Hello again! This one is one hell of a tricky problem. No wonder it’s called “Numbers”:

In this problem, you have to

find the last three digits before the decimal point for the number (3 + √5).^{n}For example, when

n= 5, (3 + √5)^{5}= 3935.73982… The answer is 935.For

n= 2, (3 + √5)^{2}= 27.4164079… The answer is 027.

Wow, that can’t be hard, eh? Indeed! It’s really just an expression, isn’t it? Of course it is! But once you try, you get to see something beautiful happening:

$ python >>> import math >>> 3+math.sqrt(5) 5.23606797749979 >>> (3+math.sqrt(5)) ** 18 8751751364607.995 >>> (3+math.sqrt(5)) ** 19 45824765067264.016 >>> (3+math.sqrt(5)) ** 20 239941584945152.1

If I was clear enough, you can see where I’m going. If not, the answers for n = 18, 19 and 20 should be 607, 263 and 151. But, it gets worse as the exponent grows:

>>> (3+math.sqrt(5)) ** 30 3.7167066517539914e+21 >>> (3+math.sqrt(5)) ** 40 5.757196418599164e+28 >>> (3+math.sqrt(5)) ** 50 8.917924847980422e+35

And those results should be 647, 551 and 247, respectively. As you can now clearly see, python can’t naturally handle those big floats, what a bummer! Well, the first thing to handle is the √5 value, since 2.23606797749979 does not give enough sampling to work with the needed precision. Looking a bit, I’ve found that python has a nice Decimal module that can deal with big numbers. So, let’s use it:

>>> from decimal import * >>> Decimal(5).sqrt() Decimal('2.236067977499789696409173669')

Hmm… not as much as I was expecting. Digging a little deeper, there’s a way to increase the precision, so things might get better:

>>> from decimal import * >>> getcontext().prec = 256 >>> Decimal(5).sqrt() Decimal('2.236067977499789696409173668731276235440618359611525724270897245410520925637804899414414408378782274969508176150773783504253267724447073863586360121533452708866778173191879165811276645322639856580535761350417533785003423392414064442086432539097252592627229')

Much better indeed! Let’s test the previous cases that way!

>>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(18) Decimal('8751751364607.992147917157027451941369199107402191688475662016738729585793189576921000417819488022521008035769458236805623355117964654519012995149203145651482621970904733675616603829159073699152262190237686134096053140643178581342535160175294280596644277408') >>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(19) Decimal('45824765067263.99400154247292878012837438073861701581979206241047287979476762412992360313819778117994498296523948378426713929520188369997784294269662668000818303924272345525577601783001726960434941938734208353272624615875651599971618368131896320524727780674') >>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(20) Decimal('239941584945151.9954175862094628730047694880020933281648497263958823604254329864718576171579087349895858656483590697583803423507394435817910056755829474974431677475727217968321896916634673228294874675631017566599732643899663816729269614472126021090970897308') >>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(30) Decimal('3716706651753989275647.999689800241818180639346264792151927705378585007996556972979408800792479361060761238505878803342888805078901553987993893381697765526226152915257781651542284398399012963211473811148141385735168384084704178517042635758936907558116236798') >>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(40) Decimal('57571964185991590695825047551.99997900148385229515959986036977505867324065830938072692611318337057286921280055552228787014811837140299777965403896929514221392361243371590607709492852574169582705605198217499438837728596606637811005617061688138760057438182845') >>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(50) Decimal('891792484798041273405092327480885247.9999985785363502863489704179083057190576105613126358868072802757567654991706904766380400266097425885364586125135675324533868405035416842283484903727056986268545269819514444772213000354546446009165968589958054059389753136')

Whoa, now that’s improvement! Since the small input ranges from 2 to 30, we’re covered, yay! To the large input! The first case is 910062006… much bigger! But we have faith in our solution!

