Re: [algogeeks] ADULT SEX PHOTOS
Plz ban him. 2011/4/14 ramya raman ramyaraman9...@gmail.com hot katrina kaif wallpapers http://cinyworld96.blogspot.com/2011/03/katrina-kaif.html new hot sruthi hassan stills http://cinyworld96.blogspot.com/2011/03/shruthi-hassan.html namitha hot sexy stills http://cinyworld96.blogspot.com/2011/04/namitha.html deepika padukone new stills http://cinyworld96.blogspot.com/2011/03/deepika-padukone.html hot tamanna sexy wallpapers http://cinyworld96.blogspot.com/2011/04/tamanna.html beautiful aishwariya rai http://cinyworld96.blogspot.com/2011/03/aishwariya-rai.html diya mirza sexy photos http://cinyworld96.blogspot.com/2011/03/diya-mirza.html stunning beauty sneha http://cinyworld96.blogspot.com/2011/03/sneha.html unbelivable iliyana photos http://cinyworld96.blogspot.com/2011/03/iliyana.html priyamani beautiful stills http://cinyworld96.blogspot.com/2011/03/priyamani.html pranitha in half saree http://cinyworld96.blogspot.com/2011/03/pranitha.html richa saree stills http://cinyworld96.blogspot.com/2011/03/richa-gangopadyay.html -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] If any one have algorithms for interviews by adnan aziz ebook... Please mail ...
Hi All , Yesterday I received an email from Author that this is *violation of Intellectual Property Ownership* ,So kindly please delete pdfs please remove all the sharing. Thanks Guys. Himanshu On Thu, Apr 14, 2011 at 10:52 AM, Harshal hc4...@gmail.com wrote: Thanks :) On Thu, Apr 14, 2011 at 9:59 AM, Rajeev Kumar rajeevprasa...@gmail.comwrote: check this link: https://docs.google.com/viewer?a=vpid=explorerchrome=truesrcid=1B5ady61W_93zq0st5FQpvzj4d6wFCdM3Vl8YGSqRt0_NVFWh3SGkNU24hIb3hl=en If you have any problem in access,please inform me On Thu, Apr 14, 2011 at 1:04 AM, Abhishek Goswami zeal.gosw...@gmail.com wrote: Hi, I tried to open this book in google docs and got message that file is not avaliable. does this file not available in google docs if yes , can anybody share this book again On Tue, Mar 22, 2011 at 10:41 PM, Himanshu Neema potential.himansh...@gmail.com wrote: Turns out that I cant send file larger than 4 MB , please download it from here , let me know if you're still unable to download: http://dl.dropbox.com/u/2681370/Algorithms%2Bfor%2BInterviews%2B%28scan%2Bocr%29%20%281%29.pdf have fun ! On Tue, Mar 22, 2011 at 10:21 PM, Himanshu Neema potential.himansh...@gmail.com wrote: Enjoy :) On Tue, Mar 22, 2011 at 10:09 PM, Saravanan T mail2sarava...@gmail.com wrote: ++ On Tue, Mar 22, 2011 at 9:51 PM, Anurag atri anu.anurag@gmail.com wrote: and me too :) On Tue, Mar 22, 2011 at 9:28 PM, Nikhil Mishra mishra00...@gmail.com wrote: count me too On Tue, Mar 22, 2011 at 11:16 AM, kunal srivastav kunal.shrivas...@gmail.com wrote: plz send it to me too On Tue, Mar 22, 2011 at 11:14 AM, D.N.Vishwakarma@IITR deok...@gmail.com wrote: -- *With Regards Deoki Nandan Vishwakarma IITR MCA Mathematics Department * -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- thezeitgeistmovement.com -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards Anurag Atri -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Thank You Rajeev Kumar -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Harshal Choudhary, III Year B.Tech CSE, NIT Surathkal, Karnataka, India. People die young because god loves them so much, I am still on earth because there is a goddess here who loves me even more. -- You received this message because you are subscribed to the
Re: [algogeeks] ADULT SEX PHOTOS
kick him plzzz On Thu, Apr 14, 2011 at 11:44 AM, taocp mojiedash...@gmail.com wrote: Plz ban him. 2011/4/14 ramya raman ramyaraman9...@gmail.com hot katrina kaif wallpapers http://cinyworld96.blogspot.com/2011/03/katrina-kaif.html new hot sruthi hassan stills http://cinyworld96.blogspot.com/2011/03/shruthi-hassan.html namitha hot sexy stills http://cinyworld96.blogspot.com/2011/04/namitha.html deepika padukone new stills http://cinyworld96.blogspot.com/2011/03/deepika-padukone.html hot tamanna sexy wallpapers http://cinyworld96.blogspot.com/2011/04/tamanna.html beautiful aishwariya rai http://cinyworld96.blogspot.com/2011/03/aishwariya-rai.html diya mirza sexy photos http://cinyworld96.blogspot.com/2011/03/diya-mirza.html stunning beauty sneha http://cinyworld96.blogspot.com/2011/03/sneha.html unbelivable iliyana photos http://cinyworld96.blogspot.com/2011/03/iliyana.html priyamani beautiful stills http://cinyworld96.blogspot.com/2011/03/priyamani.html pranitha in half saree http://cinyworld96.blogspot.com/2011/03/pranitha.html richa saree stills http://cinyworld96.blogspot.com/2011/03/richa-gangopadyay.html -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Arpit Bhatnagar (MNIT JAIPUR) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] ADULT SEX PHOTOS
