Querying maximum number of divisors that a number in a given range has

4

Given Q queries, of type: L R, for each query you must print the maximum number of divisors that a number x (L <= x <= R) has.
Examples:

L = 1 R = 10:
    1 has 1 divisor.
    2 has 2 divisors.
    3 has 2 divisors.
    4 has 3 divisors.
    5 has 2 divisors.
    6 has 4 divisors.
    7 has 2 divisors.
    8 has 4 divisors.
    9 has 3 divisors.
    10 has 4 divisors.

So the answer for above query is 4, as it is the maximum number of 
divisors a number has in [1, 10].


Pre-requisites : Eratosthenes Sieve, Segment Tree

Below are steps to solve the problem.

  1. Firstly, let’s see how many number of divisors does a number n = p1k1 * p2k2 * … * pnkn (where p1, p2, …, pn are prime numbers) has; the answer is (k1 + 1)*(k2 + 1)*…*(kn + 1). How? For each prime number in the prime factorization, we can have its ki + 1 possible powers in a divisor (0, 1, 2,…, ki).
  2. Now let’s see how can we find the prime factorization of a number, we firstly build an array, smallest_prime[], which stores the smallest prime divisor of i at ith index, we divide a number by its smallest prime divisor to obtain a new number (we also have the smallest prime divisor of this new number stored), we keep doing it until the smallest prime of the number changes, when the smallest prime factor of the new number is different from the previous number’s, we have ki for the ith prime number in the prime factorization of the given number.
  3. Finally, we obtain the number of divisors for all the numbers and store these in a segment tree that maintains the maximum numbers in the segments. We respond to each query by querying the segment tree.
// A C++ implementation of the above idea to process
// queries of finding a number with maximum divisors.
#include <bits/stdc++.h>
using namespace std;

#define maxn 1000005
#define INF 99999999

int smallest_prime[maxn];
int divisors[maxn];
int segmentTree[4 * maxn];

// Finds smallest prime factor of all numbers in
// range[1, maxn) and stores them in smallest_prime[],
// smallest_prime[i] should contain the smallest prime
// that divides i
void findSmallestPrimeFactors()
{
    // Initialize the smallest_prime factors of all
    // to infinity
    for (int i = 0 ; i < maxn ; i ++ )
        smallest_prime[i] = INF;

    // to be built like eratosthenes sieve
    for (long long i = 2; i < maxn; i++)
    {
        if (smallest_prime[i] == INF)
        {
            // prime number will have its smallest_prime
            // equal to itself
            smallest_prime[i] = i;
            for (long long j = i * i; j < maxn; j += i)

                // if 'i' is the first prime number reaching 'j'
                if (smallest_prime[j] > i)
                    smallest_prime[j] = i;
        }
    }
}

// number of divisors of n = (p1 ^ k1) * (p2 ^ k2) ... (pn ^ kn)
// are equal to (k1+1) * (k2+1) ... (kn+1)
// this function finds the number of divisors of all numbers
// in range [1, maxn) and stores it in divisors[]
// divisors[i] stores the number of divisors i has
void buildDivisorsArray()
{
    for (int i = 1; i < maxn; i++)
    {
        divisors[i] = 1;
        int n = i, p = smallest_prime[i], k = 0;

        // we can obtain the prime factorization of the number n
        // n = (p1 ^ k1) * (p2 ^ k2) ... (pn ^ kn) using the
        // smallest_prime[] array, we keep dividing n by its
        // smallest_prime until it becomes 1, whilst we check
        // if we have need to set k zero
        while (n > 1)
        {
            n = n / p;
            k ++;

            if (smallest_prime[n] != p)
            {
                //use p^k, initialize k to 0
                divisors[i] = divisors[i] * (k + 1);
                k = 0;
            }

            p = smallest_prime[n];
        }
    }
}

// builds segment tree for divisors[] array
void buildSegtmentTree(int node, int a, int b)
{
    // leaf node
    if (a == b)
    {
        segmentTree[node] = divisors[a];
        return ;
    }

    //build left and right subtree
    buildSegtmentTree(2 * node, a, (a + b) / 2);
    buildSegtmentTree(2 * node + 1, ((a + b) / 2) + 1, b);

    //combine the information from left
    //and right subtree at current node
    segmentTree[node] = max(segmentTree[2 * node],
                            segmentTree[2 *node + 1]);
}

//returns the maximum number of divisors in [l, r]
int query(int node, int a, int b, int l, int r)
{
    // If current node's range is disjoint with query range
    if (l > b || a > r)
        return -1;

    // If the current node stores information for the range
    // that is completely inside the query range
    if (a >= l && b <= r)
        return segmentTree[node];

    // Returns maximum number of divisors from left
    // or right subtree
    return max(query(2 * node, a, (a + b) / 2, l, r),
               query(2 * node + 1, ((a + b) / 2) + 1, b,l,r));
}

// driver code
int main()
{
    // First find smallest prime divisors for all
    // the numbers
    findSmallestPrimeFactors();

    // Then build the divisors[] array to store
    // the number of divisors
    buildDivisorsArray();

    // Build segment tree for the divisors[] array
    buildSegtmentTree(1, 1, maxn - 1);

    cout << "Maximum divisors that a number has "
         << " in [1, 100] are "
         << query(1, 1, maxn - 1, 1, 100) << endl;


    cout << "Maximum divisors that a number has"
         << " in [10, 48] are "
         << query(1, 1, maxn - 1, 10, 48) << endl;


    cout << "Maximum divisors that a number has"
         << " in [1, 10] are "
         << query(1, 1, maxn - 1, 1, 10) << endl;

    return 0;
}

Output:

Maximum divisors that a number has in [1, 100] are 12
Maximum divisors that a number has in [10, 48] are 10
Maximum divisors that a number has in [1, 10] are 4

Time Complexity:

For sieve: O(maxn * log(log(maxn)) )
For calculating divisors of each number: O(k1 + k2 + ... + kn) < O(log(maxn))
For querying each range: O(log(maxn))
Total: O((maxn + Q) * log(maxn))

This article is contributed by Saumye Malhotra. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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