This is a 'finite calculus' (differences summations) problem.
You can solve it using difference operator (actually its inverse which gives
you the discrete integration which is nothing but summation).
If you do not know finite calculus, Google for it (or refer Concrete
Mathematics by Knuth).
The
On Sun, Oct 11, 2009 at 6:40 PM, Gautham Muthuravichandran
gautha...@gmail.com wrote:
9.. All the factorials above 5! is divisible by 9.
Divisible by 9 does not mean exactly 9.
-Gautham
On Sun, Oct 11, 2009 at 11:54 AM, Debanjan debanjan4...@gmail.com wrote:
On Oct 11, 10:29 am,
didn't get anything plz elaborate
On Oct 10, 10:44 am, Prunthaban Kanthakumar pruntha...@gmail.com
wrote:
Sterling numbers of second kind.
On Sat, Oct 10, 2009 at 10:41 AM, vicky mehta...@gmail.com wrote:
example:
n=10,k=10
ans:1
n=30,k=7
ans:
475020
On Oct 10, 9:51
Sterling numbers of second kind.
On Sat, Oct 10, 2009 at 10:41 AM, vicky mehta...@gmail.com wrote:
example:
n=10,k=10
ans:1
n=30,k=7
ans:
475020
On Oct 10, 9:51 am, vicky mehta...@gmail.com wrote:
u have to color n similar balls with k diff. colors , such that every
color must be
Here is the right answer:
Find the sum of missing numbers. Call it S (this is a easy to do).
Now the two missing numbers are such that one is =S/2 and the other is
S/2
Have two variables S1 and S2, traverse the array and add everything = S/2
to S1 and S/2 to S2.
Now
First number = (Sum of
Try {2, 1}
On Sat, Aug 1, 2009 at 11:45 AM, Devi G devs...@gmail.com wrote:
@Vivek
I had told abt tat border case already once..
Suppose the two missin numbers are greater than n, then m==0 when exitin
the loop.
So they will be n+1 and n+2 only.
in case, one of the missin numbers is
On 3/25/07, Rajiv Mathews [EMAIL PROTECTED] wrote:
On 3/25/07, Prunthaban Kanthakumar [EMAIL PROTECTED] wrote:
If you see carefully his proof does not assume anything about sections
colored continuously. His proof assumes only one thing Half of them
are
red and half of them are white
Ouch I got the question completely wrong assuming the inner disc is
continuous.Sorry for the confusion.
On 3/25/07, Prunthaban Kanthakumar [EMAIL PROTECTED] wrote:
On 3/25/07, Rajiv Mathews [EMAIL PROTECTED] wrote:
On 3/25/07, Prunthaban Kanthakumar [EMAIL PROTECTED] wrote:
If you
not understand -
For each inner section,no matter white or black ,there is 100
color-matching events.
Can somebody explain?
~Vishal
On 3/25/07, Prunthaban Kanthakumar [EMAIL PROTECTED] wrote:
Ouch I got the question completely wrong assuming the inner disc is
continuous.Sorry
Hi,
The proof given by Vishal is correct irrespective of the arrangement (which
he himself did not realize).
If you see carefully his proof does not assume anything about sections
colored continuously. His proof assumes only one thing Half of them are
red and half of them are white
Half does not
Try googling for Josephus Permutation
On 2/22/07, Manish Garg [EMAIL PROTECTED] wrote:
hi,
I m posting a game, its like this:
Suppose N people are playing the game. All of them are numbered from 1 to
N. They all sit in a circle such that their numbering order is also
maintained, so that
Then you will get only the following numbers
(1*7)/5 = 1
(2*7)/5 = 2
(3*7)/5 = 4
(4*7)/5 = 5
(5*7)/5 = 7
What you will do to get 3 and 6?
On 2/1/07, pramod [EMAIL PROTECTED] wrote:
Why can't we simply take the 1..5 random number, multiply by 7 and
divide by 5.
XOR is the best computationally as well as space complexity.
On 1/16/07, Satish [EMAIL PROTECTED] wrote:
Hangjin Zhang wrote:
Do an XOR on all numbers. The resulte is the number which occurs only
once
HZ
On 12/30/06, Abhishek [EMAIL PROTECTED] wrote:
Hi,
Suppose I have a
Hi all,
This does not look like a 'open to all' contest.
The registration asks for college information!
Can CEG guys confirm?
Regards,
Prunthaban
On 12/26/06, Aravind Narayanan [EMAIL PROTECTED] wrote:
Hello There!
College of Engineering, Guindy announces the Kurukshetra Online
Programming
to the binary search.
regards
Arunachalam.
