Re: How do I disable issues?

2012-12-19 Thread rogerdpack
I also find it odd that under the admin - issue tracking there is no way 
to disable it...instead you have to go to a totally separate area. It 
confused me, I expected to be able to go to 'issue tracking' and click 
disable or what not :)
Cheers!

On Wednesday, May 16, 2012 11:14:53 AM UTC-6, Seth Ladd wrote:

 Apologies, found it! It's under Administer, Tabs. Perhaps you should 
 duplicate this option under each respective area, I kept searching inside 
 the Administer|Issue Tracker for the option. :)

 On Wednesday, May 16, 2012 10:11:21 AM UTC-7, Seth Ladd wrote:

 Hello,

 How do I completely turn off Issues? Our issues are tracked elsewhere, so 
 I don't want to give the wrong impression.

 Thanks!
 Seth



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atmos-c ownership

2012-12-19 Thread Jason Cwik
Hi,

I'm currently assigned as a committer on the atmos-c project.  The owner
of the project left our company about 18 months ago and abandoned the
project.  I just made a big new release of the code and realized that I
don't have ownership permissions to edit the wiki and/or add new members to
the project.

I've tried contacting Scot via gmail and through linkedIn and have not
gotten any responses back.  According to LinkedIn, he now works for a
competitor (AWS) and maybe doesn't want to help me out.  Can you promote me
to owner of this project?

Thanks,
Jason

-- 
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CTO
Connectic, Inc
Cell: 612-217-0442

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Re: atmos-c ownership

2012-12-19 Thread Chris Smith
Hey Jason,


We can make you owner of the project, but first you need to complete a few
steps as per our policy on the matter. Please see:
http://code.google.com/p/support/wiki/FAQ#Other_Questions under the heading
Requests to Take Over Another Project.

Cheers,
-Chris


On Wed, Dec 19, 2012 at 11:30 AM, Jason Cwik ja...@connecticinc.com wrote:

 Hi,

 I'm currently assigned as a committer on the atmos-c project.  The owner
 of the project left our company about 18 months ago and abandoned the
 project.  I just made a big new release of the code and realized that I
 don't have ownership permissions to edit the wiki and/or add new members to
 the project.

 I've tried contacting Scot via gmail and through linkedIn and have not
 gotten any responses back.  According to LinkedIn, he now works for a
 competitor (AWS) and maybe doesn't want to help me out.  Can you promote me
 to owner of this project?

 Thanks,
 Jason

 --
 Jason Cwik
 CTO
 Connectic, Inc
 Cell: 612-217-0442

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Re: [gcj] optimized algo for LCS0

2012-12-19 Thread Andrey Ponomarev
I think you can try the following. First, for each lowercase latin
letter prepare an array of its indexes in the second string. Overall
length is O(m) and the time is also O(m). Then, let the 'state' be two
values for each lowercase latin letter: maximal subsequence length,
ending with this letter and the leftmost position in the second string
where this maximal subsequence ends. Initially set the state for each
letter for (0,-1). Then, update the state for each prefix of the first
string from prefix of length 1 to the whole string. At each step you
have l_i - current letter in the first string, and state S. Try to
update maximal length S[l_i] based on all possible last letters by
binary searching array of indexes for allowed position of l_i with
minimal index (i.e. leftmost). It gives you complexity O(n*c*log(m)).
Where c stands for alphabet size.

2012/12/16 anupsingh anupsingh@gmail.com:
 hey can any one tell me the optimized method of solving LCS problem. I hv
 optimized still it show TLE...
 here is the link
 www.spoj.com/problems/LCS0/

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Re: [gcj] optimized algo for LCS0

2012-12-19 Thread Andrey Ponomarev
Sorry, ignore my post. Its totally broken.

2012/12/19 Andrey Ponomarev ponomarev@gmail.com:
 I think you can try the following. First, for each lowercase latin
 letter prepare an array of its indexes in the second string. Overall
 length is O(m) and the time is also O(m). Then, let the 'state' be two
 values for each lowercase latin letter: maximal subsequence length,
 ending with this letter and the leftmost position in the second string
 where this maximal subsequence ends. Initially set the state for each
 letter for (0,-1). Then, update the state for each prefix of the first
 string from prefix of length 1 to the whole string. At each step you
 have l_i - current letter in the first string, and state S. Try to
 update maximal length S[l_i] based on all possible last letters by
 binary searching array of indexes for allowed position of l_i with
 minimal index (i.e. leftmost). It gives you complexity O(n*c*log(m)).
 Where c stands for alphabet size.