* I'll if I could. * On Thu, Apr 14, 2011 at 1:24 PM, ArPiT BhAtNaGaR arpitbhatnagarm...@gmail.com wrote: kick him plzzz On Thu, Apr 14, 2011 at 11:44 AM, taocp mojiedash...@gmail.com wrote: Plz ban him. 2011/4/14 ramya raman ramyaraman9...@gmail.com hot katrina kaif wallpapers http://cinyworld96.blogspot.com/2011/03/katrina-kaif.html new hot sruthi hassan stills http://cinyworld96.blogspot.com/2011/03/shruthi-hassan.html namitha hot sexy stills http://cinyworld96.blogspot.com/2011/04/namitha.html deepika padukone new stills http://cinyworld96.blogspot.com/2011/03/deepika-padukone.html hot tamanna sexy wallpapers http://cinyworld96.blogspot.com/2011/04/tamanna.html beautiful aishwariya rai http://cinyworld96.blogspot.com/2011/03/aishwariya-rai.html diya mirza sexy photos http://cinyworld96.blogspot.com/2011/03/diya-mirza.html stunning beauty sneha http://cinyworld96.blogspot.com/2011/03/sneha.html unbelivable iliyana photos http://cinyworld96.blogspot.com/2011/03/iliyana.html priyamani beautiful stills http://cinyworld96.blogspot.com/2011/03/priyamani.html pranitha in half saree http://cinyworld96.blogspot.com/2011/03/pranitha.html richa saree stills http://cinyworld96.blogspot.com/2011/03/richa-gangopadyay.html -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Arpit Bhatnagar (MNIT JAIPUR) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- **With Regards Deoki Nandan Vishwakarma IITR MCA Mathematics Department* * -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: flux redirction
Hi Bounaim, The command for redirecting in Windows are same as available in LINUX. For example: You can redirect the output of the dir command to a file dir file.txt. I you want to know more about the redirection check out this link: http://www.microsoft.com/resources/documentation/windows/xp/all/proddocs/en-us/redirection.mspx?mfr=true . --- Naveen -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] ADULT SEX PHOTOS
@All : Please report it spam and then delete. Gmail will look into it if reported spam by many people. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] ADULT SEX PHOTOS
yeah thats the soln On Thu, Apr 14, 2011 at 6:12 PM, Kunal Patil kp101...@gmail.com wrote: @All : Please report it spam and then delete. Gmail will look into it if reported spam by many people. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Arpit Bhatnagar (MNIT JAIPUR) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] [brain teaser ] Mathematical Brain Teaser 14april
* Mathematical Equation Puzzle By moving one of the following digits, make the equation correct. 62 - 63 = 1 * *Update Your Answers at* : Click Herehttp://dailybrainteaser.blogspot.com/2011/04/mathematical-brain-teaser-14april.html?lavesh=lavesh Solution: Will be updated after 1 day -- Never explain yourself. Your friends don’t need it and your enemies won’t believe it . -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Square of Large integer
#includecstdio
#define MOD (int)1e9
using namespace std;
int main() {
int A = 33554432;
printf(%d\n,A*A);
printf(%d\n,((A*A)%MOD));
return 0;
}
How can i calculate, lets say last 9 digits of square of 33554432 ?
Thanks,
Aamir
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[algogeeks] Re: Square of Large integer
*Output:*
0
0
On Thu, Apr 14, 2011 at 7:01 PM, AAMIR KHAN ak4u2...@gmail.com wrote:
#includecstdio
#define MOD (int)1e9
using namespace std;
int main() {
int A = 33554432;
printf(%d\n,A*A);
printf(%d\n,((A*A)%MOD));
return 0;
}
How can i calculate, lets say last 9 digits of square of 33554432 ?
