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Discussion of Problem 1009. K-based NumbersPage 3 ayeritsian What is K?? [3] // Problem 1009. K-based Numbers 3 Aug 2011 13:40 Please explane what is K. What it means K=10? ayeritsian Re: What is K?? [2] // Problem 1009. K-based Numbers 3 Aug 2011 13:46 Ok now I get it. Fen9I Re: What is K?? [1] // Problem 1009. K-based Numbers 12 Jul 2013 15:24 Why then did not explain to others? gwire Re: What is K?? // Problem 1009. K-based Numbers 3 Aug 2013 13:35 Bin: K = 2, [0,1] Oct: K = 8, [0,1,2,3,4,5,6,7] (или [0..7]) Dec: K = 10, [0,1,2,3,4,5,6,7,8,9] (или [0..9]) Hex: K = 16, [0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F] (или [0..F]) То есть список цифр из которых может состоять К-ичное число = [0..K-1]. PS: В задаче K не > 10. Значит буквы не участвуют. Siroj Matchanov [TUIT] Please explain in detail [1] // Problem 1009. K-based Numbers 26 Jul 2011 13:12 1010230 is a valid 7-digit number; 1000198 is not a valid number; -------------[WHY??] 0001235 is not a 7-digit number, it is a 4-digit number. so if N=3 K=10: 100 can't be a valid number? (Because it has two continuous zeros ?) but 101 is a valid number, right? Please provide me with more examples... pyh119 Re: Please explain in detail // Problem 1009. K-based Numbers 27 Jul 2011 17:52 We define a number to be valid if its K-based notation doesn’t contain two successive zeros. 1000198 has three successive zeros,so it is invalid. Of course,100 is invalid where 101 is valid. Hrayr wrong answer on test #2 [1] // Problem 1009. K-based Numbers 8 May 2011 17:33 I think l have solved this problem,but i got WA on test #2. Moytrage Re: wrong answer on test #2 // Problem 1009. K-based Numbers 11 Jan 2014 22:45 For my solution replacing integer 32-bit with 64-bit integer helped, and all tests passed. There is a 32-bit integer overflow in test #2. scuwf Which friend can tell me why I always get the "Wrong Answer" and what is the dead test? [2] // Problem 1009. K-based Numbers 16 Nov 2010 20:33 #include<stdio.h> int power(const int a,const int b) { int k=0,num=1; for(k=0;k<b;k++) { num=num*a; } return num; } int foc(int a) { if(a==1) { return 1; } else { return foc(a-1)*a; } } int arr(int a,int b) { if(!b) { return 1; } else { return foc(a+b)/(foc(a)*foc(b)); } } int main(void) { int n=0,k=0,i=0,sum=0; scanf("%d",&n); scanf("%d",&k); for(i=0;n-i>=i;i++) { if(!i) { sum=sum+power(k-1,n); } else if(i==1) { sum=sum+(n-1)*power(k-1,n-1); } else { sum=sum+arr(i,n-2*i)*power(k-1,n-i); } } printf("%d\n",sum); return 0; } test results: N K result 3 10 891 4 10 8829 5 10 87480 6 10 866781 7 10 8588349 8 10 85096170 2 2 2 3 8 441 5 9 50688 I'll appreciate it if you find the dead test and point my bug out,thank you! By the way,I get the WA at test#2. Edited by author 16.11.2010 20:38 Enigma [UB of TUIT] There is my solution:) [1] // Problem 1009. K-based Numbers 17 Nov 2010 17:19 import java.util.Scanner; import java.util.Arrays; public class T_1009 { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int k = sc.nextInt(); int[] m = new int[n+1]; // Dynamic m[1] = k-1; m[2] = k*(k-1); for (int i = 3; i <=n; i++) { m[i] = (m[i-1]+m[i-2])*(k-1); } System.out.print(m[n]); } } bet Re: There is my solution:) // Problem 1009. K-based Numbers 24 Aug 2012 18:16 for k = 2, n = 2 Enigma [UB of TUIT] programm write 3, but right answer 2, because: 10 11 Sania i can't understand why algo is wrong // Problem 1009. K-based Numbers 26 Aug 2010 01:55 f(n) = f(n - 1) * k - f(n - 2) why it's wrong? as i understand, f(n - 2) is how many zero-ending numbers in f(n - 1). am i wrong? Aharon_Smbatyan(YSU) this is a good algorithm. [17] // Problem 1009. K-based Numbers 20 Jun 2010 04:03 n , k 1. if ( n = 0 ) , then answer = ( 1 ). 