thankx   for u r reply.in my code i implemented Sieve of Eratosthenesthat
but some type of optimization were required that i have done...got accepted.

On Wed, Apr 14, 2010 at 2:57 PM, Ashish Mishra <[email protected]>wrote:

> The Sieve of Eratosthenes is best for finding prime
>
>
> On Mon, Apr 12, 2010 at 10:58 AM, BlackdiamonD <[email protected]>wrote:
>
>> thank kartik thanks but it was with lot of optimized code i was unable to
>> understand though i have changed my code more know it is taking around 8 to
>> 8.5 seconds in my computer but still i am getting TLE on server
>>
>> #include<stdio.h>
>> #define  ten8 (100000000)
>> const int mod=32;
>> unsigned int M[ten8/mod];
>> int main()
>> {
>>      int half=10000, i,j,x=1;
>>      int y=ten8/mod;
>>    freopen("output.txt","wt",stdout);
>>     for (  i = 0;i < y;i++)
>>         M[i] = 0;
>>     printf("2\n");
>>     for ( i = 3;i < ten8;i+=2)
>>     {
>>      int a=i/mod,b=i%mod;
>>         if (((M[a]>>b)&1)==0)
>>         {
>>             x++;
>>             if(x%100==1)
>>                 printf("%d\n",i);
>>             if(i>half)
>>                 continue;
>>             for (int j = i * i;j < ten8;j += i)
>>             {
>>                 M[j/mod] =M[j/mod]|(1<<(j%mod));
>>             }
>>         }
>>     }
>> }
>>
>> On Sun, Apr 11, 2010 at 7:20 AM, Karthik Reddy 
>> <[email protected]>wrote:
>>
>>> http://wwwhomes.uni-bielefeld.de/achim/prime_sieve.html
>>>
>>> take a look at this link . The running time is less than 2 sec for 10^8.
>>>
>>> On Sun, Apr 11, 2010 at 2:38 AM, Black Diamond <[email protected]>wrote:
>>>
>>>>  i have an problem on SPOJ to find all the prime less than 10^8
>>>> https://www.spoj.pl/problems/TDPRIMES/
>>>>
>>>> i am using sieve of erastosthenes algorithm to find primes
>>>> my code is taking in my machine around 10.9 to 11.2 seconds
>>>> and time limit is 10 second how i can change my code or any diff
>>>> logic..???
>>>> //---------------------------------------------//
>>>> #include<cstdio>
>>>> using namespace std;
>>>> #define  ten8 (100000000)
>>>> bool M[ten8];
>>>> int main()
>>>> {
>>>>      int half=10000, i,j,x=0;
>>>>     for (  i = 0;i < ten8;i++)
>>>>         M[i] = false;
>>>>     for ( i = 2;i < ten8;i++)
>>>>     {
>>>>         if (M[i] == false)
>>>>         {
>>>>             x++;
>>>>             if(x%100==1)
>>>>                 printf("%d\n",i);
>>>>             if(i>half)
>>>>                 continue;
>>>>
>>>>             for (int j = i * i;j < ten8;j += i)
>>>>             {
>>>>                 M[j] = true;
>>>>             }
>>>>         }
>>>>     }
>>>> }
>>>>
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>>>
>>>
>>>
>>> --
>>> Ram Karthik Reddy Ginuga
>>> karthik.ginuga[at]gmail.com
>>> CCNA,MCP
>>> Mozilla Campus Ambassador
>>> SPOJ world rank #1088
>>> http://www.spoj.pl/users/karthu/
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>>> (91)9247818845
>>>
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>>
>>
>>
>> --
>> ~~~~BL/\CK_D!AMOND~~~~~~~~
>>
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>
>
>
> --
> Ashish Mishra
> http://ashishmishra.weebly.com/
>
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