this project violated term of service:
http://code.google.com/p/quocmy-js/
The project owner (mai0p...@gmail.com) stole the code and post it as an open
source project, he even used it as a js hosting service for his website (
http://love.123yeu.org)
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Hi!
I cloned a google code repository and wanted to add some people to the
team so that they can push code on the mercurial repo.
But we can't do that! Why isn't the option in the admin panel?
I added an issue some months ago about it, it was never answered :
On Mon, May 9, 2011 at 5:26 AM, Tri Minh dtri2...@gmail.com wrote:
this project violated term of service:
http://code.google.com/p/quocmy-js/
The project owner (mai0p...@gmail.com) stole the code and post it as an
open source project, he even used it as a js hosting service for his website
Hi,
I created this project http://code.google.com/p/forum-keyboard-heroes/
and accidentally deleted myself from it. How there is no admin on the
project. Can anyone help?
Thanks
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On Mon, May 9, 2011 at 9:52 AM, Lucian timol...@gmail.com wrote:
I created this project http://code.google.com/p/forum-keyboard-heroes/
and accidentally deleted myself from it. How there is no admin on the
project. Can anyone help?
I have restored your access.
-Nathaniel
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On May 9, 6:13 pm, Nathaniel Manista nathan...@google.com wrote:
On Mon, May 9, 2011 at 9:52 AM, Lucian timol...@gmail.com wrote:
I created this projecthttp://code.google.com/p/forum-keyboard-heroes/
and accidentally deleted myself from it. How there is no admin on the
project. Can
On Mon, May 9, 2011 at 7:38 AM, MCMic come.bernig...@gmail.com wrote:
I cloned a google code repository and wanted to add some people to the
team so that they can push code on the mercurial repo.
But we can't do that! Why isn't the option in the admin panel?
I added an issue some months ago
I was browsing by when I saw this post and remembered about one from
last week:
http://groups.google.com/group/google-code-hosting/browse_thread/thread/6a061b075ddbf3ae
- Title: Managing Users for a Cloned project?
As for the Issue 3998, there was no answer, but someone did give it a
partial
I fail to understand the logic of the proof given in the 'contest
analysis'.
For the three unsorted numbers 3 1 2, the sample set of arrangements
is-
1 2 3
1 3 2
2 1 3
2 3 1 - Not even one in correct position
3 1 2 - Not even one in correct position
3 2
Where did you read probability of getting at least one number in its
correct position is becoming 1.0? It says expected number of
elements that end up in the correct position is N * 1/N = 1..
ulzha
On May 9, 9:38 am, Eagle khirwad...@gmail.com wrote:
I fail to understand the logic of the
@Eagle
Have you even bothered to read what I wrote about the same case, just
before you?...
On May 9, 7:38 am, Eagle khirwad...@gmail.com wrote:
I fail to understand the logic of the proof given in the 'contest
analysis'.
For the three unsorted numbers 3 1 2, the sample set of arrangements
Oh yeah! OK, I guess it was me that wasn't reading carefully
enough :-)
Sam.
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@* Nzeyimana Antoine*
Large sets represent all possible cases but needs intelligent algorithm.
however ; you may solve small sets by brute force ; so it is still examines
your talent in solving problems via programming
so both are required to give the chance for the beginners to test their
ability
@Pedro,
In your last two steps,
E[n=3] = 2 + 1/3 * E[n=3]
E[n=3] = 3
there is something wrong. How are you eliminating E[n=3] on the right
hand side of the equation?
Eagle
On May 9, 12:28 am, Pedro Osório mebm.pedroso...@gmail.com wrote:
Hi Ricardo,
For your example regarding 3,
This is incorrect. The expectation for sorting 3 numbers is 3 hits, not 4.
The expected value of an event is the value of each outcome multiplied
by the probability of each outcome, all totalled.
With 3 unsorted numbers you have:
1/6 chance that 3 numbers will become correct - EV = 0.5
1/2
Hello
Can anybody kindly upload prob C output or let me know how my output is
wrong ?
Am sure my logic is right. [Comparing xor's of the 2 divided group such that
they are same and printing highest such possible sum as value]
Reg, robin.
On Sun, May 8, 2011 at 12:50 PM, vivek dhiman
Let x = E[n=3].
We have an equation of the form x = 2 + x/3.
Solve for x.
Paul Smith
p...@pollyandpaul.co.uk
On Mon, May 9, 2011 at 3:10 PM, Eagle khirwad...@gmail.com wrote:
@Pedro,
In your last two steps,
E[n=3] = 2 + 1/3 * E[n=3]
E[n=3] = 3
there is something wrong. How are
Everyone gets different inputs I think, so you probably need to post
the input as well as the output.