2. if ( n = 1 ) , then answer = ( k-1 ). 3. else answer(n) = ( k-1 ) * ( answer(n-1,k) + answer(n-2,k) ). Edited by author 20.06.2010 04:04 f309587969 Re: this is a good algorithm. // Problem 1009. K-based Numbers 27 Jul 2010 11:13 是好啊！！！！ archimede_syracuse Re: this is a good algorithm. [8] // Problem 1009. K-based Numbers 15 Aug 2010 12:34 n>=2 !!! Andrew Hoffmann aka SKYDOS [Vladimir SU] Re: this is a good algorithm. [7] // Problem 1009. K-based Numbers 15 Aug 2010 12:49 There is good O(1) memory and O(n) algo (n - number length) info: http://skydos.blogspot.com/2010/07/k.html Artem Khizha [DNU] Re: this is a good algorithm. [6] // Problem 1009. K-based Numbers 16 Aug 2010 03:25 Okay, I'll join to Captains Obvious. You can find an answer with O(1) memory and O(logN) time. Andrew Hoffmann aka SKYDOS [Vladimir SU] Re: this is a good algorithm. [5] // Problem 1009. K-based Numbers 16 Aug 2010 03:32 Can you explain this algo? Really interesting to know better one. Thx. Artem Khizha [DNU] wrote 16 August 2010 03:25 Okay, I'll join to Captains Obvious. You can find an answer with O(1) memory and O(logN) time. Artem Khizha [DNU] Re: this is a good algorithm. [4] // Problem 1009. K-based Numbers 16 Aug 2010 13:32 Sure, you're welcome. Okay, we're already know, that the answer can be found reccurently: F(N) = (K-1)*(F(K-1)+F(N-2)), where F(0) = 1, F(1) = K-1. We see, that the {F(N)} sequence is very similar to Fibonacci's one and can assume some facts. I see two different approaches to solve the problem. 1) First way is to build characteristic equation, solve it and get an exact formula for F(N). I didn't try this way, because I foresaw some problems with precision and float calculations. If you're interested, you can turn to this article as a manual: http://www.intuit.ru/department/algorithms/algocombi/8/2.html 2) I assumed, that method with fast matrix exponentiation will work as well as for Fibonacci's numbers. Just consider a matrix: L(N) = {{F(N+1), F(N)}, {F(N), F(N-1)}}. Let's try to find such R, that L(N)*R = L(N+1). If you write down this information, you'll easily see, that R = {{K-1, 1}, {K-1, 0}}. So all you need is to find L(1)*R.pow(N-2). Hope this information was helpful, I'm not sure it is the best way to describe an approach though. :-) Edited by author 16.08.2010 13:34 Artem Khizha [DNU] Re: this is a good algorithm. // Problem 1009. K-based Numbers 16 Aug 2010 13:39 Actually, I implemented only O(N)-solution before. Today I wrote a code for a fast exponentiation, but (it seems I don't know java well) my time increased in #1013, where BigInteger is needed. Strange. :-) Andrew Hoffmann aka SKYDOS [Vladimir SU] Re: this is a good algorithm. [2] // Problem 1009. K-based Numbers 16 Aug 2010 16:29 Artem Khizha [DNU] Thx for explanations and idea! I got it! Really nice approach with matrixes! dima11221122 Re: this is a good algorithm. [1] // Problem 1009. K-based Numbers 15 Sep 2010 14:10 #include <iostream> using namespace std; unsigned long stepen(unsigned long K,unsigned long N) { unsigned long i, result=1; if (N<0) return 0; else { i=1; for(; i<=N; i++) result*=K; return result; } } int main() { unsigned long