Discussion @ 1462. Uncle Scrooge's Gold @ Timus Online Judge

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2 ADMINS: ban the following user(s) for multiple accounts and flooding testing queue	Vedernikoff 'Goryinyich' Sergey (HSE: АОП)	1462. Uncle Scrooge's Gold	18 Oct 2015 14:20	2
2 ADMINS: ban the following user(s) for multiple accounts and flooding testing queue Vedernikoff 'Goryinyich' Sergey (HSE: АОП) 17 Oct 2015 21:56 Kolega FrigATE Mister IPRIT Jogn Doe Jogn Doe2 Jogn Doe3 John Doe 4 John Doe 5 Re: 2 ADMINS: ban the following user(s) for multiple accounts and flooding testing queue Vladimir Yakovlev (USU) 18 Oct 2015 14:20 Flooding submissions were cleaned out Edited by author 18.10.2015 14:20
0.046s. Who's better? (-)	Krayev Alexey (PSU)	1462. Uncle Scrooge's Gold	2 Dec 2010 12:17	9
0.046s. Who's better? (-) Krayev Alexey (PSU) 19 Aug 2006 20:30 Re: 0.046s. Who's better? (-) xurshid_n 21 Oct 2006 18:18 Please, send me your solution or algorithm!!! my email: xujand000@rambler.ru Did you use FFT? (-) Vladimir Yakovlev (USU) 23 Oct 2006 15:25 Edited by author 23.10.2006 19:00 Re: Yes (-) Krayev Alexey (PSU) 23 Oct 2006 17:45 What is FFT? (-) xurshid_n 26 Oct 2006 18:40 Re: What is FFT? (-) Todor Tsonkov 3 Nov 2006 02:41 I'm also interested in a c++ solution of this problem, please send to tabledott@gmail.com Re: What is FFT? (-) Vedernikoff Sergey 16 Nov 2006 15:30 Simple problem... FFT can be used to solve it, but for what purpose??? Re: 0.046s. Who's better? (-) [SPbSU ITMO] WiNGeR 28 Nov 2006 09:03 0.031 ;) Re: 0.046s. Who's better? (-) vindex 2 Dec 2010 12:17 [url=http://www.stratomedia.co.il]בניית אתרים[/url] [url=http://www.vindex.co.il]בעלי מקצוע[/url] [url=http://moverexpert.com]moving[/url] [url=http://www.vindex.co.il/%D7%94%D7%95%D7%91%D7%9C%D7%AA_%D7%93%D7%99%D7%A8%D7%95%D7%AA_%D7%94%D7%95%D7%91%D7%9C%D7%95%D7%AA_%D7%9E%D7%A9%D7%A8%D7%93%D7%99%D7%9D_%D7%94%D7%A2%D7%91%D7%A8%D7%AA_%D7%93%D7%99%D7%A8%D7%94/ ]הובלות[/url]
How to improve my Algo. I got TL on test 4	Todor Tsonkov	1462. Uncle Scrooge's Gold	18 Mar 2009 15:10	22
How to improve my Algo. I got TL on test 4 Todor Tsonkov 12 Aug 2006 19:35 I try to find fib(2n-2)/fib(n-2) but I get TL on test 4, my algo is linear so I couldn't pass the TL, I wonder is the direct algo for finding fib(n-2) and fib(2n-2) better, because I couldn't implement it :) Re: How to improve my Algo. I got TL on test 4 VALERO 12 Aug 2006 19:40 i have tl too, but i find that fib(2n-2)/fib(n-2)=fib(n+1)+fib(n-1) ^) but i have tl ( Re: How to improve my Algo. I got TL on test 4 Mafia {Toropov, Isenbaev, Oparin} SPb IFMO* 12 Aug 2006 19:48 fib(n) could be evaluated in O(log n) multiplication operations. Re: How to improve my Algo. I got TL on test 4 Todor Tsonkov 12 Aug 2006 19:55 Well, my idea was to evaluate fib(n-2) and fib(2n-2) using the direct formula for fib(n)- as function of n, but I couldn't implement it , while the bruteforce solution was TL, anyway I'm gonna think a bit more about it, btw congrats for the victory, guys, you did a great job, also a great competiton for ACRush as always :) Re: How to improve my Algo. I got TL on test 4 UdH-WiNGeR* 12 Aug 2006 20:20 Hint: direct formula is hard to implement, but we can rewrite it using linear algebra like this: .....................................(0, 1)^(n-1) (fib(n), fib(n+1))=(1, 1)(....) .....................................(1, 1) , so we can use integer arithmetics! Edited by author 12.08.2006 21:05* Edited by author 12.08.2006 21:05 Edited by author 12.08.2006 21:05 Edited by author 12.08.2006 21:06 Edited by author 12.08.2006 21:06 Edited by author 12.08.2006 21:06 Re: How to improve my Algo. I got TL on test 4 VALERO 13 Aug 2006 00:18 Who can you write me om email the algorithm or in what book i can find it? Thanks. adas@simnet.kiev.ua Re: How to improve my Algo. I got TL on test 4 UdH-WiNGeR 13 Aug 2006 14:37 Here is an explanation of the algorithm in Russian: http://algolist.manual.ru/maths/count_fast/fibonacci.php Edited by author 13.08.2006 14:37 Test data Anton [SUrSU] 13 Aug 2006 18:32 What is the answer