Thanks,
Aamir
--
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Re: [algogeeks] Re: Square of Large integer
search nd read chinese remainder theorem
On Thu, Apr 14, 2011 at 7:04 PM, AAMIR KHAN ak4u2...@gmail.com wrote:
*Output:*
0
0
On Thu, Apr 14, 2011 at 7:01 PM, AAMIR KHAN ak4u2...@gmail.com wrote:
#includecstdio
#define MOD (int)1e9
using namespace std;
int main() {
int A = 33554432;
printf(%d\n,A*A);
printf(%d\n,((A*A)%MOD));
return 0;
}
How can i calculate, lets say last 9 digits of square of 33554432 ?
Thanks,
Aamir
--
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Re: [algogeeks] Re: Square of Large integer
On Thu, Apr 14, 2011 at 7:40 PM, Akash Mukherjee akash...@gmail.com wrote:
search nd read chinese remainder theorem
please elaborate how can i use chinese remainder theorem.
On Thu, Apr 14, 2011 at 7:04 PM, AAMIR KHAN ak4u2...@gmail.com wrote:
*Output:*
0
0
On Thu, Apr 14, 2011 at 7:01 PM, AAMIR KHAN ak4u2...@gmail.com wrote:
#includecstdio
#define MOD (int)1e9
using namespace std;
int main() {
int A = 33554432;
printf(%d\n,A*A);
printf(%d\n,((A*A)%MOD));
return 0;
}
How can i calculate, lets say last 9 digits of square of 33554432 ?
Thanks,
Aamir
--
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[algogeeks] .Net Developer Needed in IL
Hello Folks, Hope you are doing well! Do you have any consultants for the below .Net Developer position which I have in IL?? If so, Please send me the resume and the best number to reach them. ** Looking for only local applicants * Job Title: .Net Developer Location: Warren ville, IL Duration: 12 months Rate: $48/hr on Corp to Corp * Job Description: I need a Sr .NET Developer who has expertise with ASP.NET, VB.NET, 3.0 4.0 .Net Framework, and C# Development. Thanks Leon Parker - l...@panzersolutions.com 203-983-9475 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] ADULT SEX PHOTOS
Yeah, it is very embarrassing to see this kind of post on Algorithms group. On Thu, Apr 14, 2011 at 6:13 PM, ArPiT BhAtNaGaR arpitbhatnagarm...@gmail.com wrote: yeah thats the soln On Thu, Apr 14, 2011 at 6:12 PM, Kunal Patil kp101...@gmail.com wrote: @All : Please report it spam and then delete. Gmail will look into it if reported spam by many people. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Arpit Bhatnagar (MNIT JAIPUR) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Umer -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Sites for Interview Questions
i think u shuld go thru INDIA BIX OR FRESHERSWORLD.COM On 4/11/11, Anand anandut2...@gmail.com wrote: http://anandtechblog.blogspot.com/ On Mon, Apr 11, 2011 at 10:25 AM, abhishek jain abhi.creat...@gmail.comwrote: www.careercup.com http://www.acetheinterview.com/ Thanks, Abhishek On Mon, Apr 11, 2011 at 3:25 PM, Akash Agrawal akash.agrawa...@gmail.comwrote: http://tech-queries.blogspot.com Regards, Akash Agrawal http://tech-queries.blogspot.com/ On Tue, Jan 18, 2011 at 10:09 PM, rahul rai raikra...@gmail.com wrote: http://www.ocf.berkeley.edu/~wwu/riddles/intro.shtml Rahul K Rai rahulpossi...@gmail.com On Tue, Jan 18, 2011 at 9:27 PM, Yellow Sapphire pukhraj7...@gmail.comwrote: Hi, Can someone suggest good books/websites/blogs for interview related questions. thanks-- YS -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] ADULT SEX PHOTOS
yes...ban him On Thu, Apr 14, 2011 at 11:44 AM, taocp mojiedash...@gmail.com wrote: Plz ban him. 2011/4/14 ramya raman ramyaraman9...@gmail.com hot katrina kaif wallpapers http://cinyworld96.blogspot.com/2011/03/katrina-kaif.html new hot sruthi hassan stills http://cinyworld96.blogspot.com/2011/03/shruthi-hassan.html namitha hot sexy stills http://cinyworld96.blogspot.com/2011/04/namitha.html deepika padukone new stills http://cinyworld96.blogspot.com/2011/03/deepika-padukone.html hot tamanna sexy wallpapers http://cinyworld96.blogspot.com/2011/04/tamanna.html beautiful aishwariya rai http://cinyworld96.blogspot.com/2011/03/aishwariya-rai.html diya mirza sexy photos http://cinyworld96.blogspot.com/2011/03/diya-mirza.html stunning beauty sneha http://cinyworld96.blogspot.com/2011/03/sneha.html unbelivable iliyana photos http://cinyworld96.blogspot.com/2011/03/iliyana.html priyamani beautiful stills http://cinyworld96.blogspot.com/2011/03/priyamani.html pranitha in half saree http://cinyworld96.blogspot.com/2011/03/pranitha.html richa saree stills http://cinyworld96.blogspot.com/2011/03/richa-gangopadyay.html -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Kinshuk Chandra Associate Technical Analyst, Morgan Stanley, B.Tech. CSE, IT-BHU, +91-8898008829 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Square of Large integer
@AAmir: The easiest way is to declare A as long long int. Be sure to
change the printf format string.