N, K, i, result; cin>>N; cin>>K; result=stepen(K, N)-(stepen(K,N-2)+stepen(K, N-1)-1); cout<<result<<endl; //system("Pause"); return 0; } WA #2. Why? Frankie Re: this is a good algorithm. // Problem 1009. K-based Numbers 26 Nov 2011 06:41 >>dima11221122 Firstly, I came up with something like that and got WA2 too. N = (k-1)^n + n(k-1)^(n-1) - that formula looks pretty alright; but it isn't. It assumes there can't be more than one zero in the number, while there obviously can. The actual answer is something like that - N = (k-1)^n + n * (П(i=1 to min(k,n)) of (k-i)^(n-1)) and i think it's anyways easier to solve that like original poster did. Btw, your code is sooooo bad; u might wanna do something about that. Edited by author 26.11.2011 06:42 TLaum Re: this is a good algorithm. // Problem 1009. K-based Numbers 23 Sep 2011 17:49 good job! daminus Re: this is a good algorithm. // Problem 1009. K-based Numbers 1 Mar 2012 12:15 Thanks! Good algo thefourtheye Re: this is a good algorithm. [4] // Problem 1009. K-based Numbers 29 Mar 2012 02:25 Aharon_Smbatyan(YSU) wrote 20 June 2010 04:03 n , k 1. if ( n = 0 ) , then answer = ( 1 ). 2. if ( n = 1 ) , then answer = ( k-1 ). 3. else answer(n) = ( k-1 ) * ( answer(n-1,k) + answer(n-2,k) ). Edited by author 20.06.2010 04:04 I am not sure, how the number of zero digit numbers (of course, without two zeros in it) become 1. Edited by author 29.03.2012 02:38 David Hi [3] // Problem 1009. K-based Numbers 23 May 2012 11:28 Can anybody tell me where can i get some info about this topic(k-based numbers), I really don't get from where cames this recurrent formula? Greetings Amir Aupov [MIPT] Re: Hi [1] // Problem 1009. K-based Numbers 8 Aug 2012 19:02 How can we get valid n-digit number? Trivially, we can append each of (k-1) possible digits (k digits except for 0) to the front of each of (n-1)-digit valid numbers. But we can also form valid number this way: append 0 to the front of (n-2)-digit valid number, and then append (k-1) possible digits, as in the first case. This gives us F(n) = (k-1)*F(n-1)+(k-1)*F(n-2), where F is the function to yield the amount of valid n-digit numbers. I'd recommend you to ensure this yields only the valid numbers, think why can't it produce duplicates, and whether other means to construct valid numbers are possible. Then you follow Artem Khizha's [DNU] recipe and get an AC (= Edited by author 08.08.2012 19:05 Bogatyr Re: Hi // Problem 1009. K-based Numbers 2 Oct 2012 15:31 > Amir Aupov [MIPT] I like your explanation. I approached the construction of N digit numbers from N-1 digit numbers in the typical way of constructing numbers: multiply N-1 digit number by base, and add in the new digits to the units position. This has the benefit of being clear and natural, so we can be sure that we're not missing any numbers. It has the disadvantage that you must track numbers that end in zero separately from those that don't. But this is still quite simple: Z(n): # of valid numbers ending in zero NZ(n): # of valid numbers not ending in zero Z(2) = k - 1 NZ(2) = (k - 1) * k - (k - 1) (total number of valid 2-digit numbers, minus those that end in 0) then, Z(n) = NZ(n - 1) NZ(n) = (Z(n-1) + NZ(n-1)) * (k - 1) Compute Z(n) and NZ(n) iteratively (DP), and then output F(n) at the end F(n) = Z(n) + NZ(n) Bogatyr Re: Hi // Problem 1009. K-based Numbers 2 Oct 2012 14:32 HI David, The problem statement language is confusing. A "k-based number" is a number represented