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you were not given any essay during the written round?
and if you can get your friends to tell us what kind of puzzles were asked?
just a bit of info on that.
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end at 16
not 12 so that p2 starts at 8's multiple.
This is done by padding pointer by 4bytes in both I and II struct.
declarations.
Hope i made it clear...!
Thanks.
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Shashank,
In C++ we implement it thru reference variables
in the function call we can say func(x);
in the definition we have, func(int a)
whatever changes we make to a in func() are reflected back in x in the
calling function
In C, we can simulate i t through pointers
call : func(x);
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?hl=en.
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that is exactly what i thought
thanks for the clarification
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Suppose that our rabbits never die and that the female always
produces one new pair (one male, one female) *every month from the
second month on
*
@Priyanka
You have answer 288
I added 178 to it because I got the answer that 89 females would have given
birth in the last month.
Pls clarify.
Are
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nikhilgupta2...@gmail.comwrote:
For rabbits, answer is 256.
On Sat, Jul 30, 2011 at 8:33 PM, Tushar Bindal tushicom...@gmail.comwrote:
for rabbits answer would be different from cows as they give birth to a
pair at a time, so at any time odd number of rabbits won't be there
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If this is the case, we can follow this method:
take variables
1. dir = 1
2. st_col=0
3. end_col = n-1
4. st_row = 1
5. end_row = n-1
6. row = 0
7. col = n-1
8. noOfCellsVisited = 0
dir are as follows:
1 - left ro right
2 - top to bottom
3 - right to left
4 - bottom to top
row will keep account
There will be following changes to this
while(noOfCellsVisitedmn)
{
if(dir == 1)
{
for i = st_col to end_col
{ print arr[row][i]; noOfCellsVisited++}
dir++;
end_col--;
row = n-1-row;
}
else if(dir==2)
{
for i = st_row to end_row
{ print arr[i][col]; noOfCellsVisited++}
dir++;
end_row--;
i am sorry for givin such a naive solution and making so many mistakes.
i still left a mistake in dir==4
*
col=n-1-(col+1);*
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thanks amit for this code
i tried this for the first time and am still not over finding out my
mistakes (got another one while trying it out on an array)
yours was an easy one :)
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though the code given by siddharth seems to be a bit tough to understand due
to one long statement, it gives a good idea to run the main loop fron 2^k
-1 to (2^k - 1)*(2^(n-k)) since these rae the only numbers having k digits
set
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can't we include -9 , 3 , 1, 5 , 0 as a possible sub array?
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yup
i got it now.
nice code
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we can just infer that at the point where the man met the driver, from that
point, it takes 10 minutes to reach the office (assuming that car moves at
same uniform speed all the time)
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thanks sagar for this wonderful shortcut
but can you please explain it better. in what cases can we use this
approach?
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thankyou :)
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@varun
you are contradicting your own statement
when inline replaces the code at place where it is called, then this code
will work fine.
and the other one w/o inline won't as explained by others also.
there seems to be some confusion in your statement.
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yx - xy = x0y - yx
= (10y + x) - (10x + y) = (100x + y) - (10y + x)
9y - 9x = 99x - 9y
18y = 108x
y=6x
Since, x comes in hundreds place, we know it can only be 1 as the difference
between the 3 digit number and two digit number is difference of two 2 digit
numbers only.
thus, y=6*1 = 6
a
the clock has become faster by 48 seconds in 10800 minutes
i.e., 1 sec in 225 minutes
from 8am on sunday we have to go 225*5*60 minutes behind to get time when
clock showed the correct time.
i.e., 67500 minutes or 44 days 1260 minutes
44 days 21 hours
which means clock showed correct timing at
probability that i win standing at second position: 1/365
third position : 364/365*2/365 = 1/365)*(628/365)
fourth position : 364/365*363/365*3/365
4th : 364/365*363/365*362/365*4/365
nth position:
(365-1)*(365-2)*(365-3)*(365-4)*(365-5).*(365-(n-2))*(365-(n-1))*(n)*(1/365)^n
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Sorry for the previous post
the last line was incorrect
it should have been (n+1)th position
I was just writing roughly and pressed send instead of save.
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- n +1)*(n)}
/ {365*(n-1)}
maximum probability is at nth position if at (n+1)th position,
{(365 - n +1)*(n)} / {365*(n-1)}
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Sumit,
the answer is 14
I think the example of 16 that they take on careerplus is probably confusing
you.
