找出有 4 个不同质因数的数

Question

Get the number n from the user.Then print the nth number that has at-least 4 distinct prime factors. Eg(210 prime factors(2,3,5,7) ) 210 is the first number that is having 4 distinct prime factors the next number is 330(2,3,5,11). Input:2 output:330 and Input:3 output:390.I don't know how to do this?I tried to find the prime factors for a number.

 for (int i = 2; i <= number; i++) {
     while (number % i == 0) {
        System.out.print(i + " ");
        number = number / i;
     }
 }
 if (number < 1) 
     System.out.println(number);

But i want to print the nth number that has 4 distinct prime factors.

Answer 1

You may use the following code:

public static boolean findPrimeFactors(int n) {
    Set<Integer> primeFactorSet = new HashSet<>();
    while (n % 2 == 0) {
        // here number is even so adding 2
        primeFactorSet.add(2);
        n /= 2;
    }

    // number would be odd in this loop
    for (int i = 3; i <= Math.sqrt(n); i += 2) {
        while (n % i == 0) {
            primeFactorSet.add(i);
            n /= i;
        }
    }

    if (n > 2) {
        primeFactorSet.add(n);
    }

    // true if the unique prime-factors are greater than or equal to 4
    return primeFactorSet.size() >= 4 ? true : false;
}

Now invoke it using:

public static void main(String[] args) {
    List<Integer> primeFactorList = new ArrayList<Integer>();
    // accept this from user
    int n = 2;

    for (int i = 210;; i++) {
        // find prime factors for each number
        if (findPrimeFactors(i)) {
            primeFactorList.add(i);
        }
        if (primeFactorList.size() == n) {
            System.out.println(primeFactorList.get(n - 1));
            break;
        }
    }
}

Explanation:

1. The loop iterates from 210 till the nth number that has four or more different prime factors.
2. For every number that meets the criteria, the method returns true else false.
3. Next a check is made to see if the size of the list is equal to the number ( n ) entered by the user. If its equal the n-1 th index is fetched and the loop is exited.

Answer 2

One problem with the above method is that it divides by all odd numbers up to the square root of the candidate. It would be better to divide by previously accumulated primes. That is easily done as follows:

      List<Integer> primes = new ArrayList<>(List.of(2));
      outer:
      for (int i = 3; i < n; i++) {
         int end = (int)Math.sqrt(i);
         for (int k = 0, p = primes.get(0); p <= end; p =
               primes.get(k++)) {
            if (i % p == 0) {
               continue outer;
            }
         }
         primes.add(i);
      }

And move the square root calculation out of the loop. It is evaluated on every cycle.