in base "k", that's all. Danli Why my program can not be accepted?? // Problem 1009. K-based Numbers 20 May 2009 11:16 #include<iostream.h> #include<math.h> int ji(int n); int main() { int k,n,m,c,a; int d=0; cin>>n>>k; if((k>=2&&k<=10)&&n>=2&&(n+k)<=18) { m=0; if(n%2==1) while(m<(n+1)/2) { c=ji(n-m); a=ji(m); d=d+int(pow(k-1,n-m-1)*c/a/ji(n-m-m)); m++; } else while(m<=(n+1)/2) { c=ji(n-m); a=ji(m); d=d+int(pow(k-1,n-m-1)*c/a/ji(n-m-m)); m++; } cout<<(k-1)*d<<endl; } return 0; } int ji(int n) { if(n==0) return 1; else return n*ji(n-1); } Imran Yusubov JAVA // Problem 1009. K-based Numbers 12 Apr 2009 14:34 import java.util.Scanner; public class Kb_numb { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub int i,n,k; int[] F=new int[20]; Scanner in=new Scanner(System.in); n=in.nextInt(); k=in.nextInt(); F[0]=k-1; F[1]=k*F[0]; for(i=2;i<n;i++){ F[i]=(F[i-1]+F[i-2])*(k-1); } System.out.print(F[n-1]); } } Gevorg Soghomonyan(YSU) WHY MY ALGORITM IS WRONG? [3] // Problem 1009. K-based Numbers 1 Apr 2009 17:37 #include<iostream.h> int power(int a, int b) { if(b==0) return 1; if(b%2==0) return power(a, b/2) * power(a, b/2); if(b%2!=0) return power(a, b/2) * power(a, b/2) * a; } int main() { int n, k; cin>>n>>k; int a, b; a=power(k, n-1) * (k-1); b=power(2, n-1); cout<<a-b+n<<endl; return 0; } I think that the answer of this problem is (k-1)*k^(n-1)-2^(n-1)+n why it is wrong i don't understand. bzaz Re: WHY MY ALGORITM IS WRONG? [2] // Problem 1009. K-based Numbers 2 Apr 2009 20:18 Because it is not working!!!!!!!!!!!!!!!!!!!!! Ivan (Vologda SPU) Re: WHY MY ALGORITM IS WRONG? [1] // Problem 1009. K-based Numbers 3 Oct 2009 19:41 yes it doesn't! =) Oleg Strekalovsky aka OSt [Vologda SPU] Re: WHY MY ALGORITM IS WRONG? // Problem 1009. K-based Numbers 4 Oct 2009 01:01 Ivan (Vologda SPU) wrote 3 October 2009 19:41 yes it doesn't! =) Don't use alien wrong solution!Mesh BF to precalculate [1] // Problem 1009. K-based Numbers 1 Jul 2008 22:48 I think the easiest way is to precalculate trough Brute Force a table containing the values, hard-code it, and simply give back A[ n ][ k ] :D Plus, the solution will run fast as hell XD guana Re: BF to precalculate // Problem 1009. K-based Numbers 19 Jul 2008 23:00 You can do that with problems that have small values of n and k, but it'll run just as fast with a good algorithm. If you did it that way you have not actually solved the problem. keep_myself The test data is not precise while some wrong programs can AC [1] // Problem 1009. K-based Numbers 9 Apr 2008 07:54 for n = 1, k the answer is : k while many programs give: k-1 but there is no such test data Vukasin Re: The test data is not precise while some wrong programs can AC // Problem 1009. K-based Numbers 30 Jan 2012 18:27 Because N>= 2 :)) Suphanat Chunhapanya what's output for N=5,K=10?? [4] // Problem 1009. K-based Numbers 2 Apr 2008 16:13 and also N=6,K=10 N=7,K=10 N=6,K=2 xwj1234 Re: what's output for N=5,K=10?? [3] // Problem 1009. K-based Numbers 10 Oct 2008 19:07 N=5,K=10 the answer is 87480 N=6,K=10 the answer is 866781 N=7,K=10 the answer is 8588349 N=6,K=2 the answer is 13 my program has been accepted Hrayr Re: what's output for N=5,K=10?? // Problem 1009. K-based Numbers 8 May 2011 17:37 I have solved this problem,but I got WA on test #2.When n=6,k=10 I get answer 866781 and others tests I get right answer.What is the test #2??????????????? xwj1234 wrote 10 October 2008 19:07 N=5,K=10 the answer is 87480 Ahmet