>>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(910062006) Decimal('8.919448541446674973277872746171245933275138630248130907008894180215290388336023868239234075453030387703632363584380715738200630546564329205017138690712399728915139680036686573669248915790961561461272596637583821147468503082573913305207639264660422490421272E+654339383')

Aaaaaand it fails miserably. You say increase the precision? Go!

>>> getcontext().prec = 1024 >>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(910062006) Decimal('8.919448541446674973277872746171245933275138630248130907008894180215290388336023868239234075453030387703632363584380715738200630546564329205017138690712399728915139680036686573669248915790961561461272596637583821147468503082573913305207639264660422298653727910480913903028526252731111441440780088518055499910461897770689934875196042744680956819167861668586410285893670912335938531514972687567521690574174307478919726903450127130115796805326307689705722233029462717164687160362204672599178459426195515468743692514680427999432359495315725269307474956180961814633487158821252498531057196075527072557513060268997873150760403767129373225601570326857102899047033394123502152602095967110573983281744795256688834356089540911527223279951661861220253786982201657061127666645472375405207936633238083128248926879097462312848916323276895759197636621186062828933585137465868233553517455227951169710940268113754834340838971865855889844812381822071748466341030321954673834742153256457611174551211610695251648386140804719763671673855871318598E+654339383') >>> getcontext().prec = 1000000000 >>> (Decimal(3) + Decimal(5).sqrt()) ** Decimal(910062006)

It is still computing that last command… maybe it’ll finish before Christmas! 😀 So… nothing we can do about it. Not in a natural way at least. It turns out that Code Jam organizers knew about this (that’s the whole point, indeed) and my approach, even if it’s able to solve the small input, is not the correct one. From the contest analysis:

The key in solving the problem is a mathematical concept called conjugation. In our problem, we simply note that (3 – √5) is a nice conjugate for (3 + √5). Let us define

(1) α := 3 + √5, β := 3 – √5, and X

_{n}:= α^{n}+ β^{n}.We first note that X

_{n}is an integer. This can be proved by using the binomial expansion. If you write everything down you’ll notice that the irrational terms of the sums cancel each other out.(2)

Another observation is that β

^{n}< 1, so X_{n}is actually the first integer greater than α^{n}. Thus we may just focus on computing the last three digits of X.

Whoa, there’s a ton of math symbols there. What it is trying to explain is that there’s another way around, and we can write a simple algorithm to calculate only what we need. Taking that math apart, this suggested approach is quite interesting:

Solution C. [the periodicity of 3 digits]

For this problem, we have another approach based on the recurrence (7). Notice that we only need to focus on the last 3 digits of X

_{n}, which only depends on the last 3 digits of the previous two terms. The numbers eventually become periodic as soon as we have (X_{i}, X_{i+1}) and (X_{j}, X_{j+1}) with the same last 3 digits, where i < j. It is clear that we will enter a cycle no later than 10^{6}steps. In fact, for this problem, you can write some code and find out that the cycle has the size 100 and starts at the 3rd element in the sequence. So to solve the problem we can just brute force the results for the first 103 numbers and if n is bigger than 103 return the result computed for the number (n – 3) % 100 + 3.

Hmm… so that means that 103 restarts the cycle?

