Hi All,
I am student at CSUS. I am looking some topics on database or cloud
computing for my Master's project.
It would be helpful, if anyone provides me an good idea on it.
Thanks,
Sweety
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1 from 19 makes 20
11 + 9 = 20 (take the one away from 19 and attach it to the other one and
put a + between the 2)
-Vandana
On Tue, May 10, 2011 at 12:39 PM, Lavesh Rawat lavesh.ra...@gmail.comwrote:
*MATHS TRICK TEASER
*
*
*
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*Prove that taking away 1 from 19 makes 20.
*
*Update
write test cases for the division '/' operator..
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19=10011
20=10100
therefore take leftmost one from 19 and shift second one left and
make it 10100=20
On Tue, May 10, 2011 at 12:15 AM, Vandana Bachani vandana@gmail.comwrote:
1 from 19 makes 20
11 + 9 = 20 (take the one away from 19 and attach it to the other one and
put a +
superlyk utkarsh solution..
On Tue, May 10, 2011 at 7:15 PM, UTKARSH SRIVASTAV
usrivastav...@gmail.comwrote:
19=10011
20=10100
therefore take leftmost one from 19 and shift second one left and
make it 10100=20
On Tue, May 10, 2011 at 12:15 AM, Vandana Bachani
Don't really get the question
On 10 May 2011 09:08, Akshata Sharma akshatasharm...@gmail.com wrote:
write test cases for the division '/' operator..
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cases would be:
1. division by 0 raises an appropriate Exception
2. dividing 0 by any number should result in 0
3. dividing any number by 1 should give the same number
4. a = b*q + r i.e a/b should give q
On Tue, May 10, 2011 at 7:52 PM, Carl Barton odysseus.ulys...@gmail.comwrote:
Don't really
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I would add that you should test all combinations of positive and
negative operands.
You are not clear about the type being divided.
If it is a floating point type, you can verify that (a*b) / b = a to
an acceptable tolerance for a wide variety of values of a and b.
If you are working with
Hello Anders Ma .. for inputs like iiestseig (just a random string) your
code will not produce the correct output .. cos the best possible way to
split these strings is {i,iestsei,g} .. But your code will produce
{ii,este,i,g} as output .. so when there are overlapping palindromes
your code wont
My approach :
Have a pointer to the start (smallest of the array) of each of the N
arrays.
Until all pointers reach end of respective arrays :
take the smallest value from all of the pointers
and compute the difference between the smallest and the current pointers
of each of the arrays
You may have to use a trie and also the edit distance for this problem.
Firstly , walk down the trie as you can keep matching the alphabets.
When you encounter a first mismatch , findout the edit distance for the rest
of the substring of the input with all of the strings possible from that
node
19 = XIX in Roman Numerals. Take 1 = I away and you have XX = 20.
Dave
On May 10, 2:09 am, Lavesh Rawat lavesh.ra...@gmail.com wrote:
*MATHS TRICK TEASER
*
*
*
**
*Prove that taking away 1 from 19 makes 20.
*
*Update Your Answers at* : Click
On Mon, May 9, 2011 at 8:31 PM, Don dondod...@gmail.com wrote:
That would do it if you have a 64-bit type, which most implementations
have, but the standard does not require.
I think that I can make it shorter and cleaner.
int main(int argc, char* argv[])
{
const int n=49;
char
On Mon, May 9, 2011 at 11:44 AM, Ashish Goel ashg...@gmail.com wrote:
Dave,
w.r.t statement, After all integers are processed, compress out the unused
hash table
entries and find the Kth largest element,
I could not understand the compress concept...are you saying something on
counting
check this one out:
#includeiostream
#includecstdio
#includevector
#includecstring
using namespace std;
int check_palin(string str,int *start)
{
int pos=-1,ret,size=str.size()-1;
char last_char=str[size];
while(possize)
{
ret=0;int i;
pos=str.find(last_char,pos+1);
take “aabab” for example, the result is aba, b,a; however, the
right result is aa,bab
On Wed, May 11, 2011 at 10:57 AM, shubham shubh2...@gmail.com wrote:
check this one out:
#includeiostream
#includecstdio
#includevector
#includecstring
using namespace std;
int check_palin(string
@Aamir: First, regarding overflow, 1000 = -128. Thus, the cycle
is 0, 1, 2, ..., 127, -128, -127, ..., -1, and back to 0.
Second, regarding your assertion that you only need a single character
instead of an array of them: Based on the above sequence, every time
a[0] cycles back to 0, p is
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