I want to calculate all binomial coeffficient upto a particular number say
1000.,like 1000C1 100C2
1000C3...999C1,999C2..998C12C1 1C1.
all possible binomial coefficient out of 1000*1000 modulo 17;I want
a maximum complexity of o(n^2);for 1000,i had some code of mine
use pascal's triangle
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Dear All,
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Its quite trivial..it just if there's a shorter way to reach from index j
and k by using any of the nodes as intermediate
Saurabh Singh
B.Tech (Computer Science)
MNNIT
blog:geekinessthecoolway.blogspot.com
On Sat, Mar 3, 2012 at 5:59 PM, shady sinv...@gmail.com wrote:
Can someone explain
Gene:
I am talking about file names.
On Sat, Mar 3, 2012 at 1:07 AM, Gene gene.ress...@gmail.com wrote:
It's possible you're not getting any clear answers because the
question is unclear. Linux does many different kinds of name lookup
all over the system. What names are you talking about?
check out this link :-
http://groups.google.com/group/algogeeks/browse_thread/thread/7ed957a81a426f72/fe45b6ce76e1cac5?hl=enlnk=gstq=Modular+arithmetic+%2B+Combinatorics#fe45b6ce76e1cac5
On Sat, Mar 3, 2012 at 5:56 PM, jai gupta sayhelloto...@gmail.com wrote:
use pascal's triangle
--
You
i miss type recursion it is iterative algo but recursion can be use.
On Sat, Mar 3, 2012 at 6:29 PM, amrit harry dabbcomput...@gmail.com wrote:
use recursion
c[1][1]=1;
c[2][1]=1;
c[2][2]=1;
for(i=0;in;i++)
for(j=0;(2*i)-1;j++)
{
c[i][j]=(c[i-1][j]+c[i-1][j+1])
if(c[i][j]P)
use recursion
c[1][1]=1;
c[2][1]=1;
c[2][2]=1;
for(i=0;in;i++)
for(j=0;(2*i)-1;j++)
{
c[i][j]=(c[i-1][j]+c[i-1][j+1])
if(c[i][j]P)
c[i][j]-=P;//dont use module because it is slow we know the value never
exceede 2*P because we are adding 2 variable so just use subtract operation.
}
On Sat, Mar 3,
@jai-thanks!!! , it worked...,is there any method for big numbers like
10,in my case 1000 so storing in array[10^3][10^3] worked but what if
10^5,storing will not work then??,anyway thanks a lot again
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Algorithm
consider the loop as rep(k,n) rep(i,n) rep(j,n) if(dp[i][k]
dp[k][j])dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j])
here we take one node (first for loop) and check if the cost of moving from
i-j gets reduced on choosing k as intermediate node.
On Sat, Mar 3, 2012 at 6:26 PM, saurabh singh
The Wikipedia entry is pretty good:
http://en.wikipedia.org/wiki/Floyd–Warshall_algorithm
Read the Algorithm section a few times and draw some examples and
you'll probably start seeing it.
On Mar 3, 7:56Â am, saurabh singh saurab...@gmail.com wrote:
Its quite trivial.
At least until you have
@Pankaj: You can do with a one-dimensional array since you can update the
array one row at a time. It would look like this:
int P = 17;
c[0] = 1;
for( j = 1 ; j = n ; ++j )
{
// at this point, c[i]. for i = 0, 1, 2, ..., j-1, holds (j-1) choose i
t = c[0];
for( i = 1 ; i j ;
@Don: I've looked for a description of Jonker-Volgenant-Castanon, but
haven't found one. Can you provide a reference?
Dave
On Tuesday, February 28, 2012 11:53:22 AM UTC-6, Don wrote:
Dave's answer, the Hungarian Algorithm, is correct because it does
meet the requirements of the problem.
@amrit- thnks!!
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@atul-its nice one, but its complexity is probably greater than o(n^2)
just to calculate one mCk,i want a sequence of all mCk in a
less complexity
i tried for some code which calculate all mC1,mC2..mCk in less
complexity.this works good till (100)Ck ;
if while loop can be reduced i can get a
@atul-its nice one, but its complexity is probably greater than o(n^2) just
to calculate one mCk,i want a sequence of all mCk in a
less complexity
i tried for some code which calculate all mC1,mC2..mCk in less
complexity.this works good till (100)Ck ;
if while loop can be reduced i can get a
Google is visiting our campus 4 different roles As of now IT field
technician is confirmed so how to approach 4 written test. ??
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