需要帮助计算每个数字的不同质因子的数量，直到给定数字

Question

I want to calculate the amount of different prime factors of every integer up to n. For example, 12 = 2 * 2 * 3, so it has 2 different prime factors, 2 and 3. I want to store each of these values in an array result[], of size n+1, in which result[i] contains the number of different prime factors of the integer i.

I have to use the following external method in my solution:

List<Integer> getPrimes(int n) {
    // create array to determine which numbers are prime
    boolean isPrime[] = new boolean[n+1];
    for(int i=2; i<=n; i++)
      isPrime[i] = true; // assume all are prime, we'll filter below
    
    for(int i=3; i*i<=n; i+=2) {
      if (isPrime[i]) {             // i is prime, so...
        for(int j=i*i; j<=n; j+=i)  //  remove all its multiples
          isPrime[j] = false;       //  by updating array
      }
    }
    
    // create list with only the prime numbers
    List<Integer> primes = new LinkedList<>();
    primes.add(2);
    for(int i=3; i<=n; i+=2)
      if (isPrime[i])
        primes.add(i);
    return primes;
  }

which uses the Sieve of Eratosthenes to return a List of all prime numbers up to n. This was my solution:

int[] getNumPrimeFactors(int n) {
        int[] result = new int[n + 1];
        int counter; // counts the number of different prime factors
        boolean isPrime = true;
        List<Integer> primeList = getPrimes(n);

        for (int i = 2; i < result.length; i++) {
            counter = 0;
            
            // checks if i is prime 
            if (i % 2 == 0) {
                isPrime = false;
            } else {
                for (int j = 3; j * j <= i; j += 2) {
                    if (i % j == 0)
                        isPrime = false;
                }
            }
            // if i isnt prime, counts how many different prime factors it has
            if (!isPrime) {
                for (int prime : primeList) {
                    if (i % prime == 0)
                        counter++;
                }
                result[i] = counter;
            } else {
                result[i] = 1;
            }
        }
        return result;
    }

This algorithm produces the correct results, however, I want to be able to test for n <= 5_000_000, and it isn't efficient enough. Is there any way I can improve the code for very large instances of n? Here are some example test results:

Thank you very much for your help:)

Answer 1

This is definitely not the most effective algorithm, but on my old computer (i5 3570K) it works up to 5_000_000 slightly more than 10 seconds. Sum of result array for 5_000_000 is 14838426.

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class Main {

    public static List<Integer> primeFactors(int number) {
        int n = number;
        List<Integer> factors = new ArrayList<Integer>();
        for (int i = 2; i <= n / i; i++) {
            while (n % i == 0) {
                factors.add(i);
                n /= i;
            }
        }
        if (n > 1) {
            factors.add(n);
        }
        return factors;
    }

    public static void main(String[] args) {
        int[] result = new int[5_000_001];
        for (int i = 0; i < result.length; i++) {
            result[i] = (int) primeFactors(i).stream().distinct().count();
        }
        System.out.println(Arrays.stream(result).sum());
    }
}

Answer 2

One standard "improvement" I often find useful on sieving in this way is set up least prime factors for all the values (instead of just a "true/false" value). Then the "is it prime" decision is a check on whether the least prime factor lpf is less than the value. This is basically nearly as quick as the prime sieve but then gives a more direct route to factorizing numbers in range.

In Python:

lim = 5000000+1 

# precalc all least prime factors for composites in range
lpf = [i for i in range(lim)]
for m in range(4,lim,2):
    lpf[m] = 2
k = 3
while k*k <= lim:
    if lpf[k] == k:
        for m in range(k*k, lim, 2*k):
            if lpf[m] == m:
                lpf[m] = k
    k += 2

print('lpf done',lim) ############

# find number of distinct prime factors for each
result = [0]*lim
for a in range(2,lim):
    pf = lpf[a]
    fc = 1
    res = a
    while pf < res:
        res //= pf
        if pf != lpf[res]:
            fc += 1
            pf = lpf[res]
    result[a] = fc
    
print(result[:11])
print(sum(result[:10+1]))
print(sum(result[:1234+1]))
print(sum(result[:1000000+1]))

The sieve here takes just over two seconds, which is about a quarter of the total time.