fo this test? n = 7000 Re: Test data UdH-WiNGeR 13 Aug 2006 20:23 81937319032993876558582070063011262115783024656965723338526424275071630557443673768160691777974344039339079218543821663526642065353277478431909789983160736348837929626796434540999450923786205450782143686270382692873692233405139629109704561270312322297501776980155782647708033073460077811847296059544586515736242655594916292010288630241998587835886957464588315061688674744586049802510938100726026534538639883653835446753146728656928193874140316597540202353213455487305219113003826047850156459479457447951648423579548704045234844929446845423587933723079796158760129023524040519672308404718221125336963212128205511547407802595672777585867901520427711491099757592076124797583476061864630025076181920889182653070760272884159824308542397653852400985696907858408387639956157978706489167805462854368911415071677899319871467455602850964575780531369593561207038707991166116257391616617990142018311285809472372949541001826651548109149596409807197380997932230498982610632747317000197354547509949735065278160290108373521163485352357077038417341404680343430978229222267611931184569548788576280109768575772879672619415599295926884517066063939649588253695834282602632479823971486742422068816044196848224503022646081677385555572595399717609483383203051148447238015528377198350218937442030369055848713320280710524550172256486540262008219704955151944469303730096667265538925353091481286313382299155890271045282453432557272252772046637042581378078908572780035752442042046549675078127 Re: Test data Anton [SUrSU] 13 Aug 2006 23:11 I know how to calc Fn in O(log n). But also i must use long arithmetic and TL in result. Give some hints please Re: Test data Sid 13 Aug 2006 23:25 Just use java.math.BigInteger in Java.=) Re: Test data Todor Tsonkov 14 Aug 2006 01:07 Yup, I used BigInteger but O(n) gave me TL, so does there exist a solution which doesn't use advanced maths techniques such as the described above, because I tried to use BigDecimal but it loses precision when trying to calculate directly Fib(n) Re: Test data Vasilevskiy Boris 14 Aug 2006 14:35 It`s improbable task! While I use quick power raising of matrix, I still got tl Then I tried to calculate it lineary and got AC, c++ Re: Test data Petr 31 Aug 2006 17:45 Hi Boris :) Here my sol: There is sequence a(i)=a(i-1)+a(i-2), a(1)=1,a(2)=3. Ans for this problem is a(n). I tried to calculate it O(n) and it taked ~ 17 sec in n=40000. Then I tried to calculate it o(logN) and it,of course, slower then O(n). Is there any special quick algo for long numbers?(sum or product) Edited by author 31.08.2006 17:50 Re: Test data Kaliningrad SU -J_A_MES-HeadLiner 31 Aug 2006 20:45 Of course there are fast algorithms for long multiplication. Try to find FFT or FHT. Your formula is incorrect! SPIRiT 14 Sep 2006 18:48 The actual answer is fib(2n-1)/fib(n-1). Just check the sample answer.... Re: Your formula is incorrect! Todor Tsonkov* 14 Sep 2006 23:59 Yup, but if you use some math knowledge you will find that the result is fib(n-1)+fib(n+1), during the competition as far as I got was the result you mentioned, but I think implementing it is improbable task, anyway can someone send accepted C++ solution to me at ttsonkov [at] gmail.com, I'd be very grateful, cos I implemented the algo with the matrix but I still get TL. Re: Your formula is incorrect! Todor Tsonkov 15 Sep 2006 18:21 Anyway, Java rulez!!! O(logn) got accepted in about 0.7s. ,yet it would be interesting to see some C++ solutions Re: Your formula is incorrect! SPIRiT 15 Sep 2006 19:50 What do you mean result is fib(n-1)+fib(n+1)? For sample (n=3) fib(2)+fib(4)=2+5=7, but real answer is 4. Check that yourself Re: Your formula is incorrect! Todor Tsonkov 15 Sep 2006 23:26 @spirit: It depends on how you get the Fibbonacci sequence if fib(1)=1, fib(2)=1, fib(3)=2 and etc. you get the desired result :) Oh I get it SPIRiT 18 Sep 2006 18:15 Well, then my formula is fib(n-2)+fib(n) No subject 3a[3.141592..]Jlu 18 Mar 2009 15:10 I use linear algo O(N) + long addition (10^18), but I got TL#4 :( Please help me.