Dave
On Apr 14, 8:31 am, AAMIR KHAN ak4u2...@gmail.com wrote:
#includecstdio
#define MOD (int)1e9
using namespace std;
int main() {
int A = 33554432;
printf(%d\n,A*A);
printf(%d\n,((A*A)%MOD));
return 0;
}
How can i calculate, lets say last 9 digits of square of 33554432 ?
Thanks,
Aamir
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Re: [algogeeks] [brain teaser ] Mathematical Brain Teaser 14april
On Thu, Apr 14, 2011 at 6:49 PM, Lavesh Rawat lavesh.ra...@gmail.comwrote: * Mathematical Equation Puzzle By moving one of the following digits, make the equation correct. 62 - 63 = 1 * move 6 above 2 as its power ie 2^6-63=64-63=1 -- best wishes!! Vaibhav Shukla DU-MCA -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Square of Large integer
On Thu, Apr 14, 2011 at 9:08 PM, Dave dave_and_da...@juno.com wrote:
@AAmir: The easiest way is to declare A as long long int. Be sure to
change the printf format string.
I know that. But since i want to have only the last 9 digits of the answer
that can be accommodated in int only so there is no need for long long i
think.
Dave
On Apr 14, 8:31 am, AAMIR KHAN ak4u2...@gmail.com wrote:
#includecstdio
#define MOD (int)1e9
using namespace std;
int main() {
int A = 33554432;
printf(%d\n,A*A);
printf(%d\n,((A*A)%MOD));
return 0;
}
How can i calculate, lets say last 9 digits of square of 33554432 ?
Thanks,
Aamir
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Re: [algogeeks] Re: Square of Large integer
i strongly believe that you should have tried harder . still , here goes
#includeiostream
#includecstdio
#includestdlib.h
using namespace std;
int mulmod ( int a , int b , int c )
{
int x = 0 , y = a % c ;
while ( b 0 )
{
if ( b % 2 == 1 )
{
x = ( x + y ) % c ;
}
y = ( y * 2 ) % c ;
b /= 2 ;
}
return x % c ;
}
int modulo ( int a , int b , int c )
{
int x = 1 , y = a ;
while ( b )
{
if ( b 1)
{
x = mulmod ( x , y , c ) ;
}
y = mulmod ( y ,y, c ) ;
b /= 2 ;
}
return x % c ;
}
int main ()
{
printf ( %d , modulo ( 33554432 , 2 , 10 ) ) ;
}
On Thu, Apr 14, 2011 at 9:39 PM, AAMIR KHAN ak4u2...@gmail.com wrote:
On Thu, Apr 14, 2011 at 9:08 PM, Dave dave_and_da...@juno.com wrote:
@AAmir: The easiest way is to declare A as long long int. Be sure to
change the printf format string.
I know that. But since i want to have only the last 9 digits of the answer
that can be accommodated in int only so there is no need for long long i
think.
Dave
On Apr 14, 8:31 am, AAMIR KHAN ak4u2...@gmail.com wrote:
#includecstdio
#define MOD (int)1e9
using namespace std;
int main() {
int A = 33554432;
printf(%d\n,A*A);
printf(%d\n,((A*A)%MOD));
return 0;
}
How can i calculate, lets say last 9 digits of square of 33554432 ?
