# Queries on the sum of prime factor counts in a range

• Difficulty Level : Basic
• Last Updated : 05 May, 2021

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 implementation of this approach:

## C++

 // C++ program to find sum prime factors// in given range.#include #define MAX 1000006using namespace std; // using sieve method to evaluating// the prime factor of numbersvoid 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 Programint main(){    int count[MAX];    memset(count, 0, sizeof count);    sieve(count);     cout << query(count, 6, 10) << endl         << query(count, 1, 5);     return 0;}

## Java

 // Java program to find sum prime// factors in given range. class GFG {     static final int MAX = 1000006;     // using sieve method to evaluating    // the prime factor of numbers    static 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()    static 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;    }     // Driver code    public static void main(String[] args)    {        int count[] = new int[MAX];        sieve(count);         System.out.println(query(count, 6, 10) + " " + query(count, 1, 5));    }} // This code is contributed by Anant Agarwal.

## Python3

 # Python3 program to find sum prime factors# in given range.MAX = 100006;count = [0] * MAX; # using sieve method to evaluating# the prime factor of numbersdef sieve():    i = 2;    while (i * i <= MAX):                 # if i is prime        if (count[i] == 0):            for j in range(2 * i, MAX, i):                count[j] += 1;                         # setting number of prime            # factor of a prime number.            count[i] = 1;                 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()def query(l, r):    sum = 0;     # finding the sum of number of prime    # factor of numbers in a range.    for i in range(l, r + 1):        sum += count[i];     return sum; # Driver Codesieve();print(query(6, 10), query(1, 5)); # This code is contributed by mits

## C#

 // C# program to find sum prime// factors in given range.using System; class GFG {     static int MAX = 1000006;     // using sieve method to evaluating    // the prime factor of numbers    static 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()    static 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;    }     // Driver code    public static void Main()    {                 int[] count = new int[MAX];        sieve(count);         Console.Write(query(count, 6, 10) +                " " + query(count, 1, 5));    }} // This code is contributed by nitin mittal.



## Javascript



Output:

7 4

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