You are given a range of modulo operands and find the range of modulo
result.
Expression :- *x = y % z*
where range of y is [ c1, c2 ]
c1, c2 are real numbers ( can be +ve or -ve ) and c2 c1
where range of z is [c3,c4]
c3, c4 are real numbers ( can be +ve or -ve ) and c3 c4
Range of x is
Hi all,
I am looking for a good tool or language to create graphical user interface
in linux environment. Java can be used to achieve the purpose, Can you
please let me know any other technology for such purpose.
Thanks
Sagar
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-initialized-if-i-dont-do-it-explicitly
On Sun, Sep 28, 2014 at 12:22 PM, Deepak Garg deepakgarg...@gmail.com
wrote:
Hi
In example 1, member z will have a garbage value (i.e. 0 in your case )
Thanks
Deepak
On Sep 28, 2014 11:29 AM, sagar sindwani sindwani.sa...@gmail.com
wrote:
I am working
Science)
MNNIT
blog:geekinessthecoolway.blogspot.com
On Sun, Sep 28, 2014 at 3:47 PM, sagar sindwani sindwani.sa...@gmail.com
wrote:
Thanks Deepak and Rahul for the reply.
Do you guys have any standard document or any standard book which defines
this? I totally agree with these answers
I am working on How compilers handle initialization list. I came across a
case where I am not sure what should be the compiler behaviour.
*Example 1:-*
#include iostream
class A
{
public:
int x,y,z;
};
int main()
{
A a1[2] =
{
{ 1,2 },
{ 3,4 }
};
B[n]=0;
for i=n to 1
{
if(A[i-1]=A[i])
B[i-1]=B[i];
else
B[i-1]=B[i]+1;
}
O(n) solution.
Correct me if I am wrong.
On Tue, Oct 4, 2011 at 2:07 PM, anshu mishra anshumishra6...@gmail.comwrote:
int count(int x, int tree[], int s, int e, int n)
{
tree[n]++;
if (s==e) return 0;
int cnt = 0;
if ( ten's place is != 9)
Decrement the unit place by 1 and increment the ten's place by 1.
else
Increment MSD(Most significant digit) by 1 and decrement next digit by 1.
example:-
712 721
897 987
On Sun, Oct 9, 2011 at 6:51 PM, wujin chen wujinchen...@gmail.com wrote:
@Navneet，
Use K-Median theorem with k=N
On Mon, Nov 7, 2011 at 11:33 PM, Gene gene.ress...@gmail.com wrote:
Can you explain how a heap helps get the answer? Just to put the
elements in a heap requires O ( N^2 log (N) ) time.
On Nov 7, 4:12 pm, vikas vikas.rastogi2...@gmail.com wrote:
I think the
http://stackoverflow.com/questions/3963409/interview-question-dealing-with-m-occurrences-among-n
On Tue, Sep 13, 2011 at 1:48 PM, Gene gene.ress...@gmail.com wrote:
Here's a way:
The base 2 xor operator has an obvious extension to base 3 such that
for all integers N, N ^ N ^ N = 0, just
Awesome solution karthik R
On Tue, Sep 13, 2011 at 3:02 PM, kARTHIK R k4rth...@gmail.com wrote:
Something like this.
stackchar * S;
void function(char *s) {
char *ptr=s;
while(*ptr++ !=' ') {}
*ptr = '\0';
S.push(s); // Will push
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