Thanks,
Aamir
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II year
Computer Engineering
Delhi College Of Engineering
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[algogeeks] Re: Sequence Puzzle 13april
3 1 2 2 1 1 1 3 1 1 2 2 2 1 On Apr 13, 12:32 pm, Lavesh Rawat lavesh.ra...@gmail.com wrote: * Sequence Puzzle The below is a number puzzle. It should be read left to right, top to bottom. Question 1 What is the next two rows of numbers. Question 2 How was this reached. 1 1 2 1 1 2 1 1 1 1 1 2 2 1 * *Update Your Answers at* : Click Herehttp://dailybrainteaser.blogspot.com/2011/04/sequence-puzzle-13april Solution: Will be updated after 1 day -- Never explain yourself. Your friends don’t need it and your enemies won’t believe it . -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Sites for Interview Questions
http://geeksforgeeks.org/forum/forum/interview-questions On Tue, Apr 12, 2011 at 3:32 PM, Mukesh Kumar thakur mukeshraj8...@gmail.com wrote: i think u shuld go thru INDIA BIX OR FRESHERSWORLD.COM On 4/11/11, Anand anandut2...@gmail.com wrote: http://anandtechblog.blogspot.com/ On Mon, Apr 11, 2011 at 10:25 AM, abhishek jain abhi.creat...@gmail.comwrote: www.careercup.com http://www.acetheinterview.com/ Thanks, Abhishek On Mon, Apr 11, 2011 at 3:25 PM, Akash Agrawal akash.agrawa...@gmail.comwrote: http://tech-queries.blogspot.com Regards, Akash Agrawal http://tech-queries.blogspot.com/ On Tue, Jan 18, 2011 at 10:09 PM, rahul rai raikra...@gmail.com wrote: http://www.ocf.berkeley.edu/~wwu/riddles/intro.shtml Rahul K Rai rahulpossi...@gmail.com On Tue, Jan 18, 2011 at 9:27 PM, Yellow Sapphire pukhraj7...@gmail.comwrote: Hi, Can someone suggest good books/websites/blogs for interview related questions. thanks-- YS -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Sequence Puzzle 13april
3 1 2 2 1 1 1 3 1 1 2 2 2 1 On Thu, Apr 14, 2011 at 10:24 PM, arpit.gupta arpitg1...@gmail.com wrote: 3 1 2 2 1 1 1 3 1 1 2 2 2 1 On Apr 13, 12:32 pm, Lavesh Rawat lavesh.ra...@gmail.com wrote: * Sequence Puzzle The below is a number puzzle. It should be read left to right, top to bottom. Question 1 What is the next two rows of numbers. Question 2 How was this reached. 1 1 2 1 1 2 1 1 1 1 1 2 2 1 * *Update Your Answers at* : Click Here http://dailybrainteaser.blogspot.com/2011/04/sequence-puzzle-13april Solution: Will be updated after 1 day -- Never explain yourself. Your friends don’t need it and your enemies won’t believe it . -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- **With Regards Deoki Nandan Vishwakarma IITR MCA Mathematics Department* * -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Need help on Divide and Conquer Algorithm
complexity : O(n) + O(nlogn)
Sweety wrote:
Question :Let A[1..n] be an array of integers. Design an efficient
divide and conquer algorithm to determine if A contains a majority
element, i.e an element appears more than n/2 times in A. What is the
time complexity of your algorithm?
Answer:
a[1..n] is an array
int majorityElement(int a[], int first, int last)
{
If (first = = last)
{
return a[first]; // Array has one element and its count = 1
and it is major element
}
mid= (first+last)/2;
(majorL,countL)= majorityElement(a,first,mid);
(majorR,countR)= majorityElement(a,mid
+1,last);
n = total elements in an array;
If(majorL==majorR)
return(countL+countR);
else
{
If(countLcountR)
return(majorL,countL);
elseif(countL countR)
return(majorR,countR);
else
return(majorL,majorR);
}
if(countLn/2)
temp1=majorL;
if(countRn/2)
temp2=majorR;
If(temp1 = = temp2)
return temp1;
elseif(countLcountR)
return temp1;
else (countRcountL)
return temp2;
else
return -1;
}
int main()
{
int a[8] = {2,3,2,2,4,2,2,2};
int first =1;
int last=8; //change the value of last when the array
increases or decreases in size
int x = majorityElement(a,first,last);
if(x= = -1)
printf(“No Majority Element”)
else
Majority element = x;
}
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[algogeeks] Re: Need help on Divide and Conquer Algorithm
I can post solution of this complexity if you want !!
On Apr 15, 12:19 am, vishwakarma vishwakarma.ii...@gmail.com wrote:
complexity : O(n) + O(nlogn)
Sweety wrote:
Question :Let A[1..n] be an array of integers. Design an efficient
divide and conquer algorithm to determine if A contains a majority
element, i.e an element appears more than n/2 times in A. What is the
time complexity of your algorithm?