How to more optimize~Code here...Thank you!!!	Search	1462. Uncle Scrooge's Gold	18 Mar 2009 14:59	2
How to more optimize~Code here...Thank you!!! Search 25 Sep 2008 01:10 #include <cstdio> #include <string> #include <memory> using namespace std; class BigInteger{ private: enum{MAXLENGTH=14900}; int m[MAXLENGTH]; int len; public: BigInteger(int n) { memset(m,0,sizeof(m)); len=0; if (n == 0) len++; while (n > 0) { m[len++] = n % 10; n = n / 10; } } BigInteger(void) { memset(m,0,sizeof(m)); len=1; } void print(void) { int temp = len-1; while(temp>=0) printf("%d",m[temp--]); } BigInteger operator+ (const BigInteger &a) { int i,carry = 0; BigInteger Temp(this); if (a.len > Temp.len) Temp.len = a.len; for(i=0;i<Temp.len;i++) { Temp.m[i] += (a.m[i] + carry); carry = Temp.m[i] / 10; Temp.m[i] %= 10; } if (carry > 0) {Temp.m[i] = carry; Temp.len++;} return Temp; } int operator> (const BigInteger &a) { int i; if (len > a.len) return 1; if (len < a.len) return 0; for(i=len-1;(m[i] == a.m[i]) && (i>0);i--); if (m[i] > a.m[i]) return 1; return 0; } int operator== (const BigInteger &a) { int i; if (len != a.len) return 0; for(i = len-1;i>=0;i--) if (m[i] != a.m[i]) return 0; return 1; } }; int main() { int n; scanf("%d",&n); BigInteger r1;int i; BigInteger previous = -1; BigInteger result = 1; for (i = 0; i<n; i++) { BigInteger const sum = result + previous; previous = result; result = sum; if(i==n-3){ r1=result; } } (r1+result).print(); return 0; } Edited by author 25.09.2008 01:10* Re: How to more optimize~Code here...Thank you!!! 3a[3.141592..]Jlu 18 Mar 2009 14:59 Use long ariphmetics on base 10^9 or 10^18, but not 10^1 :)
How to make it fast? Is my formula correct?	Alexey	1462. Uncle Scrooge's Gold	12 Aug 2008 05:33	3
How to make it fast? Is my formula correct? Alexey 23 Aug 2006 19:59 I use Fib(n-1)+Fib(n+1). I can find Fib(n) in O(log n). But the long arithmetics makes my programm too slow. Help, please. Re: How to make it fast? Is my formula correct? Cybernetics Team 30 Aug 2006 16:52 Read below some solutions. Your formula is correct, I used it too and recursevly computed the F[n]. I used long arithmetics in base 1.000.000 and it runs quick enough. Re: How to make it fast? Is my formula correct? Denis Koshman 12 Aug 2008 05:33 Better use long arithmetics base 2^P, and then fetch decimal answer by sequential divisions by 1'000'000 or something like that.