In fact the easiest thing to do is download someone else's answer, run
their answer, run your answer, and compare for differences.
Paul Smith
p...@pollyandpaul.co.uk
On Mon, May 9, 2011 at
Oh! thanks!
On May 9, 7:16 pm, Paul Smith p...@pollyandpaul.co.uk wrote:
Let x = E[n=3].
We have an equation of the form x = 2 + x/3.
Solve for x.
Paul Smith
p...@pollyandpaul.co.uk
On Mon, May 9, 2011 at 3:10 PM, Eagle khirwad...@gmail.com wrote:
@Pedro,
In your last two
As I understand the probability, it goes something like this- If
you toss a coin 100 times, it is 'expected' (actual result may be
different) that 50 times you will get 'Heads' 50 times 'Tails'. In
present problem also, you have to answer the 'expected' number of hits
by Goro to sort the
Hi Vivek,
Are you satisfied now..?
You got replies from good programmers.Vexorian is one of the problem writer
on TopCoder.He suggested practice.
Do practice..try to solve problems on TopCoder...
Solve all 250 points problems there and look at your confidence.And most
important do it for fun...
Thank you all! :)
Regards
Vivek Dhiman
On Mon, May 9, 2011 at 8:14 PM, Satyajit Bhadange
satyajit.bhada...@gmail.com wrote:
Hi Vivek,
Are you satisfied now..?
You got replies from good programmers.Vexorian is one of the problem writer
on TopCoder.He suggested practice.
Do practice..try
In order to make it clear, it is better to start from permutation of just 2
elements.
Let’s first prove that average number of permutations in case of 2 unsorted
elements is 2.0.
It is not so obvious, by the way.
Initial position is {2,1}. After one Goro’s hit it may turn to {1,2} or
remain
@Paul
Thanks for answering for me
@Eagle
Glad I could help =)
On May 9, 3:10 pm, Eagle khirwad...@gmail.com wrote:
@Pedro,
In your last two steps,
E[n=3] = 2 + 1/3 * E[n=3]
E[n=3] = 3
there is something wrong. How are you eliminating E[n=3] on the right
hand side of the
This was my favorite problem, since the solution is so easy
and yet many people were confused.
They also gave a complex example for the 3rd sample input even though
there is never a need to hold down any element not in it's position.
Statistically this was the hardest problem with only 58% solving
Hi,
Any ideas what should we practice ??? I mean any tips ?
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Thatz exactly what i did... :)
On May 8, 5:49 pm, ruslanb ruslan.berez...@gmail.com wrote:
For example, lets take a problem C. If you is know what is a bit xor
operand, you code a problem in 10 lines and 2min. You may be a great
developer but dont have no idea what is a xor function. In that
Dear Friends,
I need to a program to find all sorted subsets of a set. There are some
recursion solutions for generating subsets, but those are not sorted.
However I attend a solution to show me all of sorted subsets of a set.
Assume, set includes of numeric values.
For example:
input numbers
Don't forget, that professionals also take participate in Google Code Jam,
they can also have experience of 10 years in programming.
On Mon, May 9, 2011 at 8:14 PM, Satyajit Bhadange
satyajit.bhada...@gmail.com wrote:
Hi Vivek,
Are you satisfied now..?
You got replies from good
To know more about C Sharp Basics with Examples in very easy way then
please visit:
http://csharpexpress.blogspot.com/
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To
@meir
It was my favorite problem too, but the problem is not easy in the
sense that proving that the solution is correct is far from trivial.
However since in these contests one normally uses their intuition and
pattern recognition to come up with a solution fast and not full
mathematical proofs,
Hello all,
I think that adding at least one large case with the correct output would
help out clear many confusions of different problems, since in many cases
the given test cases are less than enough.
Regards,
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Hello,
here is an slightly alternative argument to see why on average a
permutation
brings one element into its correct position:
Hitting the table with n elements not pinned down is equivalent
to drawing a random permutation of length n (and it does not matter
in which order the non-pinned down
Quick and dirty Java code:
import java.util.Arrays;
public class Subset {
public static void main(String[] args) {
int[] set = { 1, 2, 3, 4 };
boolean[] mask = new boolean[set.length];
int[] picks = new int[set.length];
Hi Amir,
could you give some details about the solution you have thought so far?
regards
On Mon, May 9, 2011 at 10:56 AM, Amir Hossein Sharifzadeh
amirsharifza...@gmail.com wrote:
Dear Friends,
I need to a program to find all sorted subsets of a set. There are some
recursion solutions for
Yes, but no, I feel that it is part of the contest to test if you foresee
the possible cases based on the problem description, more than meet the
eyes, regards
On Mon, May 9, 2011 at 5:25 AM, Mohammed Mahmout raio...@gmail.com wrote:
Hello all,
I think that adding at least one large case with
Oops, pardon my previous reply (misunderstood you question correctly).
