i just want to know , when we make a program how update array after
deleting alternate element.
On Mar 21, 4:37 pm, Anurag atri anu.anurag@gmail.com wrote:
what did you not understand in this ?
On Mon, Mar 21, 2011 at 4:59 PM, rohit scofiel...@rediffmail.com wrote:
genrate prime number
cont int MAX = 10005;
bool isPrime[MAX];
void sieve()
{
int lim=sqrt(MAX);
/* Initially Mark all numbers as Prime */
for(int i=2;iMAX;i++)
isPrime[i]=1;
for(int i=2;i=lim;i++)
if(isPrime[i])/* for each Prime */
for(int j=2*i;jMAX;j+=i)/* Cross
take an array and put 0 at all positions in it (except 2 ) , then when you
mark a number as a multiple of a prime number ( 2 to start with ) put 1 in
that place ..then look for the first number in the array that has a 1 ( this
time 3 ) do the same procedure again .
On Mon, Mar 21, 2011 at 5:41
better explained by Reddy's program :)
On Mon, Mar 21, 2011 at 6:00 PM, Anurag atri anu.anurag@gmail.comwrote:
take an array and put 0 at all positions in it (except 2 ) , then when you
mark a number as a multiple of a prime number ( 2 to start with ) put 1 in
that place ..then look for
@allWhy Not Try Carziest Thing in This Question
Q.1st Prrove Timpe Complexity of Sieve of Eratosthenes is
O(Log(Log(n))) ..isn't Making U Stuck..??
Q.2nd We Need to Drease the same Number of element sieved more then
one time e.g 6,12 all the multiples of 2,3 ..then again all the
hav a look at dis one...prime no upto 10
#define N 10
bool mark [N];
memset (mark, true, sizeof (mark));
mark [0] = mark [1] = false;
for ( int i = 4; i N; i += 2 ) mark [i] = false;
for ( int i = 3; i * i = N; i += 2 ) {
if ( mark [i] ) {
for ( int j = i *