How many digits are there in the output if N=70000? (-)	Alexey	1462. Uncle Scrooge's Gold	23 May 2007 19:36	5
How many digits are there in the output if N=70000? (-) Alexey 21 Aug 2006 17:23 Re: How many digits are there in the output if N=70000? (-) Cybernetics Team 21 Aug 2006 19:17 14630 digits Thanks Alexey 22 Aug 2006 15:56 10x Re: How many digits are there in the output if N=70000? (-) Abid 7 Feb 2007 12:49 2072801145996145261520530447408850863428513339377208014386060985843763728990950560338251079604581881 2761764843963097882756899306880632339149624457792521065549662450746982954629516070098764978344151183 5995330030762779087743459391817243909019805275976633115556130331941538448665875113367934989079027834 0569811790271945906685562735304733743410753082978063360291190842638975525282371376255146251390737707 7479794770248229483843646633681833549215123470585482715472809087383941758904346522640847918233307726 9328866108345114427090779695990005117224442643471755382945481853638762026546985115627193770965426310 5394352702808301103424957405054432898553516895529167136603624447915874323780327952040118805325278847 0288847800035137267512317663342926011439333402801452136904199295820198703432837127533865033077917441 0100898022841492469103704073386054730663565822108884580528522055691701523568506284269126884159083365 2981898449962724562358921065055713479149834483505479477562321118755467916998143411260891457484332466 8332643775010924705690019102274575145613483922681596385654759233269444264241619526680808134173728034 9475872783233388450598439414037223575558750012302913355794850648784308559343577303219076693667107597 3043115850209464408267038946442563894220187739318055364751537731743362821609888945133271851672020727 6058883659104584529812229076091130643430114436305262385420314059450210042195431947096493318081320875 Re: How many digits are there in the output if N=70000? (-) {AESC USU} Evgeny Kurpilyanskij 23 May 2007 19:36 You are wrong. My AC program write answer with length equal 14630 1364009824826097138409874727232071592670236676434661436958625067025953048160195316590256042336050799 9240285236943998597755517864435051077996983326191535648244715055632646494475836054016954249495840368 9429773910248895618431459625704116289808535035391530034504604812315114781580074061315294350832017537 0075316588558621037290575768717973347211695376764733809184842518679512635211464818187929679448162171 4120419822770736842403909341535418889602295980606749453924444722749831991886615203317279082083644209 5859240299070567565865919703493001702802472903786299155673454789689666667376454899780585080004575093 5051125493829019415429695413340276560732927987519009397229769042454790699842227702861077089955276947 9789770809912699679036724558045302233243433525289535479438177687342796156957246446368380611154909281 6665987538395900320557483881627165663456662023298993638270329837346030637435609378930404319693647318 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WHY WA at test 4	SPIRiT	1462. Uncle Scrooge's Gold	25 Feb 2007 17:04	4
WHY WA at test 4 SPIRiT 19 Sep 2006 19:02 I use this formula: f(N)+f(n-2) where F(0)=1, F(1)=1, F(N)=F(N-1)+F(N-2). But wa. What is in the test 4. I checked the answer for N=7000 at the forum and it matches the answer that my program gives. Can anybody give me a test for 4. This problem does not need log(N) methods SPIRiT 25 Sep 2006 12:39 We can state that f(21)=4181f(1)+6765f(2), f(22)=6765f(1)+10946f(2). Wherefore we can calculate f(i),f(i+1) with N/20 operations. All we need is addition of two long numbers (I used base 10000 for them) and multiplication of long number and simple one, that's all. The usage of recursion formula f(2n)=f^2(n)+f^(n+1) (or something like that) is overridden here by the cost of multiplication of two long numbers. Re: This problem does not need log(N) methods Todor Tsonkov 25 Sep 2006 14:07 @Spirit: If you want to I can give you my accepted Java source, please give me your mail, but I'm interested in any accepted C++ solutions, please send to tabledott@gmail.com, cause all my C++ efforts were unsucessful, btw I used base 10^9 and I was losing accuracy :) Re: This problem does not need log(N) methods Olly 25 Feb 2007 17:04 I'll send you my C++ solution. I made it in O(n) with only + in long arithmetics, but it's well optimized. If anybode else is interested in my solution mail to me alutsenko[at]bk[dot]ru
Recursion	Cybernetics Team	1462. Uncle Scrooge's Gold	18 Sep 2006 18:33	3
Recursion Cybernetics Team 21 Aug 2006 19:38 Today I solved this problem using some nice (and easy to find) formulas for computing F[n]: F[2n] = F[n](F[n-1] + F[n+1]) and F[2n-1] = F^2[n-1] + F^2[n] with these you cand recursevly compute F[n] fast enough (~0.25 sec). I also precalculated a table with first 2000 Fibonacci and when the recursion dropped below 2000 it returned the computed value. I see some nice times on the best AC list. Will you please tell us how you did it? What formulas, optimizations... Thank you Re: Recursion Kaliningrad SU -J_A_MES-HeadLiner* 22 Aug 2006 01:32 Where can I find fast(nlog(n))long multiplication in acceptable form? Re: Recursion SPIRiT* 18 Sep 2006 18:33 From what index do you start? F(0)=1 and F(1)=1 or not? Cause I have such formulas: F(2n+1)=F(N)*(F(n-1)+F(n+1)); F(2n)=F^2(N)+F^(N-1)

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Discussion of Problem 1462. Uncle Scrooge's Gold