Problem of the Day
A new programming or logic puzzle every Mon-Fri

Comments:

• Anonymous - 10 years, 7 months ago

  Building on the code from yesterday:

  (defn nth-fibonacci [n]
    (nth (repeatedly (new-fibonacci-generator)) n))
  
  (nth-fibonacci 13)  ; => 377
  

  See yesterday's Clojure submission for "new-fibonacci-generator".

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• Nick Krichevsky - 10 years, 7 months ago

  Python. limit=int(raw_input()) num1=0 num2=1 for i in range(limit): num=num1+num2 num2=num1 num1=num print num

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• Nick Krichevsky - 10 years, 7 months ago

  Fixed formatting.

  limit=int(raw_input())
  num1=0
  num2=1
  for i in range(limit):
      num=num1+num2
      num2=num1
      num1=num
      print num
  

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• Erik Jansson - 10 years, 7 months ago

  Here's a memoized version in Haskell:

  memoizedFib :: Int -> Integer
  memoizedFib n = fiblist !! n
  
  fiblist :: [Integer]
  fiblist = [ fib x | x <- [0..] ]
      where 
          fib :: Int -> Integer
          fib 1 = 1
          fib 2 = 1
          fib n = fiblist !! (n - 1) + fiblist !! (n - 2)
  

  Running it for the 10000th number gives: *Main> memoizedFib 10000 33644764876431783266621612005107543310302148460680063906564769974680081442166662368155595513633734025582065332680836159373734790483865268263040892463056431887354544369559827491606602099884183933864652731300088830269235673613135117579297437854413752130520504347701602264758318906527890855154366159582987279682987510631200575428783453215515103870818298969791613127856265033195487140214287532698187962046936097879900350962302291026368131493195275630227837628441540360584402572114334961180023091208287046088923962328835461505776583271252546093591128203925285393434620904245248929403901706233888991085841065183173360437470737908552631764325733993712871937587746897479926305837065742830161637408969178426378624212835258112820516370298089332099905707920064367426202389783111470054074998459250360633560933883831923386783056136435351892133279732908133732642652633989763922723407882928177953580570993691049175470808931841056146322338217465637321248226383092103297701648054726243842374862411453093812206564914032751086643394517512161526545361333111314042436854805106765843493523836959653428071768775328348234345557366719731392746273629108210679280784718035329131176778924659089938635459327894523777674406192240337638674004021330343297496902028328145933418826817683893072003634795623117103101291953169794607632737589253530772552375943788434504067715555779056450443016640119462580972216729758615026968443146952034614932291105970676243268515992834709891284706740862008587135016260312071903172086094081298321581077282076353186624611278245537208532365305775956430072517744315051539600905168603220349163222640885248852433158051534849622434848299380905070483482449327453732624567755879089187190803662058009594743150052402532709746995318770724376825907419939632265984147498193609285223945039707165443156421328157688908058783183404917434556270520223564846495196112460268313970975069382648706613264507665074611512677522748621598642530711298441182622661057163515069260029861704945425047491378115154139941550671256271197133252763631939606902895650288268608362241082050562430701794976171121233066073310059947366875

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• Erik Jansson - 10 years, 7 months ago

  And yes, I'm basically just showing off Haskell's laziness with that output ;)

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• Anonymous - 10 years, 6 months ago

  #python 2.7
  def nth_fib(n):
      phi = (1+5**(0.5))/2
      nth = int(round( (phi)**n / (5**(0.5)) ))
      return nth
  

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