Find Primes with a sieve? Using 1000 sized One-Dimensional Array ,write a program that finds all primes ?

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Hi all,

Read it carefully and give the C# code for the below

Using 1000 sized One-Dimensional Array ,write a program that finds all primes.

Directions :
2 is a prime ,
Starting with 4,erase all multiples of 2 larger than 2
Erase 4,6,8,10,12,14,16,....etc.
Find next number to 2 ,that is not erased,
Similarly Scan entire array & remaining in the array should be Prime Numbers.

Thanks in advance

S.Naveen...

Posted 22-Sep-10 5:12am

nav234

Updated 23-Sep-10 12:21pm

Toli Cuturicu

v2

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Dylan Morley 22-Sep-10 11:30am

Homework

Smithers-Jones 22-Sep-10 11:32am

Lazy bum.

Toli Cuturicu 23-Sep-10 18:22pm

We know what the Eratostene's sieve is. You don't know. So, you don't have to explain it to us!

Yusuf 23-Sep-10 18:42pm

Hmmm.... unable to do your own homework, huh?

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#realJSOP · Answer 1 · 2010-09-22T07:49:00

Solution 4

Just find a web site that lists all prime numbers under 1000, stick 'em in an array, and there's your list of prime numbers.

Posted 22-Sep-10 7:49am

#realJSOP

Comments

Toli Cuturicu 23-Sep-10 18:20pm

Or you may calculate them by hand using the sieve in less than ten minutes. (any 12 year old in a normal school in my country should be able to do this)

Richard MacCutchan · Answer 2 · 2010-09-22T05:27:00

Solution 1

Quote: "Read it carefully" - OK, did that.

Quote: "give the C# code for the below" - I don't think so.

If you do not have the skills to do your own work then I suggest you take a class or read a book. If you are just too lazy then I suggest you employ someone, the jobs page is over there.

Posted 22-Sep-10 5:27am

Richard MacCutchan

shakil0304003 · Answer 3 · 2010-09-22T05:28:00

long limit,sqrLimit,i,j,n;
            long[] A = new long[1001];

            limit = 1000;
            sqrLimit = Convert.ToInt64(Math.Sqrt(1000));

            for (i = 0; i <= limit; i++)
                A[i] = 0;

                for (i = 2; i <= sqrLimit + 1; i++)
                    for (j = i + i; j <= limit; j = j + i)
                        A[j] = 1;

            n = 0;
            for (i = 2; i <= limit; i++)
                if (A[i] == 0)
                {
                    A[n] = i;
                    n++;
                }