http://www.spoj.pl/problems/PIGBANK/
can anyone give me an idea how to solve this problem...?? I dont think
the knapsack algo would be of help here as here we need to find
minimum value..please refer to the link and if anyone can help, i
would be very thankful.
regards,
aksha
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@dave..Can you please explain your logic ..
Regards,
Ashish
On Thu, Feb 24, 2011 at 6:32 AM, Dave dave_and_da...@juno.com wrote:
Try this:
int i,k,n;
long long j,nsq;
for( n = 31623 ; n 10 ; ++n )
{
nsq = (long long)n * (long long)n;
I think..
As like no are a,b,c,d,e
so sum will be
a+b,a+c,a+d,a+e,b+c,b+d,b+e,c+d,c+e,d+e;
so maximuum value will be d+e which is last element of array given
take last three value
1.c+d
2.c+e
3.d+e
eq(1)-eq(2)=d-e;
solving it with 3rd eq will give d and e
and with these value we can get other
There must be another good solution..please let me know .
Thanks
On Thu, Feb 24, 2011 at 5:09 PM, ashish agarwal
ashish.cooldude...@gmail.com wrote:
I think..
As like no are a,b,c,d,e
so sum will be
a+b,a+c,a+d,a+e,b+c,b+d,b+e,c+d,c+e,d+e;
so maximuum value will be d+e which is last
The only way to get an acute triangle this way is to connect the
diagonals of three adjacent faces. You can select 3 adjacent faces of
a cube in (6*4*2)/(3*2*1) = 8 different ways.
Regards
Priyaranjan
http://code-forum.blogspot.com
On Feb 23, 12:10 am, jalaj jaiswal jalaj.jaiswa...@gmail.com
The basic solution which is coming to the mind is to covert string first
palindrome and apply livishthein distance to both string(original one and
changed string) to check how many substiutions you require for the
palindrome.
On Wed, Feb 23, 2011 at 9:11 PM, radha krishnan
how to solve it using DP??
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Last three values could be:
1.b+e
2.c+e
3.d+e
On Thu, Feb 24, 2011 at 5:09 PM, ashish agarwal
ashish.cooldude...@gmail.com wrote:
I think..
As like no are a,b,c,d,e
so sum will be
a+b,a+c,a+d,a+e,b+c,b+d,b+e,c+d,c+e,d+e;
so maximuum value will be d+e which is last element of array given
@priyaranjan
No,Your triangle will be right angle triangle.
In fact any triangle chosen will be right angle triangle.
Proof:
suppose there are 2 planes kept at parallel with each other.(top bottom)
join corresponding vertices of up to down and form cube.
Now 3 points chosen can be on same
a small modification in normal knapsack algo ll do :)
On Thu, Feb 24, 2011 at 4:06 PM, Akshata Sharma
akshatasharm...@gmail.comwrote:
http://www.spoj.pl/problems/PIGBANK/
can anyone give me an idea how to solve this problem...?? I dont think
the knapsack algo would be of help here as here we
@Priyaranjan: Suppose that the cube has side 1 and is placed in
standard position at the origin. Then the triangle with vertices
(0,0,0), (1,0,1), and (1,1,0) is equilateral with side sqrt(2), and
therefore is not a right triangle.
Dave
On Feb 24, 8:49 am, Vikas Kumar dev.vika...@gmail.com
@Ashish: The code seems pretty straightforward, but okay. The outer
for-loop runs through the numbers that have 10-digit squares. The
squares are represented as 64 bit integers (type long long) because
some of them are larger than the maximum representable 32-bit
integers. For each number, the
How do Levenshtein distance used.. For that u need to know the palindrome
that is closest to it.. and if that is know than there is no point in
calculating the distance .. we can easily see how many changes are to be
made..!(correct me if I am wrong)
Further my approach is this :
taking the
Given a string (assume there no spaces or punctuations), write a
program that returns the max. length of the string that has repeated
more than once.
Thanks
Shashank
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@Bittu: Your statement of the problem doesn't make any sense.
Apparently, you are given a string and somehow that string is
repeated. Can you clarify it and give an example?
Dave
On Feb 24, 10:24 am, bittu shashank7andr...@gmail.com wrote:
Given a string (assume there no spaces or
hey guys... please tell me how to run OpenGl on eclipse. thanxx in
advance..
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Declare it as *static.*
On Wed, Feb 23, 2011 at 11:33 PM, Jammy xujiayiy...@gmail.com wrote:
Are you talking about IPC?
On Feb 22, 10:05 am, jaladhi dave jaladhi.k.d...@gmail.com wrote:
What do you mean by data element here ? Also by file you mean the file
where
you wrote the code ? And
hey guys.. anyone working on implementing object tracking in open cv.
please get back to me
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Hello,
Can you help me with tutorial about the segmentation of a text
document ?
I prefer the C + + but if it were in another language is no problem.
Thank you very much.
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If you had 5,623 participants in a tournament, how many games would
need to be played to determine the winner
According to me if Tournament strategy is is used then i think its
ok...
After each round, you would have half the number that started the
previous round; except if it were an odd
we have two sorted array a[]={2,6,9,60}; b[]={1,3,5,34,80}; merge the
array in such way..
a[]={1,2,3,5}; b[]={6,9,34,60,80}; ..no extra space is allowed..i.e.
In-Place merging
Many of you thinks its easy..but here is q. of minimum complexity i
have done this but min e complexity high that not
Given an ancestor matrix for an binary tree where a[i][j]=1 if i is
parent
to j, else a[i][j]=0 create the binary tree. from given ancestor
matrix
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Algorithmor program or any approach will be appreciated
Thanks Regards
Shashank
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Using Dynamic programming.
Divide array into two parts a and b. Run the minimum edit distance
solution for each pair of (a,b)(except this time for b we need to
start from the end of the array). Find the minimum among those.
On Feb 24, 10:41 am, sukhmeet singh sukhmeet2...@gmail.com wrote:
How
Simpler. Every game eliminates one participant. Since 5,622
participants must be eliminated to have one winner, it takes 5,622
games.
Dave
On Feb 24, 5:43 pm, bittu shashank7andr...@gmail.com wrote:
If you had 5,623 participants in a tournament, how many games would
need to be played to
@dave i never said it will be a right angled triangle ,it will be a
equilateral triangle and thus an acute angled triangle.
On Feb 24, 8:38 pm, Dave dave_and_da...@juno.com wrote:
@Priyaranjan: Suppose that the cube has side 1 and is placed in
standard position at the origin. Then the triangle
@vikash kumar please read my answer then give your comment..it will be
a equilateral triangle and not a right angled triangle i.e an acute
angled triangle.
On Feb 24, 7:49 pm, Vikas Kumar dev.vika...@gmail.com wrote:
@priyaranjan
No,Your triangle will be right angle triangle.
In fact any
Dave's solution is best if numerical error is possible.
If the points are precise, you can also do it in linear time. Just hash the
points on abs(y/x).
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@Priyaranjan. Right. I should have directed my comment to Vikas.
Sorry.
Dave
On Feb 24, 9:35 pm, awesomeandroid priyaranjan@gmail.com wrote:
@dave i never said it will be a right angled triangle ,it will be a
equilateral triangle and thus an acute angled triangle.
On Feb 24, 8:38 pm,
@Vikas: Suppose that the cube has side 1 and is placed in standard
position at the origin. Then the triangle with vertices (0,0,0),
(1,0,1), and (1,1,0) is equilateral with side sqrt(2), and therefore
is not a right triangle.
Dave
On Feb 24, 8:49 am, Vikas Kumar dev.vika...@gmail.com wrote:
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