finding all prime numbers in [1;N] for O(nlogn).the sieve of Eratostenes

Suppose we need to find all prime numbers between 1 and N. Usually we check if the number is prime by looking at the numbers from 1 to sqrt(K) (K is the number we want to check for). If we want to find all the prime numbers between 1 and N using this method we will have a time limit problem. It will take N*sqrt(N). There is a better way of finding the number in NlogN time. For this we will need a boolean array A[N]. A[i] will be false if the number i is prime otherwise it will be true.
 Here is the pseudocode of the sieve of Eratosfen.
    for(i=2;i<=N;i++)
    {
         if( a[i] is false)
         {
               mark all the divisors of i as true.
         }
    }
lets look at that on a small example
1 2 3 4 5 6 7 8 9 10
f  f  f  f  f  f  f  f  f   f
 number 1 is not prime but we will keep that as false.
Now our loop will continue until it meets a false value. It is the number 2.
2 is prime so our condition is met. Now we know that every number which is divisible on 2 is not a prime number . We go through the number form 2 to 10 and check if the current number is divisible on 2 we mark it as true ( not prime).
1 2 3 4 5 6 7 8 9 10
f  f  f  t  f  t  f  t  f   t
Next step is checking 3. 3 is false which means that 3 is prime so we do the same action again.
We loop through all the numbers and mark the numbers which are divisible on 3 (except 3 himself)
1 2 3 4 5 6 7 8 9 10
f  f  f  t  f  t  f  t  t   t
Now we are checking 4. 4 is true which means that we know that 4 is not prime.
We move to next step.
5 is false so we have to mark all the number which are divisible on 5
1 2 3 4 5 6 7 8 9 10
f  f  f  t  f  t  f  t  t   t
So, we got that 2,3,5,7 are primes in 5*log10


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