N = p1 ^ K1 * p2 ^ k2 ..
phi(N) = p1^(K1) * (1-1/p1) * .
phi(N) = N * (1 - 1/p1) * (1 - 1/p2) .
phi(N) = (N - N/p1) * (1 - 1/p2)..
so now result which was initally
@Mayank u can directly send hi to her on personal gmail. please dnt spam.
let it be algo group
On Thu, Aug 25, 2011 at 11:42 AM, mayank gaur gaur.mayank.9...@gmail.comwrote:
hi
On Thu, Aug 25, 2011 at 11:37 AM, S MADHU madhu.mca...@gmail.com wrote:
hi
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MADHU.
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will all u stop all this. dont put any comments on all this now.
On Wed, Aug 3, 2011 at 11:20 PM, Anil Arya anilarya...@gmail.com wrote:
@coder dumca--- Utkarsh --- google me nahi Goldman Sachs me jane
wala hai..Google ko ye saubhagya prapt nhi hoga
On Wed, Aug 3, 2011
Your code does not works proper;y for all cases
On Mon, Aug 1, 2011 at 10:42 PM, Rohit jalan jalanha...@gmail.com wrote:
Here is the recursive algo:
Rearrange(A,p,q)
1. if p is not equal to q do the following
2. r ← (p+q)/2
3. Exchange A[(p+r)/2..r] ←→ A[(p+q)/2 +1 ..(r+q)/2].
4.
i+1 * i+1
= 2 * i + 1
for i = 3, j = 7 (2 *3 + 1)
On Mon, Jul 25, 2011 at 6:06 PM, Arshad Alam alam3...@gmail.com wrote:
what should be the output of the following and how..
#includestdio.h
#includeconio.h
#define PRODUCT(x) (x*x)
void main()
{
clrscr();
int i=3,j;
Can think of a O(n^2) solution
On Sun, Jul 10, 2011 at 6:53 PM, raj singh ankurkaku...@gmail.com wrote:
@yogesh- can u explain with an example pls?
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This is mean not the median
On Mon, Mar 28, 2011 at 8:42 PM, Raunak Agrawal raunak.ra...@gmail.comwrote:
I am assuming that the median is the sum of all the values stored in the
nodes divided by 2.
So I am traversing all the nodes recursivelyand finding the median of
them.
On Mon,
beetle
On Tue, Mar 1, 2011 at 1:38 PM, Lavesh Rawat lavesh.ra...@gmail.com wrote:
Riddle Problem Solution
I am an insect. The beginning of my name is another insect's name. What am
i?
Update Your Answers at : Click Here
Solution:
Will be updated after 1 day
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@ Dave : I think this shud also give the same result
C(26, 3) * C(26, 3) * C(10, 2) * C(62, 2)
On Wed, Feb 9, 2011 at 7:16 PM, snehal jain learner@gmail.com wrote:
how many passwords can be made if
1. there should be atleast 3 capital letters
2. atleast 3 small letters
3. atleast 2
Also it shud now be multiplied with Factorial of 10
On Thu, Feb 10, 2011 at 1:14 AM, Manmeet Singh mans.aus...@gmail.comwrote:
@ Dave : I think this shud also give the same result
C(26, 3) * C(26, 3) * C(10, 2) * C(62, 2)
On Wed, Feb 9, 2011 at 7:16 PM, snehal jain learner@gmail.com
U using extra space, if you using extra space, simple C++ implementation
will be use a priority queue and do BFS.
On Wed, Feb 9, 2011 at 10:47 AM, jalaj jaiswal jalaj.jaiswa...@gmail.comwrote:
@algoseeker
mine approach is also the same but m inserting the elements in a min heap
and you in a
How to design a chess board using OOPS ? ?
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test case : a = -1, b = - 1
Result : fight successfully challenged dave's problem. Method returned 1
while it shud have returned -1.
On Sun, Feb 6, 2011 at 4:12 PM, Balaji S balaji.ceg...@gmail.com wrote:
thanks..
