Hi
i want to write a c++ webservice which should work in linux machine
with apache being the web server.
Are there any libraries which allow you to write server-side
applications that handle the HTTP requests with Apache being the
webserver on Linux?
Thanks.
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On Jul 7, 6:44 pm, rgap wrote:
> Hi.. can someone tell me about the "Computing Laboratory" you have on
> your faculty of computer science, what about the software, hardware
> there.
> Please i want to clarify my one doubts.
> Thanks in advance...
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Hi.. can someone tell me about the "Computing Laboratory" you have on
your faculty of computer science, what about the software, hardware
there.
Please i want to clarify my one doubts.
Thanks in advance...
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"Algorithm Ge
Send me links too, please
On Apr 10, 3:14 am, rahul rai wrote:
> None has been published . If you want i can give you a two links to
> video courses , and some set of solved questions . Which will be good
> for "compiler design help"
>
> On 4/10/11, Harshal wrote:
>
>
>
> > someone please share
Hi, does anybody know when/where to use
typedef long long int64;
and
const long double EPS = 1E-9;
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H
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h
Is there a maximum time complexity¿
On Mar 1, 2:50 pm, bittu wrote:
> Given an integer, print the next smallest and next largest number that
> have the same number of 1 bits in their binary representation.
>
> Example for n=12
> next smallest with same no. of set bit is 17 but to found next
>
Hi, I need help with this problem
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=38
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You re right, but this code runs slower
#include
#include
#define ABS(x) ((x<0)?(-(x)):(x))
using namespace std;
int main(){
int N,B,i,j,k,c,dif;
int R[91];
int balls[91];
while(cin>>N>>B,N,B){
memset(balls,0,sizeof(balls));
for(i=0;i>R[i];
k=c=0;
for(i=0
Here is my implementation ;)
#include
#include
using namespace std;
int balls[91]; //maximum balls
int N;
//verify if its impossible to call out every number from 0 to N
bool impossible(){
int i,j;
if(!balls[0])return true; //if ball 0 was removed
//its impos
This code prints the number of digits 0,1,2,... 9sbetween A and B
How does it works?
#include
using namespace std;
int pot10(int i) {
int n = 1;
while (i--) n *= 10;
return n;
}
int digits(int n, int d) {
if (n < 10 && n > d - 1) return d ? 1 : 0;
int s = 0, r = 0, i =
How can i count the number of digits 0s,1s,2s,3s,... 9s
in a number n
for example if n=33902
number of 0s=1
number of 3s=2
number of 9s=1
number of 2s=1
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What is the time complexity of sqrt function of c++?
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please, how can i prove the correctness of an algorithm, for example
this one:
#include
#include
#include
using namespace std;
#define MAXN 12
int P[MAXN][20];
int T[MAXN];
int L[MAXN];
long long int C[MAXN];
int query(int p, int q){//lca
int tmp,log,i,c=0;
if(L[p]=0;--i)
The problem is from:
http://acmicpc-live-archive.uva.es/nuevoportal/data/problem.php?p=4813
Can someone explain me please what does this code do? why the GCD?
#include
#include
#include
#include
using namespace std;
int a[15],b[15],n,m,s,t;
int gcd(int c,int d)
{
if (d==0) return
Thanks, i finished it. :)
it wasdist[i] + dist[j] - 2*dist[lca(i, j)]
"this problem is difficult" :(
http://acmicpc-live-archive.uva.es/nuevoportal/data/problem.php?p=4809
I dont have idea how to solve it.
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I have solved the last one :)
how can i solve this one:
http://acmicpc-live-archive.uva.es/nuevoportal/data/problem.php?p=4805
using LCA, I used this algorithm:
http://www.topcoder.com/tc?module=Static&d1=tutorials&d2=lowestCommonAncestor
but i dont know how to "count the weights of the edges".
Hi.. thanks for your response.
The number of kids:
3 <= K <= 10^9
I cant declare an array: long long A[10];
and how does dfs or bfs finds the components of the graph?
because i have to verify if there is a cycle in all the components.
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Hi, I need some help solving this problem from ICPC regionals, 2010,
South America.
http://acmicpc-live-archive.uva.es/nuevoportal/region.php?r=sa&year=2010
Problem K - Kid's Wishes
Each kid may wish to sit down next to at most two other kids, because
each kid has just two neighbors in the circle.
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