>>> getcontext().prec = 1024 >>> str((Decimal(3) + Decimal(5).sqrt()) ** Decimal(3)) '143.5541752799932702850935573994008395340997875075688231766687118531366696204097567812612610681210327990242616368247610721361045671823063636347635238890704866837369015421401333059608526503244754105771443632133610811201095485572500621467658412511120829640713240415845568781018115768290851039517558845033513247514268365293660766111722141944646319577243016028721322545978616614879992775936489738916811506453866403999620857004610560179974673677933515377954018029702474909866234538867118832726784429359876544843489516123361404699859563411214642284786110581426213620665031024800019372553994881602971764760169511058165201173143835473121683945817118282695117541643928648522370457633421861064234721885791403592429823579671023036790457862064776462707851047085181514532031422824259553334558658255398529105918381922193776593783203964365329603283979620022727823878012615898529089431909156945831424556056508947493929464193144955007377329721849300605964711369201940159834308432696751549585124838734830577613026312919174891129200931587878779' >>> str((Decimal(3) + Decimal(5).sqrt()) ** Decimal(103)) '114167750723954075118506909152243184891636055812752354180398291472449798143.9999999999990991848908257663970808178511998207346190405521007125450976083864218743136022578058573724307353453054670910157005516822203809922099339999229730921853127935894210481666862007195197029531387123303773842565764868785959347563566969625431650825652899007545905552208934231424236695732218626707231897771500030743392334437529104775115629703470534218197873009424685334262220812905849999596760216536020750239094398005209138812883762761775627540601688798157222970582315547073108170909501918275307282209081948400177828835529550843580664303363677674984334542177398232468766365174239306558484619915875621147584977655598495728316487075831729616962367129848658475956289730636910101894486413362652617806007673400475350422005999251984236600004538944250397180445073308202368906957039259759659759171424708223454172257556499138474886346739987815229848744135302653447948550170944394443810710713254841099059181243617896686076589736820366175425243225873342031095' >>> str((Decimal(3) + Decimal(5).sqrt()) ** Decimal(4)) '751.6594202199646689967411763468544075540238844147363216775107372289675155071512231016216206076354221948773735933299956287145489777071084090825085004176200550896187330962356998562944764142034959055300079068701456758805751299255628262705206665683384355613744512183189236100345107783526967957467183936425944549449908917791719022086541245209393177780525834150786943366387737228119962073666571129313260408882798620998009499274205440944867036809150955734258594655937993276797731329052373871815618254139351860428319959647647374674262707908876871995127080552487621508491412880200101705908473128415601764990889933055367306159005136233888840715539870984149367093630625404742444902575464770587232289900404868860256573793272870943149903775840076429216217997197202951293164969827362655006432955840842277806071505091517327117361820812917980417240893005119321075359566233467277719517523073965614978919296671974343129687014011013788730981039708828181314734688310185839130119271657945635321905403357860532468388142825668178428304890836363588' >>> str((Decimal(3) + Decimal(5).sqrt()) ** Decimal(104)) '597790103628874365020709731601266597564608080186061135989016966589686218751.9999999999993118384917497799741823573220669410860125985943311312283719331038685561426361821548229163694601416604360587022213664498323168042300721157774333881227779113388111518769679330555901306111571209984513448987672187738925735173356845673100783866660482880393505457702913561140292994651458434392515606071335149084501581993111115430324514143366418268949755088875955878162140123915031739484539675175716477662141640774955323553027076342238845088259277403081453735076960587935422087441774083950451065986208934485511159638170169172967026214058721899896878428807562496126156550849295831738165499903388396019937912975536800221981964584521002823218293972856046263138381804404409768650023404607326059948575617153491987096854714709481244777860640664620489694413768067522712403081007060606347733537888589375255218262079999570621589681646821848243321335219275048192774559691932753194312696801857829138795089010920052048921614192197989676705158844531382420174'

Yay! To the git-pushed solution!

from decimal import * # Improved math getcontext().prec = 1024 # Adjust the precision base = (Decimal(3) + Decimal(5).sqrt()) # Calculate the base for case in range(1,int(raw_input())+1): # For all test cases exp = (int(raw_input()) - 3) % 100 + 3 # Get the recurrent exponent num = str(base ** Decimal(exp)) # Calculate ans = num[:num.find('.')] # Get only the integer part print "Case #%d: %03d" % (case, int(ans[-3:])) # Zero-lead the last 3 digits

One thing that’s really interesting about this is that I actually solved this (of course, not from scratch) while posting! I was already OK with the idea of only solving the small input, but reading the analysis brought some new light into this problem. Blogging FTW! See ya!