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8, 32, 96
On Sat, Feb 5, 2011 at 2:46 PM, priya mehta priya.mehta...@gmail.comwrote:
#include stdio.h
#define PrintInt(expr) printf(%s : %d\n,#expr,(expr))
*int* FiveTimes(*int* a)
{
*int* t;
t *=* a**2 *+* a;
*return* t;
}
*int* main()
{
*int*
because u not thinking of operator precedence :P :P
On Sat, Feb 5, 2011 at 2:52 PM, priya mehta priya.mehta...@gmail.comwrote:
why is this happening?
On Sat, Feb 5, 2011 at 2:51 PM, Manmeet Singh mans.aus...@gmail.comwrote:
8, 32, 96
On Sat, Feb 5, 2011 at 2:46 PM, priya mehta
@ Shubham : Well Said :) :)
On Fri, Feb 4, 2011 at 8:01 PM, shubham singh
shubhamsisodia0...@gmail.comwrote:
if (u r calling dp of spoj as basic)
{
i would say u are a champ already ;
}
else
{
do practise more on spoj
}
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C/C++ user : Take input in a char array or string(if using C++) : a
now do string addition. : a = a + a;
now do string substraction. : a = a - 2.
Java user : Use BigInt class
On Fri, Feb 4, 2011 at 8:12 PM, Logic King crazy.logic.k...@gmail.comwrote:
please help me solve the problem on SPOJ
What is incorrect in the given question, except the constraints not given.
On Mon, Jan 24, 2011 at 4:03 PM, juver++ avpostni...@gmail.com wrote:
DP and clarify your incorrect question.
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Also has any one solved https://www.spoj.pl/problems/MONONUM/
with only DP. I would like to know the solution, as my solution works for
small nos. and then the ratio reduces to a simple mathematical formula.
On Mon, Jan 24, 2011 at 4:09 PM, Manmeet Singh mans.aus...@gmail.comwrote:
What
ok, i thought he wanted the both, but he means ratio i guess. So, its the
same problem.
On Mon, Jan 24, 2011 at 4:21 PM, juver++ avpostni...@gmail.com wrote:
@fight incorrect -
how to calculate the number of the decreasing *n*-digit integers to the
increasing *n*-digit integers
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@JUVER++ : One personal question. Is this algo groups paying you something
or you among the admins :) :). As in every problem I find your name and with
a superb solution.
On Mon, Jan 24, 2011 at 4:26 PM, Manmeet Singh mans.aus...@gmail.comwrote:
ok, i thought he wanted the both, but he means
n / 9.0 + 1 for n 20 works, before that I apply a simple DP. even i want
to know a gud DP based solution for the problem, with no formula used, as I
have done.
On Mon, Jan 24, 2011 at 11:43 PM, Logic King crazy.logic.k...@gmail.comwrote:
hey, but what's the solution to the problem...how
Number exponentiation
On Fri, Jan 21, 2011 at 1:05 PM, snehal jain learner@gmail.com wrote:
Given M, find if M = 3^x for some positive integer x efficiently. DO
NOT think of log, pow etc
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this will be O(log(n) * log(n)) solution
On Fri, Jan 21, 2011 at 4:29 PM, abhijith reddy abhijith200...@gmail.comwrote:
Below is code for modular exponentation in general
// (a^b)%c
int modexp(int a,int b,int c)
{
int ans=1;
while(b)
{
if(b1) ans=(ans*a)%c;
On Fri, Jan 21, 2011 at 5:24 PM, Manmeet Singh mans.aus...@gmail.comwrote:
this will be O(log(n) * log(n)) solution
On Fri, Jan 21, 2011 at 4:29 PM, abhijith reddy
abhijith200...@gmail.com wrote:
Below is code for modular exponentation in general
// (a^b)%c
int modexp(int a,int b,int c
@juver++ : the above is a user defined function. but its possible to the
problem using bit wise operators.
On Fri, Jan 21, 2011 at 7:29 PM, juver++ avpostni...@gmail.com wrote:
@above it's a user-defined function using fast exponentiation
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write n, n-1 to base 3
check if thr gives 0, if it gives no. is of form 3^n
On Fri, Jan 21, 2011 at 8:04 PM, snehal jain learner@gmail.com wrote:
@manmeet
how?
