Re: Using a different license
Well, although thats not the main issue, CPAL is actually one of the simplest licenses I found, identical to the MPL with 2 important additions, especially the one pertaining to SaaS as a form of distribution. Nothing complicated or special, even when compared to traditional open source licenses, which are usually not clear cut on some issues, thus the progress in additional licenses. Like you said, we can use other hosting facilities. Thanks for the response mike On Oct 11, 4:22 pm, Ben Collins-Sussman suss...@google.com wrote: On Sun, Oct 11, 2009 at 8:56 AM, MikeT mikete...@gmail.com wrote: Is there any chance you will enable use of other approved OSI licenses or any chosen license ? it would be a shame to stop using Google Hosting just for this reason. Chris DiBona is really the final word on this, but I can reiterate the party line fairly well: that by choosing the relatively obscure CPAL license, the open source community is being incrementally harmed. When somebody using a common license (like GPL, Apache, BSD) wants to re-use your software, they now have to consult lawyers to figure out what's compatible and what's not. However, by rejecting your license, Google is not really hurting your project. You have lots of other places to host. Our belief is that for every project we lose (due to our anti-proliferation policy), we also probably persuade several others to switch from some made-up license to something standardized. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Hosting at Google Code group. To post to this group, send email to google-code-hosting@googlegroups.com To unsubscribe from this group, send email to google-code-hosting+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code-hosting?hl=en -~--~~~~--~~--~--~---
How can i change the description of an issue?
I find that once I created an issue i can't change its description. I checked all the places and still haven't managed to do that. Is this true? How can you make sure you're always right at the first time? --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Hosting at Google Code group. To post to this group, send email to google-code-hosting@googlegroups.com To unsubscribe from this group, send email to google-code-hosting+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code-hosting?hl=en -~--~~~~--~~--~--~---
Re: How can i change the description of an issue?
Hi, It is true, there is no way to change the description of an issue. Instead, you just add comments. You can think of the description as just the initial report. As you learn more about the problem, you add comments that describe what you have learned, even if you learn that part of the initial report was mistaken. There is an enhancement request to allow editing of issue descriptions. You may want to star this issue: http://code.google.com/p/support/issues/detail?id=919 Thanks, jason! On Mon, Oct 12, 2009 at 3:02 AM, LiYanhui whi...@gmail.com wrote: I find that once I created an issue i can't change its description. I checked all the places and still haven't managed to do that. Is this true? How can you make sure you're always right at the first time? --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Hosting at Google Code group. To post to this group, send email to google-code-hosting@googlegroups.com To unsubscribe from this group, send email to google-code-hosting+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code-hosting?hl=en -~--~~~~--~~--~--~---
Re: Cannot read issue tracker on small screens
On how many different platforms and browsers can you reproduce this issue? Your screenshot appears to show Opera on Puppy Linux, but do you still see the problem in, say, Firefox or Konqueror?-Nathaniel On Sat, Oct 10, 2009 at 3:19 AM, technosaurus b...@openplatformeducation.org wrote: Text entered into the issue tracker is unreadable on lower res monitors. Here is an example http://www.murga-linux.com/puppy/viewtopic.php?mode=attachid=22502 Changing the window so that it has a scrollbar at the bottom does not fix the problem, it still gets cut off at the end of the last rounded tab. Maybe wordwrap would be good? --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Hosting at Google Code group. To post to this group, send email to google-code-hosting@googlegroups.com To unsubscribe from this group, send email to google-code-hosting+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code-hosting?hl=en -~--~~~~--~~--~--~---
Re: Please increase repository quota
I've increased your quota to 4GB. Happy hacking! -- Jacob Lee artd...@gmail.com On Mon, Oct 12, 2009 at 11:15 AM, kevinwan...@gmail.com kevinwan...@gmail.com wrote: Could you please increase quota for this project? http://code.google.com/p/cyberaide/ Currently we have used around 97% of its 1G space. We are a large group as you can see and actively working on a bunch of projects related to Grid/Cloud computing. I'm not on the owner list, but the owner delegated the request to me. Thanks a lot! Fugang --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Hosting at Google Code group. To post to this group, send email to google-code-hosting@googlegroups.com To unsubscribe from this group, send email to google-code-hosting+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code-hosting?hl=en -~--~~~~--~~--~--~---
Re: What is If using Google Groups, add the address directly with no email delivery.
