Queries on the sum of prime factor counts in a range

1

There are Q queries. Each query is of the form of L and R. The task is to output sum of number of prime factors of each number in the given range of each query.

Examples:

Input : Q = 2
        L = 6, R = 10
        L = 1, R = 5        
Output : 7 
         4
For query 1,
6 => 2  [Prime factors are 2 and 3]
7 => 1
8 => 1
9 => 1
10 => 2
Sum = 2 + 1 + 1 + 1 + 2 = 7
For query 2,
1 => 0
2 => 1
3 => 1
4 => 1
5 => 1
Sum = 0 + 1 + 1 + 1 + 1 = 4.

Method 1 (brute force):
The idea is to traverse from L to R for each query, and for each number find the number of prime factor and add to the answer.

Method 2 (efficient approach):
The idea is to use the Sieve of Eratosthenes method for counting the number of prime factor of composite numbers. Just like, the inner loop of Sieve of Eratosthenes is used to mark composite number. We can use it for incrementing the prime factor of numbers. Instead of marking each array cell as 0 or 1, we can store the number of prime number of that index. And then for each query, find the sum of array from L to R.

Below is the C++ implementation of this approach:

// C++ program to find sum prime factors
// in given range.
#include <bits/stdc++.h>
#define MAX 1000006
using namespace std;

// using sieve method to evaluating
// the prime factor of numbers
void sieve(int count [])
{
    for (int i = 2; i*i <= MAX; i++)
    {
        // if i is prime
        if (count[i] == 0)
        {
            for (int j = 2*i; j < MAX; j+=i)
                count[j]++;

            // setting number of prime
            // factor of a prime number.
            count[i] = 1;
        }
    }
}

// Returns sum of counts of prime factors in
// range from l to r. This function mainly
// uses count[] which is filled by Sieve()
int query(int count[], int l, int r)
{
    int sum = 0;

    // finding the sum of number of prime
    // factor of numbers in a range.
    for (int i = l; i <= r; i++)
        sum += count[i];

    return sum;
}

// Driven Program
int main()
{
    int count[MAX];
    memset(count, 0, sizeof count);
    sieve(count);

    cout << query(count, 6, 10) << endl
         << query(count, 1, 5);

    return 0;
}

Output:

7 4

This article is contributed by Anuj chauhan. 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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