N,M], the quantity of votes to N and M

*Output:* [V], the ratio of N always leading sequences to total sequences

*Examples:* [N,M] [V]; [2,1] [0.33333333]

For the case [2,1] there are 6 sequences [N1N2M1,N2N1M1,N1M1N2,N2M1N1,M1N1N2,M1N2N1] with only the first two always having N in the lead.

*Theory:* Brute force permutations and counting will not succeed in a timely manner for 0<=M<N<=2000 as 2000! may be a bit large. Determining the inherent mathematical pattern is usually the best way to succeed in GJam. <http://code.google.com/codejam/contest/6304486/scoreboard#vf=1 GJam Kickstart solutions(C++,Python)>.

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Last Solution submitted on Feb 08, 2023

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Rafael S.T. Vieira on 10 Oct 2020

Tip: This kind of problem that seems incredibly hard usually have a formula (someone already researched this). Find the formula, and solve the problem.

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