This problem is there even now. I wonder how all other projects are wroking? --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Hosting at Google Code group. To post to this group, send email to google-code-hosting@googlegroups.com To unsubscribe from this group, send email to google-code-hosting+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code-hosting?hl=en -~--~~~~--~~--~--~---
[gcj] Re: Round 3, Problem D analysis
My replies inline ... -Bharath 2009/10/12 Hawston LLH haws...@gmail.com Can someone explain point 2? And there are only O(sqrt(N)) palindromes up to N - so the number of groups of consecutive zeros or ones in the first N characters is O(sqrt(N)). consider N is some x digit number ... divide the digits into 2 parts .. first x/2 and the last x/2. Now assume that the first x/2 digits can be filled in k ways. For being a palindrome, the second x/2 is automatically decided. Otherwise, the second x/2 digits have their own k ways to be filled. Hence, there are roughly k palindromes in k^2 numbers. Hence O(sqrt(N)) palindromes up to N. All groups except maybe two boundary ones fit into [L-1,R] segment entirely what is the meaning of All groups except maybe two boundary ones fit into [L-1,R] segment entirely, is the group refers to group of ones or zeros mentioned earlier? Yes. consider [L-1,R] as here, you have 2 groups that fit entirely into the range. And the 2 boundary ones are not fitting. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups google-codejam group. To post to this group, send email to google-code@googlegroups.com To unsubscribe from this group, send email to google-code+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code?hl=en -~--~~~~--~~--~--~---
[gcj] Re: Round 3, Problem D analysis
another question, in point 1, it deals with [0, X], where X from L-1 to R. In point 2 here, it use [L-1,R], what is the relation or typo error? On Mon, Oct 12, 2009 at 5:48 PM, Bharath Raghavendran rbharat...@gmail.comwrote: My replies inline ... -Bharath 2009/10/12 Hawston LLH haws...@gmail.com Can someone explain point 2? And there are only O(sqrt(N)) palindromes up to N - so the number of groups of consecutive zeros or ones in the first N characters is O(sqrt(N)). consider N is some x digit number ... divide the digits into 2 parts .. first x/2 and the last x/2. Now assume that the first x/2 digits can be filled in k ways. For being a palindrome, the second x/2 is automatically decided. Otherwise, the second x/2 digits have their own k ways to be filled. Hence, there are roughly k palindromes in k^2 numbers. Hence O(sqrt(N)) palindromes up to N. All groups except maybe two boundary ones fit into [L-1,R] segment entirely what is the meaning of All groups except maybe two boundary ones fit into [L-1,R] segment entirely, is the group refers to group of ones or zeros mentioned earlier? Yes. consider [L-1,R] as here, you have 2 groups that fit entirely into the range. And the 2 boundary ones are not fitting. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups google-codejam group. To post to this group, send email to google-code@googlegroups.com To unsubscribe from this group, send email to google-code+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code?hl=en -~--~~~~--~~--~--~---
[gcj] Re: Round 3, Problem D analysis
here are the statements that show the relation : [L,R] can be represented as [0,R] minus [0,L-1]; thus it contains an even number of palindromes if and only if [0,L-1] and [0,R] both contain even or both contain odd number of palindromes. We can also reduce the number of boundary groups to consider from two to one relying on the fact that the number of zeroes/ones in [L-1,R] is the number of zeroes/ones in [0,R] minus the number of zeroes/ones in [0,L-2]. I didn't read the analysis properly and hence not sure why its [L-1, R] and not [L,R] and so cant comment on that. Is this what you were looking for? 2009/10/12 Hawston LLH haws...@gmail.com another question, in point 1, it deals with [0, X], where X from L-1 to R. In point 2 here, it use [L-1,R], what is the relation or typo error? On Mon, Oct 12, 2009 at 5:48 PM, Bharath Raghavendran rbharat...@gmail.com wrote: My replies inline ... -Bharath 2009/10/12 Hawston LLH haws...@gmail.com Can someone explain point 2? And there are only O(sqrt(N)) palindromes up to N - so the number of groups of consecutive zeros or ones in the first N characters is O(sqrt(N)). consider N is some x digit number ... divide the digits into 2 parts .. first x/2 and the last x/2. Now assume that the first x/2 digits can be filled in k ways. For being a palindrome, the second x/2 is automatically decided. Otherwise, the second x/2 digits have their own k ways to be filled. Hence, there are roughly k palindromes in k^2 numbers. Hence O(sqrt(N)) palindromes up to N. All groups except maybe two boundary ones fit into [L-1,R] segment entirely what is the meaning of All groups except maybe two boundary ones fit into [L-1,R] segment entirely, is the group refers to group of ones or zeros mentioned earlier? Yes. consider [L-1,R] as here, you have 2 groups that fit entirely into the range. And the 2 boundary ones are not fitting. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups google-codejam group. To post to this group, send email to google-code@googlegroups.com To unsubscribe from this group, send email to google-code+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code?hl=en -~--~~~~--~~--~--~---
[gcj] how to solve this?
