i wouldn't suggest using fmod() because it is going to use the value returned by pow() which is double in this case and my intuition says that it is bound to take longer than simply taking the mod (%) of 2 integer numbers as it involves significant conversions. moreover in this particular problem it is more advisable to use 1<<n instead of (long int)pow(2,n). i used pow() because i am simply used to.
On May 17, 4:19 pm, Manish kumar Singh <[email protected]> wrote: > tonka, > > is using this statement a good idea? > if (fmod(k + 1, pow(2, n)) == 0)... > It worked. > > > > > > On Sat, May 15, 2010 at 11:45 PM, tonka <[email protected]> wrote: > > why the heck will you write such a clumsy and lengthy code??? take a > > look at my code: > > #include<iostream.h> > > #include<fstream.h> > > #include<conio.h> > > #include<math.h> > > #include<stdio.h> > > #include<string.h> > > > void main() > > { > > int t,flag=0,n; > > long int k; > > ifstream fi("A-large.in",ios::binary|ios::in); > > ofstream fo("outputlar.out",ios::out); > > fi>>t; > > for(int t_c=0;t_c<t;t_c++) > > { > > fi>>n>>k; > > if(((k+1)%(long int)pow(2,n))==0) > > fo<<"Case #"<<(t_c+1)<<": "<<"ON"<<endl; > > else > > fo<<"Case #"<<(t_c+1)<<": "<<"OFF"<<endl; > > } > > fi.close(); > > fo.close(); > > getch(); > > } > > > This works for the large input also. > > > On May 14, 7:49 pm, Bharath Raghavendran <[email protected]> wrote: > > > I found one mistake : If b < PD, you are returning 0. Why is that? > > > Lets say you have 4 switches, PD = 16. If b = 15, the answer should be > > ON. > > > > Other than that, there are many ways you can improve your code. > > > 1. > > > dont use double. It will remove accuracy in many places. 10^8 will come > > > within 4 bytes (usually int suffices this. else use long int). > > > > 2. > > > To calculate 2^n, you can use the left shift operator. "int PD = 1<<n" > > will > > > put 2^n into PD. > > > > 3. > > > To check if b gives reminder PD -1, you don't need to run a loop. You can > > do > > > "if (b%PD == PD-1)" > > > > On 13 May 2010 10:47, Aamir Khan <[email protected]> wrote: > > > > > #include <cmath> > > > > #include <cstdio> > > > > #include <iostream> > > > > using namespace std; > > > > > #define _CRT_SECURE_NO_WARNINGS > > > > #define For(i,a,b) for (long i(a),_b(b); i <= _b; ++i) > > > > > char buf[1024*1024]; > > > > > bool isEven(long a,long b) > > > > { bool result=0; > > > > if(b%2==0) > > > > result=0; > > > > else > > > > { > > > > double ad=0.00; > > > > for(int j=0;j<=30;j++) > > > > { > > > > if(int(ad)==a) > > > > { break;} > > > > ad=ad+1.00; > > > > } > > > > long PD=long(pow(2.00,ad)); > > > > if(b<PD) > > > > result=0; > > > > > else > > > > { > > > > for(long i=PD;i<=100000000;i+=PD) > > > > { > > > > if(b==i-1) > > > > { > > > > result=1; > > > > break; > > > > } > > > > } > > > > } > > > > } > > > > return result; > > > > } > > > > > int main() { > > > > freopen("A-large.in", "rt", stdin); > > > > freopen("output_Snapper_Chain.txt", "wt", stdout); > > > > > gets(buf); > > > > long size=atoi(buf); > > > > > For(test, 1, size) > > > > { > > > > // cout<<"No. of loops"<<test; > > > > long N; > > > > long K; > > > > > scanf("%d%d", &N, &K); > > > > // cout<<"N"<<N<<"K"<<K; > > > > > bool res=isEven(N,K); > > > > if(res) > > > > printf("Case #%d: ON\n", test); > > > > else > > > > printf("Case #%d: OFF\n", test); > > > > } > > > > exit(0); > > > > } > > > --------------------------------------------------------------------------- > > ------------------------------------------------ > > > > SNAPPER CHAIN: > > > > Can anybody tell me the errors in the code.....i have done this > > > > problem using the simple concept which i know....it is not giving > > > > right output.... > > > > > Brief explanation of code: > > > > i have first converted the long taken from file into the double using > > > > for loop..then using pow function i have found 2^N and then found > > > > whether k==2^N-1; > > > > > _____ > > > > Aamir > > > > > -- > > > > You received this message because you are subscribed to the Google > > Groups > > > > "google-codejam" group. > > > > To post to this group, send email to [email protected]. > > > > To unsubscribe from this group, send email to > > > > [email protected]<google-code%2bunsubscr...@googlegr > > > > oups.com> > > <google-code%2bunsubscr...@googlegr oups.com> > > > > . > > > > For more options, visit this group at > > > >http://groups.google.com/group/google-code?hl=en. > > > > -- > > > You received this message because you are subscribed to the Google Groups > > "google-codejam" group. > > > To post to this group, send email to [email protected]. > > > To unsubscribe from this group, send email to > > [email protected]<google-code%2bunsubscr...@googlegr > > oups.com> > > . > > > For more options, visit this group athttp:// > > groups.google.com/group/google-code?hl=en. > > > -- > > You received this message because you are subscribed to the Google Groups > > "google-codejam" group. > > To post to this group, send email to [email protected]. > > To unsubscribe from this group, send email to > > [email protected]<google-code%2bunsubscr...@googlegr > > oups.com> > > . > > For more options, visit this group at > >http://groups.google.com/group/google-code?hl=en. > > -- > You received this message because you are subscribed to the Google Groups > "google-codejam" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group > athttp://groups.google.com/group/google-code?hl=en. -- You received this message because you are subscribed to the Google Groups "google-codejam" group. 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