_(n,1,n,0) is true if n is prime.
I set out to create an O(n^n) algorithm. It essentially computes the
product of every possible set of n integers in the range (1..n-1). If
any of those products equal n, the number is composite. You will
notice that the program does not use the * operator to perfo
@Don : can you plz explain it ?
Sanjay Kumar
B.Tech Final Year
Department of Computer Engineering
National Institute of Technology Kurukshetra
Kurukshetra - 136119
Haryana, India
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I wrote it. Can you figure out how it works?
Don
On Aug 17, 1:25 am, Nitin Nizhawan wrote:
> Hi Dod,
>
> Could you pls expalin what this algorithm is doing and from where you got
> it.
>
> Thanks
> Nitin
>
> On Wed, Aug 17, 2011 at 2:56 AM, Don wrote:
> > I wrote a program to print prime numbe
Do take a look at this also
http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes#cite_note-7
Source code to list all prime numbers till number n
#include
#include
#define bool int
using namespace std;
int main(){
int n;
cin>>n;
int i,j;
bool primeValues[1000]={0};
for(i=4;i<=n;i=i+2
thanq u very much
On Jul 23, 10:13 pm, arun kumar wrote:
> hope this link will help
> youhttp://www.topcoder.com/tc?module=Static&d1=tutorials&d2=primalityTes...
>
>
>
>
>
>
>
> On Sat, Jul 23, 2011 at 10:39 PM, frank wrote:
> > what is the efficient algorith to find the prime numbers or to ch
Okay I got it myself
@OP
This can be done in O(n*k*logn)
where k is of order 10^3 at the very max , when u need a prime like 1
trillion
On Jan 28, 6:44 pm, sankalp srivastava
wrote:
> Correct me if I'm wrong , but according to you , will this be the
> approach ?
>
> boolean array[10];
> int
Correct me if I'm wrong , but according to you , will this be the
approach ?
boolean array[10];
int primes[100];
void seive()
{
int index=0;
for(int i=2;i*i<10;i++)
{
if(isprime[i])
{
primes[index++]=i;
for(int j=i*2;j<100;j+=i)
#include
#include
#include
#define MAX 15485863
#define LMT 100
unsigned flag[MAX>>6];
#define ifc(n) (flag[n>>6]&(1<<((n>>1)&31)))
#define isc(n) (flag[n>>6]|=(1<<((n>>1)&31)))
void sieve() {
unsigned i, j, k;
for(i=3; i>6);
}
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@above One million is 10^6.
Problem wants 1 million of prime numbers. Not the prime numbers in range
1..1000.
So, you should use two sieves.
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@rahul, juver++,sankalp
Though this can be solved using incrementing the range in sieve. But due to
this incremental approach how much the efficiency can be improved, any
guesses :)
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Use a seive . it's complexity is O(nlog n)
1 million is 1*10^7 , this will run within time limit ... assuming u
optimize ur code enough !!
On Jan 26, 5:48 pm, juver++ wrote:
> Generate primes numbers for the range 1..10^7 using sieve.
> Than apply sieve bigger range using these prime numbers.
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Generate primes numbers for the range 1..10^7 using sieve.
Than apply sieve bigger range using these prime numbers.
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I think one can use an elimination method . List out all the numbers
. Keep on eliminating the multiples of 2{excluding 2} , then multiples
of 3 , then multiples of 5 , then 7 , the denseness of the numbers
eliminated will get less . And obviouslly you will get the numbers .
On 1/25/11, siddharth
Hey Dave
On 25 January 2011 18:17, Dave wrote:
> The most efficient approach is to google "millionth prime number" and
> select the first hit.
>
> Good one. But it was asked to me in an interview.
The trivial approach would be to check for every number to be a prime an
continue till
the count of
The most efficient approach is to google "millionth prime number" and
select the first hit.
Dave
On Jan 25, 6:00 am, siddharth srivastava wrote:
> Hi
>
> Its an easy one but still I am looking for the most efficient approach.
>
> Find first 1 million prime numbers.
>
> --
> Siddharth Srivastava
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