Count number of non-prime pairs that when multiplied form a given number N,

问题

A non-prime pair which forms N is 2 different non-prime numbers where the product of the numbers is N.

1<=N<=10^6

For example For N = 24 there are 2 good pairs (non-prime pairs that form N) (4,6), (1,24), but (2,12), (3,8) are not good.

Note: for any 2 numbers a and b pair(a,b) = pair(b,a).

There is another condition which states that if the number is a special number, so output = -1 otherwise count the number of non-primes.

Number is called special number if it can be represented as a product of three prime numbers. Example: 12 is a special number because 12=2*2*3;

I tried brute-force approach using https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes , which takes O(N*log(log(N)).

"Is there any more optimized way to solve it except brute-force?"

Any idea will be appreciated.

Thanks in advance.

回答1:

First of all, Eratosthenes' sieve is O(N*log(log(N)) to list all primes below or equal N (when well implemented).

Second: suppose you factored your number in Q primes with multiplicity which, without sieving, is a process of O(sqrt(N)) at worst (worst: your number is prime). So you have a map of:

p₀ -> multiplicity m₀
p₁ -> multiplicity m₁
...
p_Q -> multiplicity m_Q

How many divisors made from multiplying at least 2 prime factors?

Well, there will be m0*m1*...mq of them [correction here]. Why? Well, prepare a list of all the divisors generated wit the powers of each factor (including p_i⁰==1), but cross out the ones with a power of 1.

{1, p₀, p₀², ...p₀^m0} are m0 ways of generating divisors with the powers of p0 except p0
{1, p₁, p₁², ...p₁^m1} are m1 ways of generating divisors with the powers of p1 except p1
...
{1, ~~p_Q~~, p1^Q, ...p1^mQ} are mQ ways of generating divisors with the powers of pQ

The number of all combinations with non-prime divisors (as 1 is already included in each set and each prime factors by itself is excluded ) will be the cardinality of the cartesian product of all the above subsets - thus the product of the individual cardinalities, therefore m0*m1*...mq

Code - Java

import java.util.HashMap;
import java.util.Map;

class Example {

  static void factor(long N, Map<Long, Short> primesWithMultiplicity) {
    // some arg checking and trivial cases
    if(N<0) N=-N;
    if(0==N) {
       throw new IllegalArgumentException(
         "Are you kidding me? Every number divides 0, "+
         "you really want them all listed?"
       );
    }
    if(1==N) {
      primesWithMultiplicity.put(1L,(short)1);
      return;
    }

     // don't try divisors higher than sqrt(N), 
    // if they would have been detected by their composite-complement 
    for(long div=2; div*div < N; ) {
      short multiplicity=0;
      while((N % div)==0) {
        multiplicity++;
        N /= div; // reduce N
      }
      if(multiplicity>0) {
        primesWithMultiplicity.put(div, multiplicity);
      }
      div+= (div == 2 ? 1 : 2); // from 2 to 3, but then going only on odd numbers
    }
    // done.. well almost, if N is prime, then 
    // trying to divide up to sqrt(N) will lead an empty result. But,
    // in this case, N will still be at original value (as opposed 
    // to being 1 if complete factored)
    if(N>1) {
      primesWithMultiplicity.put(N, (short)1);
    }
  }

  static int countDistinctCompositePairs(long N) {
    HashMap<Long, Short> factoringResults=new HashMap<>();
    factor(N, factoringResults);
    int ret=1;
    for(short multiplicity : factoringResults.values()) {
      ret*=multiplicity;
    }
    return ret/2;
  }

  static public void main(String[] args) {
    System.out.println(countDistinctCompositePairs(24));
  }
}

回答2:

Since you haven't mentioned the constraints I assume 1<=N<=10^6 Implement the sieve and simultaneously store the greatest prime factor of the numbers up to 10^6.

Here's the code for same.

int a[1000001];
a[1]=1;
for(int i=2;i*i<1000001;i++)
{
    if(a[i]==0)
    {
        for(int j=2*i;j<1000001;j+=i)
        a[j]=i;
    }
}

Now if the number is prime your answer is 0.

if(a[n]==0) 
cout<<'0';

If it is semi prime(product of two primes) your answer will be 1.

if(a[n/a[n]]==0)
cout<<"1";

If it is special then

int x=n/a[n];
if(a[x/a[x]]==0)
cout<<"-1";

If it doesn't satisfy any of the aforementioned conditions then calculate all non prime divisors.

int c=0;
for(int i=1;i*i<n;i++)
    {
        if(n%i==0)
        {
            if(a[i]!=0&&a[n/i]!=0)
            c++;

        }
    }

Hope this helps!

来源：https://stackoverflow.com/questions/40692681/count-number-of-non-prime-pairs-that-when-multiplied-form-a-given-number-n

标签

java

arrays

algorithm

time-complexity

primes