Hey .. here is a problem I saw somewhere on net. Can anyone give me some ideas on its algo ? There is a tray that contains a row of s slots where biscuits may be placed. You can have atmost one biscuit per slot. You and your friend play a game with it. Each of you get to pick biscuits in turns. You can pick atmost b biscuits in a single turn provided they are in consecutive slots with no gaps (empty slots) in between (but you have to pick atleast one in your turn). Whoever picks the last biscuit wins the game. You are given a snapshot of the game and its your turn. Assuming that from this point onwards, both the players play optimally. Find out if you can win the game. And if yes, what is your next move? If there are multiple options for the next move, choose the one where you pick the highest number of biscuits in the next turn. If still there are multiple options, choose the one where you take the leftmost biscuit among them. As input, you will be given 't' which indicates the number of test cases. For each case, you will be given s ( 1=s=2000 ), b ( 1=b=20 ) seperated by space and the next line will contain s characters that are either '0' or '1' representing each of the s slots. '0' indicates that a biscuit is not present at that particular slot and '1' indicates that a biscuit is present. As output, print yes or no indicating if you can win and if the answer is yes, print a string of s characters of either '0' or '1' that indicates whether a biscuit is absent or present at the particular slot after you have played your turn. Sample input: 3 3 2 111 8 3 10110111 8 4 Sample output: yes 101 no yes 1111 --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups google-codejam group. To post to this group, send email to google-code@googlegroups.com To unsubscribe from this group, send email to google-code+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code?hl=en -~--~~~~--~~--~--~---
[gcj] Re: how to solve this?
hello one request if any one know how to test the Topcoder problems offline please help me currently marathin56 is open i wanted to test my solution Regards Jawahar On Mon, Oct 12, 2009 at 4:41 PM, Brats rbharat...@gmail.com wrote: Hey .. here is a problem I saw somewhere on net. Can anyone give me some ideas on its algo ? There is a tray that contains a row of s slots where biscuits may be placed. You can have atmost one biscuit per slot. You and your friend play a game with it. Each of you get to pick biscuits in turns. You can pick atmost b biscuits in a single turn provided they are in consecutive slots with no gaps (empty slots) in between (but you have to pick atleast one in your turn). Whoever picks the last biscuit wins the game. You are given a snapshot of the game and its your turn. Assuming that from this point onwards, both the players play optimally. Find out if you can win the game. And if yes, what is your next move? If there are multiple options for the next move, choose the one where you pick the highest number of biscuits in the next turn. If still there are multiple options, choose the one where you take the leftmost biscuit among them. As input, you will be given 't' which indicates the number of test cases. For each case, you will be given s ( 1=s=2000 ), b ( 1=b=20 ) seperated by space and the next line will contain s characters that are either '0' or '1' representing each of the s slots. '0' indicates that a biscuit is not present at that particular slot and '1' indicates that a biscuit is present. As output, print yes or no indicating if you can win and if the answer is yes, print a string of s characters of either '0' or '1' that indicates whether a biscuit is absent or present at the particular slot after you have played your turn. Sample input: 3 3 2 111 8 3 10110111 8 4 Sample output: yes 101 no yes 1111 --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups google-codejam group. To post to this group, send email to google-code@googlegroups.com To unsubscribe from this group, send email to google-code+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code?hl=en -~--~~~~--~~--~--~---
[gcj] Re: how to solve this?
