Write a program to swap odd and even bits in an unsigned integer with as few
instructions as possible (e.g., bit 0 and bit 1 are swapped, bit 2 and bit 3
are swapped, etc).
eg[?]
input---
1234567
7654321
888
output--
411
12109682
948
--
regards
*Sudhir Mishra*
MNNIT ALLAHABAD
--
For 32-bit integers:
result = ((x 0x) 1) | ((x 0x) 1);
Dave
On Nov 12, 2:12 am, sudhir mishra sudhir08.mis...@gmail.com wrote:
Write a program to swap odd and even bits in an unsigned integer with as few
instructions as possible (e.g., bit 0 and bit 1 are swapped, bit 2
Mask all odd bits with 10101010 in binary (which is 0xAA), then shift
them left to put them in the even bits. Then, perform a similar
operation for even bits. This takes a total 5 instructions.
int swapOddEvenBits(int x)
{
return ( ((x 0x) 1) | ((x 0x) 1) );
}
On Nov 12,
#includestdio.h
int main()
{
int a,b=0x,c=0x;
scanf(%d,a);
a=((ab)1)+((ac)1);
printf(%d,a);
}
From: sudhir mishra sudhir08.mis...@gmail.com
To: algogeeks@googlegroups.com
Sent: Fri, 12 November, 2010 1:42:46 PM
Subject: Re: [algogeeks] Learn
I am not sure if this what you are looking for.
Assuming that the arrays are sorted in ascending order.
Choose one of the 2 arrays as a Reference Array.
for each element element in reference array, do a binary search to find all
elements that are less than the current element in reference and
*One possible solution is:
|4|6|3|5|
|2|8|1|7|
*Regards,
Janani
On Thu, Nov 11, 2010 at 1:26 PM, Amod gam...@gmail.com wrote:
We have a rectangle
It is divided in eight parts by three vertical and one horizontal line
so that there are 8 chambers.
Now we have numbers from 1-8 to be filled
Good point. Here is one step further:
3. Consider the center 4 numbers(regardless of order): {1,8,a,b},
containing 2 even and 2 odd.
and {a,b} can be chosen from {3,4,5,6}. Then there is only one case:
{1,8,3,6}.
The rest is trivial.
On 2010-11-12 13:26, Amod wrote:
Let me try to
(((x 0x)1) | ((x 0x)1))
5 bit operations in total.
On 2010-11-12 16:12, sudhir mishra wrote:
Write a program to swap odd and even bits in an unsigned integer with
as few instructions as possible (e.g., bit 0 and bit 1 are swapped,
bit 2 and bit 3 are swapped, etc).
first do logical AND of 10101010 with number for masking odd bit shift it
right 1 times suppose it comes n1
first do logical AND of 01010101 with number for masking even bit shift it
left 1 times suppose it come n2
now do n1 || n2
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result=((n ((128)/3))1)|((n ((129)/3))1);
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can anyone explain me how exactly servlet object is created
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