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UPCOMING CHALLENGES:

CURRENT CHALLENGES:

Palladium 2020

PAST CHALLENGES

Rhodium 2019

Ruthenium 2019

Technetium 2019

Molybdenum 2019

Niobium 2019

Zirconium 2019

Yttrium 2019

Strontium 2019

Rubidium 2018

Arsenicum 2018

Krypton 2018

Bromum 2018

Future Mobility

Grand Challenge

Digital Gold

Selenium 2018

Germanium 2018

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Cutting Complexity

Nickel 2018

Cobaltum 2018

Ferrum 2018

Manganum 2017

Chromium 2017

Vanadium 2016

Titanium 2016

Scandium 2016

Calcium 2015

Kalium 2015

Argon 2015

Chlorum 2014

Sulphur 2014

Phosphorus 2014

Silicium 2014

Aluminium 2014

Magnesium 2014

Natrium 2014

Neon 2014

Fluorum 2014

Oxygenium 2014

Nitrogenium 2013

Carbo 2013

Boron 2013

Beryllium 2013

Lithium 2013

Helium 2013

Hydrogenium 2013

Omega 2013

Psi 2012

Chi 2012

Phi 2012

Upsilon 2012

Tau 2012

Sigma 2012

Rho 2012

Pi 2012

Omicron 2012

Xi 2012

Nu 2011

Mu 2011

Lambda 2011

Kappa 2011

Iota 2011

Theta 2011

Eta 2011

Zeta 2011

Epsilon 2011

Delta 2011

Gamma 2011

Beta 2010

Alpha 2010

Programming language:

The Fibonacci sequence is defined by the following recursive formula:

F(0) = 0

F(1) = 1

F(N) = F(N−1) + F(N−2) for N ≥ 2

Write a function:

class Solution { public int solution(int N, int M); }

that, given two non-negative integers N and M, returns a remainder of F(N^{M}) modulo 10,000,103.

Note: 10,000,103 is a prime number.

For example, given N = 2 and M = 3, the function should return 21, since 2^{3} = 8 and F(8) = 21.

Write an ** efficient** algorithm for the following assumptions:

- N and M are integers within the range [0..10,000,000].

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