Faruk Ozkan OTTOMAN(Devlet - i Âliye)) Re: what's output for N=5,K=10?? [1] // Problem 1009. K-based Numbers 11 Oct 2012 12:16 350975 350974 350973 350971 350970 350967 350966 350965 350959 350958 350957 350955 350954 350943 350942 350941 350939 350938 for n = 6 , k = 2 this values are true more than 13 Michael Re: what's output for N=5,K=10?? // Problem 1009. K-based Numbers 19 Aug 2020 13:21 You are not right because k = 2 => all numerals < 2. Ok? And now I show you this 13 numbers: 111111 111110 111101 111011 110111 101111 111010 110110 110101 101110 101101 101011 101010 sorry for my English Edited by author 19.08.2020 13:21 Edited by author 19.08.2020 13:21 Suphanat Chunhapanya What's output for N=3 , K=10? [2] // Problem 1009. K-based Numbers 2 Apr 2008 14:42 KIRILL(ArcSTUpid coder:) Re: What's output for N=3 , K=10? [1] // Problem 1009. K-based Numbers 2 Apr 2008 14:44 891 Suphanat Chunhapanya Re: What's output for N=3 , K=10? // Problem 1009. K-based Numbers 2 Apr 2008 14:58 ThankZ หลายๆเด้อ Svyatoslav Help please // Problem 1009. K-based Numbers 16 Feb 2008 00:53 my code crashed on 1 test {*****************} type integer=longint; var n,k:integer; obr:string; str:string; otv:int64; procedure p(s:string); var i,l:integer; begin l:=length(s)-1; if l=n then begin inc(otv); writeln(s); exit; end; if l>n then begin exit; end; for i:=1 to k do begin if (s[l]='0')and(obr[i]='0') then continue; if (l=0)and(obr[i]='0') then continue; p(s+obr[i]); end; end; begin assignfile(input,'in.txt'); reset(input); assignfile(output,'out.txt'); rewrite(output); readln(n); readln(k); otv:=0; obr:='0123456789'; str:='_'; p(str); writeln(otv); end. {*******************} Please help! MSDN AC algo - f[i]=(k-1)*(f[i-1]+f[i-2]) // Problem 1009. K-based Numbers 26 Jan 2008 17:04 #include <stdio.h> typedef unsigned long int INT; void main() { INT N,K,i; scanf("%lu%lu",&N,&K); INT *F=new INT[N]; F[0]=K-1; F[1]=K*F[0]; for(i=2;i<N;i++) F[i]=(K-1)*(F[i-1]+F[i-2]); printf("%lu",F[N-1]); } Alexander (201 - P TNU) HELP!!! Why my algo is wrong??? [1] // Problem 1009. K-based Numbers 23 Dec 2007 15:27 help me, i can't find error in my algo. for (n = 3, k = 10) we have 1 case x00, where x can take (k-1) digits (anyway first digit can take k - 1) so answer is 9*10^2 - 9 = 891 - right! for (n = 4, k = 10) we have 2 cases: x00y - y can take k digits - 9*10 xx00 - 9*9 answer - 9*10^3 - 90 - 81 = 8829 - right for(n = 5, k = 10) we have 3 cases x00yy (9*10^2) xx00y (9^2*10) xxx00 (9^3) we have 87561, but right answer is 87480 Edited by author 23.12.2007 15:29 AndreyM Re: HELP!!! Why my algo is wrong??? // Problem 1009. K-based Numbers 28 Sep 2008 00:14 for(n = 5,k = 10) You are wrong Right answer x00yy(9*10^2) xx00y(9^2*10) xyx00(9^2*10) //!!!xYx00 87480 ₦¯ίμ®@£ Are my tests true? Can someone cheeck them and give me true answer? [4] // Problem 1009. K-based Numbers 10 Dec 2007 23:07 Here are my tests not very important but please abswer them too. 2 2> 2 3 2> 3 4 2> 5 5 2> 9 6 2> 17 7 2> 33 16 2> 16385 more important tests for me: 2 10> 90 3 10> 891 4 10> 8829 5 10> 87561 6 10> 869049 7 10> 8631441 8 10> 85782969 Thanks very much. ¥@§@® Δδδ@$☺√ Re: Are my tests true? Can someone cheeck them and give me true answer? [2] // Problem 1009. K-based Numbers 10 Dec 2007 23:38 No. some of them are wrong.I mean tests are ok but answers are not. Here are the true ones: 2 2 > 2 3 2 > 3 4 2 > 5 5 2 > 8 6 2 > 13 7 2 > 21 16 2 > 1597 2 10> 90 3 10> 891 4 10> 8829 