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Sum of largest prime factor of each number less than equal to n
  • Difficulty Level : Easy
  • Last Updated : 17 Dec, 2018

Given a non-negative integer n. The problem is to find the sum of the largest prime factor of each number less than equal to n.

Examples:

Input : n = 10
Output : 32
Largest prime factor of each number
Prime factor of 2 = 2
Prime factor of 3 = 3
Prime factor of 4 = 2
Prime factor of 5 = 5
Prime factor of 6 = 3
Prime factor of 7 = 7
Prime factor of 8 = 2
Prime factor of 9 = 3
Prime factor of 10 = 5

Sum = (2+3+2+5+3+7+2+3+5) = 32

Input : n = 12
Output : 46

Algorithm:

sumOfLargePrimeFactor(n)
    Declare prime[n+1] and initialize all value to 0
    Initialize sum = 0
    max = n / 2
 
    for p = 2 to max
        if prime[p] == 0 then
            i = p*2
            while i <= n
                prime[i] = p
                i = i + p
        
    for p = 2 to n
        if prime[p] then
               sum = sum + prime[p]
           else
              sum = sum + p
    
    return sum

C++




// C++ implementation to find sum of largest prime factor
// of each number less than equal to n
#include <bits/stdc++.h>
  
using namespace std;
  
// function to find sum of largest prime factor
// of each number less than equal to n
int sumOfLargePrimeFactor(int n)
{
    // Create an integer array "prime[0..n]" and initialize
    // all entries of it as 0. A value in prime[i] will
    // finally be 0 if 'i' is a prime, else it will 
    // contain the largest prime factor of 'i'.
    int prime[n+1], sum = 0;
    memset(prime, 0, sizeof(prime));
    int max = n / 2; 
   
    for (int p=2; p<=max; p++)
    {
        // If prime[p] is '0', then it is a 
        // prime number
        if (prime[p] == 0)
        {
            // Update all multiples of p
            for (int i=p*2; i<=n; i += p)
                prime[i] = p;
        }
    }
   
    // Sum up the largest prime factor of all
    // the numbers
    for (int p=2; p<=n; p++)
    {
        // if 'p' is a non- prime number then
        // prime[p] gives its largesr prime
        // factor
        if (prime[p])
             sum += prime[p];
             
        // 'p' is a prime number         
        else
             sum += p;
    }
      
    // required sum
    return sum;      
}
  
// Driver program to test above
int main()
{
    int n = 12;
    cout << "Sum = "
         << sumOfLargePrimeFactor(n);
    return 0;         
}
Java



// Java implementation to find sum
// of largest prime factor of each
// number less than equal to n
import java.io.*;
import java.util.*;
  
class GFG {
      
    // function to find sum of largest
    // prime factorof each number 
    // less than equal to n
    static int sumOfLargePrimeFactor(int n)
    {
        // Create an integer array "prime[0..n]" 
        // and initialize all entries of it as 0.
        // A value in prime[i] will finally be 0 
        // if 'i' is a prime, else it will contain
        // the largest prime factor of 'i'.
        int prime[] = new int[n + 1], sum = 0;
        Arrays.fill(prime, 0);
        int max = n / 2; 
      
        for (int p = 2; p <= max; p++)
        {
            // If prime[p] is '0', then it is a 
            // prime number
            if (prime[p] == 0)
            {
                // Update all multiples of p
                for (int i = p * 2; i <= n; i += p)
                    prime[i] = p;
            }
        }
      
        // Sum up the largest prime factor of all
        // the numbers
        for (int p = 2; p <= n; p++)
        {
            // if 'p' is a non- prime number then
            // prime[p] gives its largesr prime
            // factor
            if (prime[p] != 0)
                sum += prime[p];
                  
            // 'p' is a prime number         
            else
                sum += p;
        }
          
        // required sum
        return sum;     
    }
      
    // Driver program 
    public static void main(String args[])
    {
        int n = 12;
        System.out.println("Sum = " 
                           + sumOfLargePrimeFactor(n));
    }
}
  