On 10/10/06, Prunthaban Kanthakumar
[EMAIL PROTECTED] wrote:
hi,
Giving a short thought...
I think an O(n^2) greedy solution will work.
Regards,
Prunthaban
On 10/10/06, Amal [EMAIL PROTECTED]
wrote:
http://www.spoj.pl/problems/BOOKS1/ For this problem
= 42 (This is lesser than previous one)
So the optimum one including 4 is,is 4+7+8
Now appy the same O(n) procedure to both small half arrays...
On 9/26/06, GoCooL [EMAIL PROTECTED] wrote:
Prunthaban Kanthakumar wrote: Hi, The proposed solution does not talk about multiplication. It talks about
Hi,
The proposed solution does not talk about multiplication. It talks about the sum alone.
The sum is simply a[m].
Of course you need to muliply later.But that does not affect the correctness of the solution!
Regards,
Prunthaban
On 9/24/06, GoCooL [EMAIL PROTECTED] wrote:
This is the original
I didn't look into thecorrectness of the algorithm.
But I can tell you the reason for that compilation error
Visual Studio 6.0 does not follow ANSI standards!
Especially the scope of the loop variable 'i' is the thing which causes compilation error for you.
In ANSI C (or gcc for that matter),
if
When you have a nice algorithm from Googmeister in O(n log n) why try something else...?
On 7/5/06, mg [EMAIL PROTECTED] wrote:
#includestdio.hstruct key {int min;doublesum;};
int main(int argc,char **argv){int n,i,j,k;int start=0,end=0;double currentValue=0,eValue=0;struct key
You can't sort using push-pop-findmin.
@rajivmat
push --- 4, 5, 6, 1, 3, 2- minarray = {0,3}pop -- 4, 5, 6, 1, 3 - minarray = {0,3}pop -- 4, 5, 6, 1- minarray = {0,3}find_min = 3rd element = 1pop -- 4, 5, 6 - minarray{0} =(Now top = mintop so minarray is also popped)
find_min = 0th element = 4pop
Hi All,One of my friends asked me this questions:Can you frame a Stack like Data structure where Push,Pop and find_min takes O(1) time?Well...A Stack is already doing a good job by giving O(1) for Push and Pop.
We need a method to make it support O(1) find_min too.Let me share my ideas. Comments
Hi All,If I have understood the question correctly, then this will work! DFS is recursive. So even though our tree has no parent pointers, we will get such a 'pointer for free' , due to recursion. (Remember recursion allows you to pass information from child to parent)
So a single highly pruned
Hi all,I think there can be one more solution... where a BFS is folowed by DFS works.Algo 2:IsLockable(Node N){ queue Q; if(root is not locked) Q.insert(root); while(!Q.empty()) {
Node n = Q.removeHead(); if(n == N) return DFS(n); for(each child a of N) If(a is not locked) Q.insert(a); } return
Hi all,The third solution is... (I hope there will not be any fourth ;) )Two BFSs...1. The first BFS will traverse all unlocked nodes (same as solution one)...2. The second BFS will traverse until one locked node is encountered.
Hey, guys... this is boring...It is always possible to convert
No,
dp[2] is
100
110
010
011
This is3 * dp[1] + 1 = 3 * 1 + 1 = 4 steps
Regards,
Prunthaban
Use console application
System calls are calls which modify the process,etc (Like fork() in unix)
Win32 calls - (Accessing win32 api - Don't include windows.h)
On 2/3/06, prick [EMAIL PROTECTED] wrote:
hi everyoneRules for round 3 code4billf.Headers/Libraries: You can use the standard libraries
Hi all,
Sorry for the late reply. Anyway here is the solution...
I want to note down 2 facts...
1. With 1 one the maximum reachable is 1
2. Any sequence of move is exactly reversible. You just have to toggle the bits in reverse order.
Now the solution.
With n ones..
1. Use first n-1 ones
Inmost of the good opcs(Including Abacus)only the best code survives the tests! You can't write bubble sort when asked for a sorting algorithm! It will simply time out!
So provided only the best algo survives, the next criteria for evaluation will obviously be How quick u came out with the idea
Hi Guys,
I havea question...
Integer partitioning is NP-complete. So partitioning a set into 'n' number of equal sets is also in NP (When n=2 we get the first one). But Integer partitioning has a nice DP solution (though not in P)... Similarly does Integer partitioning into 'n' equal sets also
Hi guys,
I think most of you knew it... Incase some of you don't know...
Bitwise is BACK!
http://www.bitwise.iitkgp.ernet.in/
Check out on Jan 1st to register...
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