5 10 > 87480 6 10 > 866781 7 10 > 8588349 8 10 > 85096170 ₦¯ίμ®@£ Re: Are my tests true? Can someone cheeck them and give me true answer? [1] // Problem 1009. K-based Numbers 10 Dec 2007 23:40 спасибо ₦¯ίμ®@£ Re: Are my tests true? Can someone cheeck them and give me true answer? // Problem 1009. K-based Numbers 11 Dec 2007 10:33 ₦¯ίμ®@£ wrote 10 December 2007 23:40 спасибо то есть, thanks. =) I've got AC. Alias (Alexander Prudaev) Re: Are my tests true? Can someone cheeck them and give me true answer? // Problem 1009. K-based Numbers 10 Dec 2007 23:58 5 10 87480 Ciprian new kind of algo [2] // Problem 1009. K-based Numbers 5 Dec 2007 23:57 I found a new kind of algo for this problem.I wish if someone could tell me if it's good or bad. I got wa on test 2 .Bad algo or bad implementation? !!11.First we find numbers made from n-1 numbers that can begin with 0, but don't have 2 zeros one after the other. We find them like this.for(i=0;i<=n-2;i++) we put the two zeros on i position and now we have 2 numbers one in the left of 2 zeros and one in the right.the number of numbers in right is k^((n-1)-2-i).In left we do !!11. for i-1.(if n<=4)else the number is (k-1)*k^(n-1-2-i)*(k-1)^(i-1) Bobur Re: new kind of algo // Problem 1009. K-based Numbers 6 Dec 2007 18:57 I've too new algo, but I've CRASh(over flowler steck-check) here is my code, function f(n, k : integer) : integer; begin f := 0; if n = 2 then f := 0 else begin if n = 3 then f := k-1 else if n > 3 then f := TRUNC(f(n-1, k)+ (n-2)*exp((n-2)*ln(k-1))); end; end; begin read(n, k); s := TRUNC((k-1)*exp((n-1)*ln(k)) - f(n, k)); write(s); end. Bobur Re: new kind of algo // Problem 1009. K-based Numbers 17 Dec 2007 19:15 first I find how many numbers have 00; for example: k = 6; n = 3 ---- 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 5 n = 4 ---- 0 1 2 3 4 5 5*6+5*5 n = 5 ---- 0 1 2 3 4 5 5+25+25+125+125+125 n = 6 ---- 0 1 2 3 4 5 5+25+25+125+125+125+625+625+625+625 n = 7 ---- 0 1 2 3 4 5 5+25+25+125+125+125+625+625+625+625+3125+3125+3125+3125+3125 f = f(n-1)+(n-2)*exp((n-2)*ln(k-1)); 4 = 5 + 2*5*5 5 = 55 + 3*5*5*5 6 = 330+4*5*5*5*5 ... then k^n all numbers k^(n-1) all numbers till n digits then we've f=k^n-k^(n-1)-(n-2)*(k-1)^(n-2); i can't find what's wrong in this Bobur I can't find wrong!! help me!! // Problem 1009. K-based Numbers 2 Dec 2007 21:08 here is code of program! var n, k : integer; begin read(n, k); n := TRUNC(k*exp((n-1)*ln(k-1))); writeLn(n); end. if k = 10 and n = 3 then we can write that: 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 and 3 number maybe one of them. first number maybe all of them, second 9. third 9 too, Kolesnikov_Aleksey***Barnaul*** WRONG AnSwEr # 2!!!! [1] // Problem 1009. K-based Numbers 18 Sep 2007 09:55 Where is wrong? This is MY SOLVe!!! program MEGA; {$APPTYPE CONSOLE} uses SysUtils; var i,n,k:integer; a:array[1..100] of int64; m:int64; begin read(n,k); if n=2 then begin writeln((k-1)*k); halt(0); end; if n=3 then begin writeln((k-1)*k*k-k+1); halt(0); end; a[3]:=k-1; m:=k-1; for i:=4 to n do begin m:=m*(k-1); a[i]:=a[i-1]+(i-2)*m; end; m:=k-1; for i:=1 to n-1 do m:=m*k; writeln(m-a[n]); end. Moytrage Re: WRONG AnSwEr # 2!!!! // Problem 1009. K-based Numbers 11 Jan 2014 22:43 I had a wrong answer in test #2 due to integer overflow. Replacing 32-bit integers with 64-bit integers helped and all tests passed (only in my solution, I didn't test yours).
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