使用筛查发现小于长数的质数

Question

I have to find number of prime numbers less than or equal to a given number n . I am using Sieve of Eratosthenes for finding primes.

This is the code I've written:

static int find_prime(int n) {
    boolean[] arr = new boolean[n+1];
    Arrays.fill(arr, true);

for(int i=2;i<=n;i++) {
    if(arr[i]==true) {
        for(int j=2*i; j<=n; j+=i)
            arr[j] = false;
    }
}
int count=0;
for(int i=2;i<=n;i++) {
    //System.out.print(arr[i]+" ");
    if (arr[i] == true)
        count++;
    }
    return count;
}

This code works well for integer values, but I have to implement it for long numbers. And Java doesn't allow creating long size array so boolean[] arr = new boolean[n+1]; doesn't work. Can someone suggest me a way to solve this?

Answer 1

You wouldn't have enough memory to represent the entire sieve, because it is in exabytes. Even BitSet is not going to fit all the indexes that you need, because ultimately it uses an array of long s to store the bits.

You can find your answer with a mixed approach:

• Build and run a sieve up to Integer.MAX_VALUE
• Harvest primes up to Integer.MAX_VALUE into an array of long s
• Observe that you can stop checking for divisors upon reaching square root of candidate number c
• This means that you already have all potential divisors for values above Integer.MAX_VALUE
• Use array of divisors to filter numbers between Integer.MAX_VALUE and n

Since you do not need to store additional primes after Integer.MAX_VALUE , your code would not run out of memory.