I think this one would suffice:
import java.util.Arrays;
public class Subset {
public static void main(String[] args) {
int[] set = { 1, 2, 3, 4 };
int[] picks = new int[set.length];
acrush is not ACRush, GCJ nickname is case sensitive. :)
On Sun, May 8, 2011 at 8:33 PM, ravi shanker ravishanker@gmail.comwrote:
I dont know about others.But AC RUSH is participating this time
too...u caN locate him if u aarange the contestent country
wise(china)...with the help of
I don't agree that being a professional programmer for many years can help
you be a better coder at a programming contest. I think that the only way of
being good at programming contests is practicing and participating in
contests.
For those who know TopCoder, last year I solved a lot of problems
This is a C++ implementation assuming that vectorint numbersOfTheSet has
the elements of the set sorted and your result will be each vector in
vectorvectorint solution. If you want to solve it with sets instead of
vector it's the same, you only have to copy the elements of the set to a
vector and
Thanks,
I thought both small and large sets represent all possible cases, the
only difference being the data size, thus affecting the program
running time.
My idea is that an algorithm that solves the small sets, should be
able to solve the large sets as well if only the running time and
memory
I'd like to make a discussion about possible reasons for failing the Large
Data Set after passing the Small one for preventing people of making
mistakes that lead to a time expired or a wrong answer for the Large
input.
Some possible reasons I have in mind:
Time Limit: Your code is not so fast
There is a beautiful technique of generating subsets for a set of not more
than 20 elements (Why , u'll understand later as u read the code ) .
This code will just generate all the subsets of a given set . The sorted
condition and other stuff can b done easily .
for(int i=0;ipow(2,n);i++)
{
Another reason may be that, regardless of n, the small input doesn't
include a certain tricky case where your solution fails, but the
large input does.
On May 9, 8:51 pm, Leopoldo Taravilse ltaravi...@gmail.com wrote:
I'd like to make a discussion about possible reasons for failing the Large
I understood how to solve this, but *how does one come to this solution ?*
*
If the xor of all numbers is zero, you can pick any candy, and the xor to
this number is going to be equal to the xor from the rest of them.*
I get this, if I have 9 numbers with XOR 3, XORing it with 3 will get me
zero.
XOR is the bread and butter of Symmetric Cryptography.
~mhb
On Mon, May 9, 2011 at 4:44 PM, Marcelo Ramires
marcelo.f.rami...@gmail.com wrote:
I understood how to solve this, but how does one come to this solution ?
If the xor of all numbers is zero, you can pick any candy, and the xor to
For most of problems I would agree with you, but there are a few examples
that can't be put in these conditions.
A good example is this problem:
http://code.google.com/codejam/contest/dashboard?c=635102#s=p3
Here, the small data set is a special case that can be solved by a simpler
algorithm
u can do a lot of stuff with the bitwise operators , u just gotta experiment
with it
On Tue, May 10, 2011 at 2:21 AM, Morgan Bauer bauer.mor...@gmail.comwrote:
XOR is the bread and butter of Symmetric Cryptography.
~mhb
On Mon, May 9, 2011 at 4:44 PM, Marcelo Ramires
An alternative use for XORing numbers greater than 1: The game of Nim. I
think this is where I first learned about the operation.
On 9 May 2011 21:44, Marcelo Ramires marcelo.f.rami...@gmail.com wrote:
I understood how to solve this, but *how does one come to this solution ?*
*
If the xor of
Hi Marcelo
i guess you already answered part of your question how does one come
to this solution?
as you have already mentioned
If the xor of all numbers is zero, you can pick any candy, and the
xor to
this number is going to be equal to the xor from the rest of them.
one can come to this solution
You are seriously mistaken. Several kinds of experience help a great
deal on facing programming contest. I agree that training has the most
impact on your performance, but experience in coding, theoretical
knowledge, organizational skills, academical experience with different
kinds of proofs and
On Mon, May 9, 2011 at 17:44, Marcelo Ramires
marcelo.f.rami...@gmail.comwrote:
I understood how to solve this, but *how does one come to this solution ?*
*
If the xor of all numbers is zero, you can pick any candy, and the xor to
this number is going to be equal to the xor from the rest of
.. for the line, Algorithms are not Goro's strength; strength is
Goro's strength.