Answer:
a[1..n] is an array
int majorityElement(int a[], int first, int last)
{
If (first = = last)
{
return a[first]; // Array has one element and its count = 1
and it is major element
}
mid= (first+last)/2;
(majorL,countL)= majorityElement(a,first,mid);
(majorR,countR)= majorityElement(a,mid
+1,last);
n = total elements in an array;
If(majorL==majorR)
return(countL+countR);
else
{
If(countLcountR)
return(majorL,countL);
elseif(countL countR)
return(majorR,countR);
else
return(majorL,majorR);
}
if(countLn/2)
temp1=majorL;
if(countRn/2)
temp2=majorR;
If(temp1 = = temp2)
return temp1;
elseif(countLcountR)
return temp1;
else (countRcountL)
return temp2;
else
return -1;
}
int main()
{
int a[8] = {2,3,2,2,4,2,2,2};
int first =1;
int last=8; //change the value of last when the array
increases or decreases in size
int x = majorityElement(a,first,last);
if(x= = -1)
printf(“No Majority Element”)
else
Majority element = x;
}
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Re: [algogeeks] Re: Need help on Divide and Conquer Algorithm
Yes , I also need the same...Thanks for the help .
On Fri, Apr 15, 2011 at 12:52 AM, vishwakarma
vishwakarma.ii...@gmail.comwrote:
I can post solution of this complexity if you want !!
On Apr 15, 12:19 am, vishwakarma vishwakarma.ii...@gmail.com wrote:
complexity : O(n) + O(nlogn)
Sweety wrote:
Question :Let A[1..n] be an array of integers. Design an efficient
divide and conquer algorithm to determine if A contains a majority
element, i.e an element appears more than n/2 times in A. What is the
time complexity of your algorithm?
Answer:
a[1..n] is an array
int majorityElement(int a[], int first, int last)
{
If (first = = last)
{
return a[first]; // Array has one element and its count = 1
and it is major element
}
mid= (first+last)/2;
(majorL,countL)= majorityElement(a,first,mid);
(majorR,countR)= majorityElement(a,mid
+1,last);
n = total elements in an array;
If(majorL==majorR)
return(countL+countR);
else
{
If(countLcountR)
return(majorL,countL);
elseif(countL countR)
return(majorR,countR);
else
return(majorL,majorR);
}
if(countLn/2)
temp1=majorL;
if(countRn/2)
temp2=majorR;
If(temp1 = = temp2)
return temp1;
elseif(countLcountR)
return temp1;
else (countRcountL)
return temp2;
else
return -1;
}
int main()
{
int a[8] = {2,3,2,2,4,2,2,2};
int first =1;
int last=8; //change the value of last when the array
increases or decreases in size
int x = majorityElement(a,first,last);
if(x= = -1)
printf(“No Majority Element”)
else
Majority element = x;
}
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IIIT Allahabad (Amethi Capmus),
6th Sem.
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[algogeeks] Re: Need help on Divide and Conquer Algorithm
Let A be the input array;
Now algorithm is follows;
struct leave{
int cand;
int count;
};
struct leave tree[120];
void build(int s,int e,int node ,int *A)
{
if(s == e){
tree[node].cand = A[s];
tree[node].count = 1;
return;
}
else{
int mid = (s+e)1;
build(s,mid,2*node,A);
build(mid+1,e,2*node+1,A);
if(tree[2*node].cand == tree[2*node+1].cand){
tree[node].cand = tree[2*node].cand;
tree[node].count = tree[2*node].count +
tree[2*node+1].count;
}
else{
if(tree[2*node].count tree[2*node+1].count){
tree[node].cand = tree[2*node].cand;
tree[node].count = tree[2*node].count -
tree[2*node+1].count;
}
else{
tree[node].cand = tree[2*node+1].cand;
tree[node].count = tree[2*node+1].count -
tree[2*node].count;
}
}
}
}
int main()
{
//read Array A
build(1,n,1);
//sort array A
int val = Tree[1].cand;
//perform binary search on sorted array A
//if its count is strictly greater than n/2 then yes else no
On Apr 15, 12:28 am, LALIT SHARMA lks.ru...@gmail.com wrote:
Yes , I also need the same...Thanks for the help .
On Fri, Apr 15, 2011 at 12:52 AM, vishwakarma
vishwakarma.ii...@gmail.comwrote:
I can post solution of this complexity if you want !!
On Apr 15, 12:19 am, vishwakarma vishwakarma.ii...@gmail.com wrote:
complexity : O(n) + O(nlogn)
Sweety wrote:
Question :Let A[1..n] be an array of integers. Design an efficient
divide and conquer algorithm to determine if A contains a majority
element, i.e an element appears more than n/2 times in A. What is the
time complexity of your algorithm?