Some observations: Every position is either a win for you or your opponent - the game does not contain draws. Furthermore, the sub-game played from one consecutive set of biscuits is entirely independent of the other sub-games. A single '1' is always winnable (b = 1) A string of 1s is winnable if its length is less than b. A string of 0s is identical to a single 0 : 111111 is the same game as 1110111 A position is winnable if it contains an even number of unwinnable sub-positions. I think you can DP it from there. On Mon, Oct 12, 2009 at 12:11 PM, Brats rbharat...@gmail.com wrote: Hey .. here is a problem I saw somewhere on net. Can anyone give me some ideas on its algo ? There is a tray that contains a row of s slots where biscuits may be placed. You can have atmost one biscuit per slot. You and your friend play a game with it. Each of you get to pick biscuits in turns. You can pick atmost b biscuits in a single turn provided they are in consecutive slots with no gaps (empty slots) in between (but you have to pick atleast one in your turn). Whoever picks the last biscuit wins the game. You are given a snapshot of the game and its your turn. Assuming that from this point onwards, both the players play optimally. Find out if you can win the game. And if yes, what is your next move? If there are multiple options for the next move, choose the one where you pick the highest number of biscuits in the next turn. If still there are multiple options, choose the one where you take the leftmost biscuit among them. As input, you will be given 't' which indicates the number of test cases. For each case, you will be given s ( 1=s=2000 ), b ( 1=b=20 ) seperated by space and the next line will contain s characters that are either '0' or '1' representing each of the s slots. '0' indicates that a biscuit is not present at that particular slot and '1' indicates that a biscuit is present. As output, print yes or no indicating if you can win and if the answer is yes, print a string of s characters of either '0' or '1' that indicates whether a biscuit is absent or present at the particular slot after you have played your turn. Sample input: 3 3 2 111 8 3 10110111 8 4 Sample output: yes 101 no yes 1111 -- Paul Smith http://www.nomadicfun.co.uk p...@pollyandpaul.co.uk --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups google-codejam group. To post to this group, send email to google-code@googlegroups.com To unsubscribe from this group, send email to google-code+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code?hl=en -~--~~~~--~~--~--~---
[gcj] Re: how to solve this?
2009/10/12 Paul Smith paulsmithena...@gmail.com Some observations: Every position is either a win for you or your opponent - the game does not contain draws. Furthermore, the sub-game played from one consecutive set of biscuits is entirely independent of the other sub-games. A single '1' is always winnable (b = 1) A string of 1s is winnable if its length is less than b. A string of 0s is identical to a single 0 : 111111 is the same game as 1110111 A position is winnable if it contains an even number of unwinnable sub-positions. are you sure of this ? suppose there are just 2 symmetric subpositions .. its impossible to win this because whatever move i do, the opponent can do the same in the other sub-position and hence winning. this means irrespective of whether the subpositions are winnable or not, a problem having 2 symmetric subpositions is unwinnable. infact, just because a sub-position is unwinnable doesnt mean you will definitely lose the sub-position. the opponent may play it in such a way that you may still win the sub-position if that leads to his victory. this is where i don't know how to proceed. would i need to use 2 flags for winnable and losable ? I think you can DP it from there. On Mon, Oct 12, 2009 at 12:11 PM, Brats rbharat...@gmail.com wrote: Hey .. here is a problem I saw somewhere on net. Can anyone give me some ideas on its algo ? There is a tray that contains a row of s slots where biscuits may be placed. You can have atmost one biscuit per slot. You and your friend play a game with it. Each of you get to pick biscuits in turns. You can pick atmost b biscuits in a single turn provided they are in consecutive slots with no gaps (empty slots) in between (but you have to pick atleast one in your turn). Whoever picks the last biscuit wins the game. You are given a snapshot of the game and its your turn. Assuming that from this point onwards, both the players play optimally. Find out if you can win the game. And if yes, what is your next move? If there are multiple options for the next move, choose the one where you pick the highest number of biscuits in the next turn. If still there are multiple options, choose the one where you take the leftmost biscuit among them. As input, you will be given 't' which indicates the number of test cases. For each case, you will be given s ( 1=s=2000 ), b ( 1=b=20 ) seperated by space and the next line will contain s characters that are either '0' or '1' representing each of the s slots. '0' indicates that a biscuit is not present at that particular slot and '1' indicates that a biscuit is present. As output, print yes or no indicating if you can win and if the answer is yes, print a string of s characters of either '0' or '1' that indicates whether a biscuit is absent or present at the particular slot after you have played your turn. Sample input: 3 3 2 111 8 3 10110111 8 4 Sample output: yes 101 no yes 1111 -- Paul Smith http://www.nomadicfun.co.uk p...@pollyandpaul.co.uk --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups google-codejam group. To post to this group, send email to google-code@googlegroups.com To unsubscribe from this group, send email to google-code+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code?hl=en -~--~~~~--~~--~--~---
[gcj] Re: how to solve this?
