Labels

Project Euler #47: Distinct primes factors

Question
The first two consecutive numbers to have two distinct prime factors are:
14 = 2 × 7
15 = 3 × 5
The first three consecutive numbers to have three distinct prime factors are:
644 = 2² × 7 × 23
645 = 3 × 5 × 43
646 = 2 × 17 × 19.
Find the first four consecutive integers to have four distinct prime factors each. What is the first of these numbers?

Answer : 134043

Hacker Rank Problem

Solution

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
import java.util.*;

public class Solution {
    private static boolean[] prime = null;
    
    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        int limit = 2500000;
        int root = (int) Math.ceil(Math.sqrt(limit));
        prime = generatePrimes(root, limit);

        for(int i=2;i<=n;i++)
        {
            ArrayList<TreeSet> list = new ArrayList<TreeSet>();
            if(!prime[i])
            {
                for(int j=0;j<k;j++)
                {
                    if(!prime[i+j])
                    {
                        TreeSet s = primeFactorize(i+j);
                        if(s.size()==k)
                            list.add(s);
                        else break;
                    }
                    else break;
                }
            }
            if(list.size()==k)
                System.out.println(i);
        }
    }
    
    public static TreeSet<Integer> primeFactorize(Integer n)
    {
        try
        {
            TreeSet<Integer> set = new TreeSet<Integer>();
            for(int i=2;i<=n/i;i++)
            {
                while(n%i==0)
                {
                    set.add(i);
                    n/=i;
                }
            }
            if(n>1) set.add(n);
            return set;
        }
        catch(Exception e)
        {
            throw e;
        }
    }
    
    public static boolean[] generatePrimes(int root, int limit)
    {
        boolean[] prime = new boolean[limit+1];
        prime[2] = true;
        prime[3] = true;

        //Sieve of Atkin for prime number generation
        for (int x = 1; x < root; x++)
        {
            for (int y = 1; y < root; y++)
            {
                int n = 4 * x * x + y * y;
                if (n <= limit && (n % 12 == 1 || n % 12 == 5))
                    prime[n] = !prime[n];

                n = 3 * x * x + y * y;
                if (n <= limit && n % 12 == 7)
                    prime[n] = !prime[n];

                n = 3 * x * x - y * y;
                if ((x > y) && (n <= limit) && (n % 12 == 11))
                    prime[n] = !prime[n];
            }
        }

        for (int i = 5; i <= root; i++)
        {
            if (prime[i])
            {
                for (int j = i * i; j < limit; j += i * i)
                {
                    prime[j] = false;
                }
            }
        }

        return prime;
    }
}
Labels:

Comments

Post a Comment

Popular Posts