Answer:
a[1..n] is an array
int majorityElement(int a[], int first, int last)
{
If (first = = last)
{
return a[first]; // Array has one element and its count = 1
and it is major element
}
mid= (first+last)/2;
(majorL,countL)= majorityElement(a,first,mid);
(majorR,countR)= majorityElement(a,mid
+1,last);
n = total elements in an array;
If(majorL==majorR)
return(countL+countR);
else
{
If(countLcountR)
return(majorL,countL);
elseif(countL countR)
return(majorR,countR);
else
return(majorL,majorR);
}
if(countLn/2)
temp1=majorL;
if(countRn/2)
temp2=majorR;
If(temp1 = = temp2)
return temp1;
elseif(countLcountR)
return temp1;
else (countRcountL)
return temp2;
else
return -1;
}
int main()
{
int a[8] = {2,3,2,2,4,2,2,2};
int first =1;
int last=8; //change the value of last when the array
increases or decreases in size
int x = majorityElement(a,first,last);
if(x= = -1)
printf(“No Majority Element”)
else
Majority element = x;
}
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IIIT Allahabad (Amethi Capmus),
6th Sem.
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[algogeeks] Re: Need help on Divide and Conquer Algorithm
example : let the array be { a,b,a,b,c,d,e,d,d,e,f};
now...
step 1 :pick any 2 different element and remove from the array till
array contains only same elements or any single element ...// dis
is implemented wid the above mentioned funtion build()
if there is any element whose occurence is greater than n/2 than u ll
always get an unique value left after step 1 , it wont depend on the
way u select 2 different element n removing them.
On Apr 15, 12:43 am, vishwakarma vishwakarma.ii...@gmail.com wrote:
Let A be the input array;
Now algorithm is follows;
struct leave{
int cand;
int count;};
struct leave tree[120];
void build(int s,int e,int node ,int *A)
{
if(s == e){
tree[node].cand = A[s];
tree[node].count = 1;
return;
}
else{
int mid = (s+e)1;
build(s,mid,2*node,A);
build(mid+1,e,2*node+1,A);
if(tree[2*node].cand == tree[2*node+1].cand){
tree[node].cand = tree[2*node].cand;
tree[node].count = tree[2*node].count +
tree[2*node+1].count;
}
else{
if(tree[2*node].count tree[2*node+1].count){
tree[node].cand = tree[2*node].cand;
tree[node].count = tree[2*node].count -
tree[2*node+1].count;
}
else{
tree[node].cand = tree[2*node+1].cand;
tree[node].count = tree[2*node+1].count -
tree[2*node].count;
}
}
}
}
int main()
{
//read Array A
build(1,n,1);
//sort array A
int val = Tree[1].cand;
//perform binary search on sorted array A
//if its count is strictly greater than n/2 then yes else no
On Apr 15, 12:28 am, LALIT SHARMA lks.ru...@gmail.com wrote:
Yes , I also need the same...Thanks for the help .
On Fri, Apr 15, 2011 at 12:52 AM, vishwakarma
vishwakarma.ii...@gmail.comwrote:
I can post solution of this complexity if you want !!
On Apr 15, 12:19 am, vishwakarma vishwakarma.ii...@gmail.com wrote:
complexity : O(n) + O(nlogn)
Sweety wrote:
Question :Let A[1..n] be an array of integers. Design an efficient
divide and conquer algorithm to determine if A contains a majority
element, i.e an element appears more than n/2 times in A. What is the
time complexity of your algorithm?
Answer:
a[1..n] is an array
int majorityElement(int a[], int first, int last)
{
If (first = = last)
{
return a[first]; // Array has one element and its count =
1
and it is major element
}
mid= (first+last)/2;
(majorL,countL)= majorityElement(a,first,mid);
(majorR,countR)= majorityElement(a,mid
+1,last);
n = total elements in an array;
If(majorL==majorR)
return(countL+countR);
else
{
If(countLcountR)
return(majorL,countL);
elseif(countL countR)
return(majorR,countR);
else
return(majorL,majorR);
}
if(countLn/2)
temp1=majorL;
if(countRn/2)
temp2=majorR;
If(temp1 = = temp2)
return temp1;
elseif(countLcountR)
return temp1;
else (countRcountL)
return temp2;
else
return -1;
}
int main()
{
int a[8] = {2,3,2,2,4,2,2,2};
int first =1;
int last=8; //change the value of last when the array
increases or decreases in size
int x = majorityElement(a,first,last);
if(x= = -1)
printf(“No Majority Element”)
else
Majority element = x;
}
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[algogeeks] Re: Square of Large integer
@Aamir: Okay. How about this:
Write A = 1 * ahi + alo, where 0 alo 1; i.e.
ahi = A/1 and alo = A%1.