2009/10/12 Bharath Raghavendran rbharat...@gmail.com 2009/10/12 Paul Smith paulsmithena...@gmail.com Some observations: Every position is either a win for you or your opponent - the game does not contain draws. Furthermore, the sub-game played from one consecutive set of biscuits is entirely independent of the other sub-games. A single '1' is always winnable (b = 1) A string of 1s is winnable if its length is less than b. A string of 0s is identical to a single 0 : 111111 is the same game as 1110111 A position is winnable if it contains an even number of unwinnable sub-positions. are you sure of this ? suppose there are just 2 symmetric subpositions .. its impossible to win this because whatever move i do, the opponent can do the same in the other sub-position and hence winning. this means irrespective of whether the subpositions are winnable or not, a problem having 2 symmetric subpositions is unwinnable. infact, just because a sub-position is unwinnable doesnt mean you will definitely lose the sub-position. the opponent may play it in such a way that you may still win the sub-position if that leads to his victory. this is where i don't know how to proceed. would i need to use 2 flags for winnable and losable ? here is an example : say b = 1 and the problem is 11011 ... both subpositions are unwinnable but yet the main position is also unwinnable. Infact, while trying to find this example, i realized that a problem with only '1's without gaps is unwinnable only if its even number of 1s and b=1. As any other case, you can split the problem into 2 symmetric sub-problems and mirror whatever move your opponent does. I think you can DP it from there. On Mon, Oct 12, 2009 at 12:11 PM, Brats rbharat...@gmail.com wrote: Hey .. here is a problem I saw somewhere on net. Can anyone give me some ideas on its algo ? There is a tray that contains a row of s slots where biscuits may be placed. You can have atmost one biscuit per slot. You and your friend play a game with it. Each of you get to pick biscuits in turns. You can pick atmost b biscuits in a single turn provided they are in consecutive slots with no gaps (empty slots) in between (but you have to pick atleast one in your turn). Whoever picks the last biscuit wins the game. You are given a snapshot of the game and its your turn. Assuming that from this point onwards, both the players play optimally. Find out if you can win the game. And if yes, what is your next move? If there are multiple options for the next move, choose the one where you pick the highest number of biscuits in the next turn. If still there are multiple options, choose the one where you take the leftmost biscuit among them. As input, you will be given 't' which indicates the number of test cases. For each case, you will be given s ( 1=s=2000 ), b ( 1=b=20 ) seperated by space and the next line will contain s characters that are either '0' or '1' representing each of the s slots. '0' indicates that a biscuit is not present at that particular slot and '1' indicates that a biscuit is present. As output, print yes or no indicating if you can win and if the answer is yes, print a string of s characters of either '0' or '1' that indicates whether a biscuit is absent or present at the particular slot after you have played your turn. Sample input: 3 3 2 111 8 3 10110111 8 4 Sample output: yes 101 no yes 1111 -- Paul Smith http://www.nomadicfun.co.uk p...@pollyandpaul.co.uk --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups google-codejam group. To post to this group, send email to google-code@googlegroups.com To unsubscribe from this group, send email to google-code+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code?hl=en -~--~~~~--~~--~--~---
[gcj] Re: how to solve this?
On Mon, Oct 12, 2009 at 1:13 PM, Bharath Raghavendran rbharat...@gmail.com wrote: 2009/10/12 Bharath Raghavendran rbharat...@gmail.com 2009/10/12 Paul Smith paulsmithena...@gmail.com Some observations: Every position is either a win for you or your opponent - the game does not contain draws. Furthermore, the sub-game played from one consecutive set of biscuits is entirely independent of the other sub-games. A single '1' is always winnable (b = 1) A string of 1s is winnable if its length is less than b. A string of 0s is identical to a single 0 : 111111 is the same game as 1110111 A position is winnable if it contains an even number of unwinnable sub-positions. are you sure of this ? No, in fact I am wrong :) -- Paul Smith http://www.nomadicfun.co.uk p...@pollyandpaul.co.uk --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups google-codejam group. To post to this group, send email to google-code@googlegroups.com To unsubscribe from this group, send email to google-code+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code?hl=en -~--~~~~--~~--~--~---
[gcj] Re: how to solve this?