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Sory once again for that incomplete answer.
The complete one is here.
probability that i win standing at second position: 1/365
probability that i win standing at third position : 364/365*2/365 =
1/365)*(628/365)
probability that i win standing at fourth position : 364/365*363/365*3/365
Evenly divisible simply means that a number should be completely divisible
by the given numbers, i.e., it should give a whole number as an answer when
divided by that particular number. Evenly divisible doesn't mean that
quotient should be an even number. It just needs to be a whole number.
.
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- win-win situation
For a change the easier method is faster as well
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, Jul 6, 2011 at 10:16 PM, saurabh singh saurab...@gmail.com
wrote:
I have proposed my solution in one of the previous posts.Check the
solution
there
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good job
but how can this be done in one traversal as asked
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@Navneet
Didn't get your point
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the solution is given
herehttp://www.thecareerplus.com/?page=resourcescat=150subCat=10qNo=2
but can anyone lease explain it better
please give a original solution
and stop making rude comments about answers posted genuinely.
If you have an original solution, please post it.
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I think you are getting it wrong.
Surender, your solution says that numbers divisible by all of the first 5
prime numbers will be taken into account whereas the question says that only
the numbers *not* evenly divisible by *any* of the first 5 prime numbers are
to be added.
Shiv,
you are making
If my interpretation is right, following should be the code.
int main()
{
int userInteger = 0;
cout Enter A Number endl;
cin userInteger; // Ask For a number from the user
if (userInteger 0) // Is the number valid?
{
int result = 0;
int prime[5] = { 2, 3, 5, 7, 11 };
int a,b, count = 0;
that theorem is for odd primes
9 is not an odd prime
so we can;t apply this theorem
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Manish is right.
This would start from the left.
In cout also processing starts from left.
The processing is from right in the printf statement.
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I have the book
but I hope there are no issues with ths book being shared here. I don't have
the link now.
would it be fine if I share the book directly?
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is there anyone who has this book?
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searching in the group did not help
please give the link to your previous post
or better give the link once again
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Is this one fine?
http://www.megaupload.com/?d=YSSE195C
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Plz mail the book to me too
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Arpit
The formula that you have written is wrong. You would have approached it
correctly but the formula written here is incorrect.
m= n + (n**1/6* + (n-1)) + ((n**7/6* + (n-1))**1/6* + n-2) +..
If you add all the terms there should be 1/6 in these terms as the terms
give the number of
a[i][j]=1 if 'i' is a parent to 'j'
But in your matrix,
a[1][2] = 1
a[1][3] = 1
a[2][3] = 1
a[4][2] = 1
a[4][3] = 1
You have taken 'j' to be a parent to 'i'
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xxx, yyy are considered as cases?
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jus a promotional activity. We thot here ppl discuss on
technical topics so it wud b gr8 to enlighten ppl vd such fest whr ppl
compete n exchange knowledge.
Sorry if we broke ne rule for the group.
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how can you send
***DELHI TECHNOLOGICAL UNIVERSITY*
*(Formerly DELHI COLLEGE OF ENGINEERING)**
CSI-DTUPHOENIX'11
presents
FIREWALL*
*The Art Of Hacking
*
*
*
Do you swear by having the key to every single lock?
Do you think you possess the innate talent of finding loopholes in anything
and
It would be
*vertcal_sum(root-right,level+1);
*in the last line
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I think the following algo should work:
1. Create a DLL of the inorder traversal of the tree
2. for each node, check whether P of that node points to the previous node
in the DLL or not.
3. If not, assign it value NULL
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bittu,
you have written this code at end of main()
*printGivenOrder(start);
printList(rslt);*
so the BT in spiral order should be printed twice
But you are getting a extra 4 only in place of the whole tree because of
these lines
*temp=current;
current=head;
rslt=appnd(temp,current);
* There
Is it necessary that the two lists are merging at their ends??
Do we have to find whether they merge at the end into same lists or wheter
they are just intersecting??
On Thu, Jan 6, 2011 at 10:04 PM, Aditya adit.sh...@gmail.com wrote:
There are two aspects here for second question.
1. to
have one pointer to the
next. So, if they intersect they will definitely be merging.
On Thu, Jan 6, 2011 at 11:01 PM, Tushar Bindal tushicom...@gmail.comwrote:
Is it necessary that the two lists are merging at their ends??
Do we have to find whether they merge at the end into same lists
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