  
// This code is contributed 
// by Nikita Tiwari.
Python3



# Python3 code implementation to find
# sum of largest prime factor of 
# each number less than equal to n
  
# function to find sum of largest 
# prime factor of each number less
# than equal to n
def sumOfLargePrimeFactor( n ):
  
    # Create an integer array "prime[0..n]"
    # and initialize all entries of it 
    # as 0. A value in prime[i] will 
    # finally be 0 if 'i' is a prime, 
    # else it will contain the largest
    # prime factor of 'i'.
    prime = [0] * (n + 1)
    sum = 0
    max = int(n / 2)
    for p in range(2, max + 1):
          
        # If prime[p] is '0', then 
        # it is a prime number
        if prime[p] == 0:
              
            # Update all multiples of p
            for i in range(p * 2, n + 1, p):
                prime[i] = p
                  
    # Sum up the largest prime factor
    # of all the numbers
    for p in range(2, n + 1):
          
        # if 'p' is a non- prime 
        # number then prime[p] gives 
        # its largesr prime factor
        if prime[p]:
            sum += prime[p]
          
        # 'p' is a prime number
        else:
            sum += p
      
    # required sum
    return sum
      
# Driver code to test above function
n = 12
print("Sum =", sumOfLargePrimeFactor(n))
  
# This code is contributed by "Sharad_Bhardwaj".
C#



// C# implementation to find sum
// of largest prime factor of each
// number less than equal to n
using System;
  
class GFG {
  
    // function to find sum of largest
    // prime factorof each number
    // less than equal to n
    static int sumOfLargePrimeFactor(int n)
    {
        // Create an integer array "prime[0..n]"
        // and initialize all entries of it as 0.
        // A value in prime[i] will finally be 0
        // if 'i' is a prime, else it will contain
        // the largest prime factor of 'i'.
        int[] prime = new int[n + 1];
        int sum = 0;
  
        for (int i = 1; i < n + 1; i++)
            prime[i] = 0;
  
        int max = n / 2;
  
        for (int p = 2; p <= max; p++) 
        {
            // If prime[p] is '0', then it is a
            // prime number
            if (prime[p] == 0)
            {
                // Update all multiples of p
                for (int i = p * 2; i <= n; i += p)
                    prime[i] = p;
            }
        }
  
        // Sum up the largest prime factor of all
        // the numbers
        for (int p = 2; p <= n; p++)
        {
            // if 'p' is a non- prime number then
            // prime[p] gives its largesr prime
            // factor
            if (prime[p] != 0)
                sum += prime[p];
  
            // 'p' is a prime number
            else
                sum += p;
        }
  
        // required sum
        return sum;
    }
  
    // Driver program
    public static void Main()
    {
        int n = 12;
        Console.WriteLine("Sum = " + sumOfLargePrimeFactor(n));
    }
}
  
// This code is contributed by Sam007
PHP



<?php
// PHP implementation to find sum of largest prime factor
// of each number less than equal to n
  
// function to find sum of largest prime factor
// of each number less than equal to n
function sumOfLargePrimeFactor($n)
{
    // Create an integer array "prime[0..n]" and initialize
    // all entries of it as 0. A value in prime[i] will
    // finally be 0 if 'i' is a prime, else it will 
    // contain the largest prime factor of 'i'.
    $prime=array_fill(0,$n+1,0);
    $sum = 0;
    $max = (int)($n / 2); 
  
    for ($p=2; $p<=$max; $p++)
    {
        // If prime[p] is '0', then it is a 
        // prime number
        if ($prime[$p] == 0)
        {
            // Update all multiples of p
            for ($i=$p*2; $i<=$n; $i += $p)
                $prime[$i] = $p;
        }
    }
  
    // Sum up the largest prime factor of all
    // the numbers
    for ($p=2; $p<=$n; $p++)
    {
        // if 'p' is a non- prime number then
        // prime[p] gives its largesr prime
        // factor
        if ($prime[$p])
            $sum += $prime[$p];
              
        // 'p' is a prime number         
        else
            $sum += $p;
    }
      
    // required sum
    return $sum;     
}
  
// Driver program to test above
  
    $n = 12;
    echo "Sum = ".sumOfLargePrimeFactor($n);
  
// This code is contributed by mits.
?>

Output:
Sum = 46

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