You should use Sprague-Grundy theory to solve this problem. On Oct 12, 6:11 pm, Brats rbharat...@gmail.com wrote: Hey .. here is a problem I saw somewhere on net. Can anyone give me some ideas on its algo ? There is a tray that contains a row of s slots where biscuits may be placed. You can have atmost one biscuit per slot. You and your friend play a game with it. Each of you get to pick biscuits in turns. You can pick atmost b biscuits in a single turn provided they are in consecutive slots with no gaps (empty slots) in between (but you have to pick atleast one in your turn). Whoever picks the last biscuit wins the game. You are given a snapshot of the game and its your turn. Assuming that from this point onwards, both the players play optimally. Find out if you can win the game. And if yes, what is your next move? If there are multiple options for the next move, choose the one where you pick the highest number of biscuits in the next turn. If still there are multiple options, choose the one where you take the leftmost biscuit among them. As input, you will be given 't' which indicates the number of test cases. For each case, you will be given s ( 1=s=2000 ), b ( 1=b=20 ) seperated by space and the next line will contain s characters that are either '0' or '1' representing each of the s slots. '0' indicates that a biscuit is not present at that particular slot and '1' indicates that a biscuit is present. As output, print yes or no indicating if you can win and if the answer is yes, print a string of s characters of either '0' or '1' that indicates whether a biscuit is absent or present at the particular slot after you have played your turn. Sample input: 3 3 2 111 8 3 10110111 8 4 Sample output: yes 101 no yes 1111 -- Vladislav Isenbaev (winger) --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups google-codejam group. To post to this group, send email to google-code@googlegroups.com To unsubscribe from this group, send email to google-code+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code?hl=en -~--~~~~--~~--~--~---
[gcj] Re: Invitation for getting a visa
Our visa specialists have now sent email to the top 30 contestants (the top 25, plus backups in case some of them can't make it) in order to get things started. Please send me an email if you haven't been contacted. Note that we still haven't made the results of the round official, pending a search for rule-breakers. Cheers, Bartholomew On Sun, Oct 11, 2009 at 8:22 PM, Jeru jeru.sh...@gmail.com wrote: So is there a timetable for things like the date the invitation letter should arrive, the due to confirm VISA availability or something else? On Oct 11, 8:06 am, Bartholomew Furrow fur...@gmail.com wrote: We'll start contacting people shortly. On Sat, Oct 10, 2009 at 1:04 PM, e-maxx e-m...@inbox.ru wrote: How could I receive an invitation, which is a first document necessary to start obtaining a visa? --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups google-codejam group. To post to this group, send email to google-code@googlegroups.com To unsubscribe from this group, send email to google-code+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code?hl=en -~--~~~~--~~--~--~---
[gcj] Re: Invitation for getting a visa
Hi, Code Jam team, Can you please advise us on what to write in these fields of visa application form (from https://evisaforms.state.gov/ds156.asp): 22. When Do You Intend To Arrive In The U.S.? 24. At What Address Will You Stay in The U.S.? 25. Name and Telephone Numbers of Person in U.S. Who You Will Be Staying With or Visiting for Tourism or Business 26. How Long Do You Intend To Stay in The U.S.? 27. What is The Purpose of Your Trip? [Pariticipating in Google Code Jam - right?] 28. Who Will Pay For Your Trip? [Google, Inc.] (Assuming that I'm applying for a visa only to go to Code Jam.) On Oct 13, 12:13 am, Bartholomew Furrow fur...@gmail.com wrote: Our visa specialists have now sent email to the top 30 contestants (the top 25, plus backups in case some of them can't make it) in order to get things started. Please send me an email if you haven't been contacted. Note that we still haven't made the results of the round official, pending a search for rule-breakers. Cheers, Bartholomew On Sun, Oct 11, 2009 at 8:22 PM, Jeru jeru.sh...@gmail.com wrote: So is there a timetable for things like the date the invitation letter should arrive, the due to confirm VISA availability or something else? On Oct 11, 8:06 am, Bartholomew Furrow fur...@gmail.com wrote: We'll start contacting people shortly. On Sat, Oct 10, 2009 at 1:04 PM, e-maxx e-m...@inbox.ru wrote: How could I receive an invitation, which is a first document necessary to start obtaining a visa? --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups google-codejam group. To post to this group, send email to google-code@googlegroups.com To unsubscribe from this group, send email to google-code+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/google-code?hl=en -~--~~~~--~~--~--~---