Then A*A = 1 * ahi * ahi + 2 * ahi * alo + alo * alo,
so A*A%Mod = (1 * ((ahi * ahi) % (Mod/1)) + 1 *
((2 * ahi * alo) % (Mod/1)) + alo * alo % Mod) % Mod
This simplifies to (1*(ahi*ahi)%10) + 1*((2*ahi*alo)
%10 + (alo*alo)%Mod)%Mod
Dave
On Apr 14, 11:09 am, AAMIR KHAN ak4u2...@gmail.com wrote:
On Thu, Apr 14, 2011 at 9:08 PM, Dave dave_and_da...@juno.com wrote:
@AAmir: The easiest way is to declare A as long long int. Be sure to
change the printf format string.
I know that. But since i want to have only the last 9 digits of the answer
that can be accommodated in int only so there is no need for long long i
think.
Dave
On Apr 14, 8:31 am, AAMIR KHAN ak4u2...@gmail.com wrote:
#includecstdio
#define MOD (int)1e9
using namespace std;
int main() {
int A = 33554432;
printf(%d\n,A*A);
printf(%d\n,((A*A)%MOD));
return 0;
}
How can i calculate, lets say last 9 digits of square of 33554432 ?
Thanks,
Aamir
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[algogeeks] Re: Dp problem
Here i proposed an algorithm. correct me if i am wrong !!
int main()
{
int N; //number of boards
int W; // number of workers
int smax = 0;
cin N W;
int *A = new int[N];
for(int i=0;iN;i++){
cin A[i];
smax += A[i];
}
int m = *max_element(A ,A+N);
for(int i=m;i=smax;i++){
int sum = 0;
int w = 1;
for(int j=0;jN;j++){
if(sum+A[j] = i){
sum += A[j];
}
else{
sum = A[j];
w++;
}
}
if(w=W){
cout i \n;
break;
}
}
return 0;
}
rajat ahuja wrote:
You have to paint N boards of length {B1, B2, B3… BN}. There are K painters
available and you are also given how much time a painter takes to paint 1
unit of board. You have to get this job done as soon as possible under the
constraints that any painter will only paint continuous sections of board,
say board {2, 3, 4} or only board {1} or nothing but not board {2, 4, 5}.
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[algogeeks]
char *x(int c,char *k,char *s)
{
if(!k)
{
return *s-36?x(0,0,s+1):s;}
if(s)
if(*s)
c=10+(c?(x(c,k,0),x(c,k+=*s-c,s+1),*k):(x(*s,k,s+1),0));
else
c=10;
printf(x(~0,0,k)[c-~-c+1[~c-c]],c);
}
main()
{
x(0,^[kXc6]dn_eaoh$%c,-34*1'.+(,03#;+,)/'///*);
}
can any body tell me why its output is given below ? and how i should
evaluate?
--
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Re: [algogeeks] Re: Square of Large integer
Probably the best would be to use long long int int this case.Anyhow a few
bytes of memory wont hurt you,rather it will significantly improve your
time.
For much larger numbers than this try Exponentiation by squaring
methodThis does the same in o(log n)
timehttp://en.wikipedia.org/wiki/Exponentiation_by_squaring
On Fri, Apr 15, 2011 at 1:27 AM, Dave dave_and_da...@juno.com wrote:
@Aamir: Okay. How about this:
Write A = 1 * ahi + alo, where 0 alo 1; i.e.
ahi = A/1 and alo = A%1.
Then A*A = 1 * ahi * ahi + 2 * ahi * alo + alo * alo,
so A*A%Mod = (1 * ((ahi * ahi) % (Mod/1)) + 1 *
((2 * ahi * alo) % (Mod/1)) + alo * alo % Mod) % Mod
This simplifies to (1*(ahi*ahi)%10) + 1*((2*ahi*alo)
%10 + (alo*alo)%Mod)%Mod
Dave
On Apr 14, 11:09 am, AAMIR KHAN ak4u2...@gmail.com wrote:
On Thu, Apr 14, 2011 at 9:08 PM, Dave dave_and_da...@juno.com wrote:
@AAmir: The easiest way is to declare A as long long int. Be sure to
change the printf format string.
I know that. But since i want to have only the last 9 digits of the
answer
that can be accommodated in int only so there is no need for long long i
think.
Dave
On Apr 14, 8:31 am, AAMIR KHAN ak4u2...@gmail.com wrote:
#includecstdio
#define MOD (int)1e9
using namespace std;
int main() {
int A = 33554432;
printf(%d\n,A*A);
printf(%d\n,((A*A)%MOD));
return 0;
}
How can i calculate, lets say last 9 digits of square of 33554432 ?
Thanks,
Aamir
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MNNIT ALLAHABAD
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