It still makes me laugh every time I think about it!
:-)
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I agree, that line had me laughing for a long time.
Carlos Guía
On Mon, May 9, 2011 at 9:54 PM, royappa roya...@gmail.com wrote:
.. for the line, Algorithms are not Goro's strength; strength is
Goro's strength.
It still makes me laugh every time I think about it!
:-)
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for fast processing we used the buffer.
you know the speed of I/O device is very slow in comparison to CPU
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Hi guys,
I've seen that there are three rounds 1A, 1B and 1C and the top1000
goes to round 2.
Does it means that the top1000 counting the three rounds goes or the
top1000 of each one?
Thanks a lot
Lara
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CodeChef.com is also a good programming contest platform.
On Tue, May 10, 2011 at 12:20 AM, Leopoldo Taravilse
ltaravi...@gmail.comwrote:
I don't agree that being a professional programmer for many years can help
you be a better coder at a programming contest. I think that the only way of
Thanks :)
On Sun, May 8, 2011 at 10:38 PM, Vexorian vexor...@gmail.com wrote:
In c++:
string s =ABCXDEF;
sort(s.begin(), s.end());
do {
cout s;
} while (next_permutation(s.begin(), s.end() );
For languages that don't have the library built in, or if you really
want to code it
A few thoughts:
It says Goro is sorting N integers, but not that the integers are
1...N. Are the GCJ question setters being kind by only using arrays
of 1...N in their test cases?
It says Goro wants to sort the integers, but doesn't require the
sorted order to be increasing. How would the
Hi,
One could come to this solution by understanding the Sean adding skills,
1100
+ 0101
--
1001
compare this with the XOR truth table and you will have a match.
As others posted before practice will help us to see these patterns while
reading the problems.
(Personally I didn't use an
In 2008 someone asked if sources could be submitted as .tar.gz files. The
answer at the time was zip
onlyhttp://groups.google.com/group/google-code/browse_thread/thread/d13febe8b36b3d1c/005d1940b1a01276?lnk=gstq=zip#005d1940b1a01276.
Is that still the case?
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It says Goro is sorting N integers, but not that the integers are
1...N. Are the GCJ question setters being kind by only using arrays
of 1...N in their test cases?
The second line of each test case will contain a permutation of the
*N* smallest
positive integers.
Did anyone else go through
Now D(n,m) is also known as Rencontres number (see
e.g.http://en.wikipedia.org/wiki/Rencontres_numbers) and when one
looks at one of the references in this article (http://oeis.org/
A008290)
one sees in the 'formula' section that the above expression
is indeed 1 (no proof is given there
hello,
i am now 3rd year BE ,information technology, from bengal engineering and
science university,shibpur that will be upgraded to
indian institute of engineering , science and technology(IIEST).
i want to learn the mathematical tools for progamming, i have done some
problem from project euler
For those of you who are interested, here is a little more background
on Linearity of Expectation, which is what the analysis uses.
A random variable is any measurement you make after some random
events have occurred. The average value discussed in this problem is
officially called expected
And me.
I still think that this line was written by Goro.
- Gennadiy gennad.zlo...@gmail.com
On Tue, May 10, 2011 at 11:38 AM, Carlos Guia zyx3d...@gmail.com wrote:
I agree, that line had me laughing for a long time.
Carlos Guía
On Mon, May 9, 2011 at 9:54 PM, royappa roya...@gmail.com
Top 1000 of each round...totally 3000
On Mon, May 9, 2011 at 11:42 AM, alara a.lara0...@gmail.com wrote:
Hi guys,
I've seen that there are three rounds 1A, 1B and 1C and the top1000
goes to round 2.
Does it means that the top1000 counting the three rounds goes or the
top1000 of each one?
In the general case, all you need to do is find the bijection between the
current array and the desired array, and then work out the cycle structure
of this bijection.
This yields the cycles that are out of place, and you add these up (which
will always total the number of non-fixed elements).
I wonder why his name was choosen to be Goro.
Anyway, he isn't that bad, since his worst case average run time is
n(if he always freeze any number that gets into the right position),
which would be nlgn for comparison based approaches.
On 5/10/11, Gennadiy Zlobin gennad.zlo...@gmail.com wrote:
Mortal Kombat. Goro is a character with four arms.
~mhb
On Tue, May 10, 2011 at 1:36 AM, Nzeyimana Antoine anthonz...@gmail.com wrote:
I wonder why his name was choosen to be Goro.
Anyway, he isn't that bad, since his worst case average run time is
n(if he always freeze any number that gets
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