 2012/12/16 anupsingh anupsingh@gmail.com:
 hey can any one tell me the optimized method of solving LCS problem. I hv
 optimized still it show TLE...
 here is the link
 www.spoj.com/problems/LCS0/

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Re: [gcj] SPOJ - ONEZERO - WA :(

2012-12-19 Thread ZHANG Xiongqi, Parker
Bignumber operation can be avoid, but you should not use unsigned long 
long anyway.


http://ideone.com/3nlO2d

Above is another accepted solution, the idea is still BFS, but without 
BigInteger.


Parker

On 2012/12/19 22:14, thefourtheye dIVi wrote:

Hi Parker,

Thanks for the Suggestion... I am new to the Big Integer stuff, so 
working on it... Moreover, 
http://www.spoj.com/forum/viewtopic.php?f=3t=3555p=19155


This thread has a solution similar to what I have implemented but no 
one replied in the thread was worried about the Big Integer, so might 
it be a bug in my implementation which gets me WA?


But 19998 is a very valid input... :(


On Mon, Dec 17, 2012 at 3:40 AM, ZHANG Xiongqi, Parker 
zhangxion...@gmail.com mailto:zhangxion...@gmail.com wrote:


Hi thefourtheye dIVi,

Basically your idea is correct and this problem can be solved
using BFS.

However, it is not sufficient to use unsigned long long to solve
this problem because the answer to 19998 is
0 which is much larger than
the maximum number that could be represented using unsigned long long.

One more thing to note, you can keep track of all the possible
remainder and no need to process the remainder that has appeared
before. The reason for that is left for your exercise. :)

Here is the accepted code which I adapted from yours.

http://ideone.com/7ziUxy

If you have any more questions, fell free to ask.

Parker


On 2012/12/16 22:42, thefourtheye dIVi wrote:

I am trying to solve http://www.spoj.com/problems/ONEZERO/

I referred so many internet posts about this, and they all
talk about storing reminders and building a tree. I am running
a simple BFS, nothing else... But this gets me WA :(

http://ideone.com/SZDn5T

Dont know whats wrong with this code. Please help me fix it.
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Re: [gcj] SPOJ - ONEZERO - WA :(

2012-12-19 Thread thefourtheye dIVi
Cool. Got ACed... But still, it was almost at the verge of TLE :( I see
other solutions getting ACed within 0.5 minutes. Wondering what they would
be doing...


On Wed, Dec 19, 2012 at 8:33 PM, ZHANG Xiongqi, Parker 
zhangxion...@gmail.com wrote:

 Bignumber operation can be avoid, but you should not use unsigned long
 long anyway.

 http://ideone.com/3nlO2d

 Above is another accepted solution, the idea is still BFS, but without
 BigInteger.

 Parker


 On 2012/12/19 22:14, thefourtheye dIVi wrote:

 Hi Parker,

 Thanks for the Suggestion... I am new to the Big Integer stuff, so
 working on it... Moreover, http://www.spoj.com/forum/**
 viewtopic.php?f=3t=3555p=**19155http://www.spoj.com/forum/viewtopic.php?f=3t=3555p=19155

 This thread has a solution similar to what I have implemented but no one
 replied in the thread was worried about the Big Integer, so might it be a
 bug in my implementation which gets me WA?

 But 19998 is a very valid input... :(


 On Mon, Dec 17, 2012 at 3:40 AM, ZHANG Xiongqi, Parker 
 zhangxion...@gmail.com mailto:zhangxion...@gmail.com** wrote:

 Hi thefourtheye dIVi,

 Basically your idea is correct and this problem can be solved
 using BFS.

 However, it is not sufficient to use unsigned long long to solve
 this problem because the answer to 19998 is
 11**110 which is much larger than
 the maximum number that could be represented using unsigned long long.

 One more thing to note, you can keep track of all the possible
 remainder and no need to process the remainder that has appeared
 before. The reason for that is left for your exercise. :)

 Here is the accepted code which I adapted from yours.

 http://ideone.com/7ziUxy

 If you have any more questions, fell free to ask.

 Parker


 On 2012/12/16 22:42, thefourtheye dIVi wrote:

 I am trying to solve 
 http://www.spoj.com/problems/**ONEZERO/http://www.spoj.com/problems/ONEZERO/

 I referred so many internet posts about this, and they all
 talk about storing reminders and building a tree. I am running
 a simple BFS, nothing else... But this gets me WA :(

 http://ideone.com/SZDn5T

 Dont know whats wrong with this code. Please help me fix it.
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