On Fri, Jan 21, 2011 at 7:36 PM, Manmeet Singh mans.aus...@gmail.comwrote:
@juver++ : the above is a user defined function
@ juver++ : fight is my tc handle. here u can call me manmeet :) :)
On Fri, Jan 21, 2011 at 9:35 PM, juver++ avpostni...@gmail.com wrote:
@fight good solution.
3^n contains only one 1 in the 3-base representation.
So write number M in base-3, and check if it contains only one digit(1).
how merge sort ?/
On Thu, Jan 20, 2011 at 11:23 PM, nishaanth nishaant...@gmail.com wrote:
Ya as Ashish said hashing is the best solution :)
On Fri, Jan 14, 2011 at 6:00 PM, Ashish Goel ashg...@gmail.com wrote:
ideally, a hashMap would be preferred
walk through one array and set the
whts complexity for this movegen()
On Fri, Jan 21, 2011 at 9:52 PM, nishaanth nishaant...@gmail.com wrote:
Ok lets define the following functions.
movegen()- gives the list of next move given the state. This is basically
all the 1 distance words given the current word.
heuristic()- this
It can be done in O(n) space too using DP :) :).
U dont need that flag, but that solution u said is absolutely correct.
On Thu, Jan 20, 2011 at 7:27 PM, Decipher ankurseth...@gmail.com wrote:
There is a row of houses in which each house contains some amount of money.
Write an algorithm that
; i++)
{
S[i] = Math.Max(S[i - 2], S[i - 3]) + A[i];
maxloot = Math.Max(maxloot, S[i]);
}
return maxloot;
}
On Jan 20, 7:30 pm, Manmeet Singh mans.aus...@gmail.com wrote:
It can be done in O(n) space too using DP
1-D simple dp with state current index you are at
On Wed, Jan 5, 2011 at 7:41 AM, Akash Agrawal akash.agrawa...@gmail.comwrote:
didn't get the question correct...
Can u cite an example...
Regards,
Akash Agrawal
http://tech-queries.blogspot.com/
On Wed, Jan 5, 2011 at 12:36 AM, Decipher
Can you elaborate how to put segment tree into picture
On Wed, Jan 5, 2011 at 2:57 PM, juver++ avpostni...@gmail.com wrote:
Right. It can be solved simply in O(n^2). To optimize use segment trees.
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Algorithm Geeks
int solve(int id) {
if(id==n)
return 0;
int d = dp[id];
if(~d) return d;
d = 0;
for(int i=id+1;in;i++) {
if(dis[i]-dis[id]=k) {
d ?= val[id] + solve(i);
}
}
return d;
}
Its O(n^2), and Juver++, is very
I am telling the DP O(n^2) solution, whts wrong in it??
On Wed, Jan 5, 2011 at 10:36 PM, juver++ avpostni...@gmail.com wrote:
Unfortunately, you are wrong.
We need one loop for i and that is all. All other things for searching max
p[j] is solved by segment tree in O(log n).
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ok :):)
On Wed, Jan 5, 2011 at 10:43 PM, juver++ avpostni...@gmail.com wrote:
I was replying to the @Decipher post. Not yours. :)
Your algorithm seems to be ok.
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In main just pass
answerIs = max(val[0], solve(0));
On Wed, Jan 5, 2011 at 10:49 PM, juver++ avpostni...@gmail.com wrote:
Sorry, disregard my first proposition about index i.
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Go through gates in Electronics, code will be easy to visualize.
On Fri, Dec 31, 2010 at 6:33 PM, juver++ avpostni...@gmail.com wrote:
Yeah, add bits one by one from right to left, as for ordinary addition of
two numbers in 10-th radix
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class BadNeighbors {
public:
int maxDonations(vector int);
};
int dp[51][2][2];
int l;
vectorint v;
int solve(int id, int taken, int first) {
if(id==l)
return 0;
int d = dp[id][taken][first];
if(~d) return d;
d = 0;
if(id!=l-1) {
if(taken==0) {
d
When you pass an array as argument, it gets broken into a pointer. So you
get size as 4 for 32 bit machine.
On Tue, Dec 14, 2010 at 10:59 PM, Divesh Dixit
dixit.coolfrog.div...@gmail.com wrote:
#define SIZE 10
void size(int arr[SIZE])
{
printf(size of array is:%d\n,sizeof(arr));
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