Find the sum of all the numbers between the range l and r. Here each number is represented by the sum of its distinct prime factors.
Examples:
Input: l = 1, r = 6
Output: 17
Explanation: For 1, sum of prime factors = 0
For 2, Sum of Prime factors = 2
For 3, Sum of Prime factors = 3
For 4, Sum of Prime factors = 2
For 5, Sum of Prime factors = 5
For 6, Sum of Prime factors = 2 + 3 = 5
So, Total sum of all prime factors for the given range = 2 + 3 + 2 + 5 + 5 = 17Input: l = 11, r = 15
Output: 46
Explanation: For 11, sum of prime factors = 11
For 12, Sum of Prime factors = 2 + 3 = 5
For 13, Sum of Prime factors = 13
For 14, Sum of Prime factors = 2 + 7 = 9
For 15, Sum of Prime factors = 3 + 5 = 8
So, Total sum of all prime factors for the given range = 11 + 5 + 13 + 9 + 8 = 46
Approach: To solve the problem follow the below steps:
- Create a function to find out all prime factors of a number and sum all prime factors which will represent that number.
- Sum all the modified numbers in the range [l, r] numbers and return that as the total sum.
Below is the implementation of the above approach.
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Function to check prime
bool isPrime(int x)
{
if (x == 1)
return false;
if (x == 2)
return true;
for (int i = 2; i * i <= x; i++) {
// If X has factor that is i,
// then return not Prime
if (x % i == 0)
return false;
}
// If we reach here means no factor
// found so return prime
return true;
}
// This function is to represent a number
// as a sum of all its prime factors
int SumOfPrimeFactors(int l, int r)
{
int sum = 0, ans = 0;
for (int i = l; i <= r; i++) {
sum = 0;
for (int j = 1; j * j <= i; j++) {
// If num has factor i and that
// also prime
if (i % j == 0) {
if (isPrime(j))
sum += j;
if (i / j != j and isPrime(i / j))
sum += i / j;
}
}
ans += sum;
}
return ans;
}
// Driver Code
int main()
{
int l = 11, r = 15;
// Function call
cout << SumOfPrimeFactors(l, r);
return 0;
}
// Java code to implement the approach
import java.io.*;
class GFG {
// Function to check prime
static boolean isPrime(int x)
{
if (x == 1)
return false;
if (x == 2)
return true;
for (int i = 2; i * i <= x; i++) {
// If X has factor that is i,
// then return not Prime
if (x % i == 0)
return false;
}
// If we reach here means no factor
// found so return prime
return true;
}
// This function is to represent a number
// as a sum of all its prime factors
static int SumOfPrimeFactors(int l, int r)
{
int sum = 0, ans = 0;
for (int i = l; i <= r; i++) {
sum = 0;
for (int j = 1; j * j <= i; j++) {
// If num has factor i and that
// also prime
if (i % j == 0) {
if (isPrime(j)) {
sum += j;
}
if (i / j != j && isPrime(i / j)) {
sum += i / j;
}
}
}
ans += sum;
}
return ans;
}
public static void main(String[] args)
{
int l = 11, r = 15;
// Function call
System.out.print(SumOfPrimeFactors(l, r));
}
}
// This code is contributed by lokeshmvs21.
# python code to implement the approach
import math
# Function to check prime
def isPrime(x):
if (x == 1):
return False
if (x == 2):
return True
for i in range(2,int(math.sqrt(x))+1):
# If X has factor that is i,
# then return not Prime
if (x % i == 0):
return False
# If we reach here means no factor
# found so return prime
return True
# This function is to represent a number
# as a sum of all its prime factors
def SumOfPrimeFactors(l,r):
sumvar = 0
ans = 0
#for ( i = l i <= r i++)
for i in range(l,r+1):
sumvar = 0
# for (int j = 1; j * j <= i; j++) {
for j in range(1,int(math.sqrt(i))+1):
# If num has factor i and that
# also prime
if (i % j == 0):
if (isPrime(j)):
sumvar += j
if (int(i / j) != j and isPrime(int(i / j))):
sumvar += math.floor(i / j)
ans += sumvar
return ans
l = 11
r = 15
# Function call
print(SumOfPrimeFactors(l, r))
# This code is contributed by ksam24000
// C# implementation
using System;
public class GFG {
// Function to check prime
public static bool isPrime(int x)
{
if (x == 1)
return false;
if (x == 2)
return true;
for (int i = 2; i * i <= x; i++) {
// If X has factor that is i,
// then return not Prime
if (x % i == 0)
return false;
}
// If we reach here means no factor
// found so return prime
return true;
}
// This function is to represent a number
// as a sum of all its prime factors
public static int SumOfPrimeFactors(int l, int r)
{
int sum = 0, ans = 0;
for (int i = l; i <= r; i++) {
sum = 0;
for (int j = 1; j * j <= i; j++) {
// If num has factor i and that
// also prime
if (i % j == 0) {
if (isPrime(j) == true)
sum += j;
if ((int)(i / j) != j
&& isPrime((int)(i / j)))
sum += i / j;
}
}
ans += sum;
}
return ans;
}
static public void Main()
{
int l = 11, r = 15;
// Function call
Console.WriteLine(SumOfPrimeFactors(l, r));
}
}
// this code is contributed by ksam24000
// Javascript code to implement the approach
// Function to check prime
function isPrime(x)
{
if (x == 1)
return false;
if (x == 2)
return true;
for (let i = 2; i * i <= x; i++) {
// If X has factor that is i,
// then return not Prime
if (x % i == 0)
return false;
}
// If we reach here means no factor
// found so return prime
return true;
}
// This function is to represent a number
// as a sum of all its prime factors
function SumOfPrimeFactors(l, r)
{
let sum = 0, ans = 0;
for (let i = l; i <= r; i++) {
sum = 0;
for (let j = 1; j * j <= i; j++) {
// If num has factor i and that
// also prime
if (i % j == 0) {
if (isPrime(j))
sum += j;
if (i / j != j && isPrime(i / j))
sum += i / j;
}
}
ans += sum;
}
return ans;
}
// Driver Code
let l = 11, r = 15;
// Function call
console.log(SumOfPrimeFactors(l, r));
// This code is contributed by Samim Hossain Mondal.
Output
46
Time Complexity: O(N * sqrt(r) * sqrt(r)) where n is the number of elements in the range and it takes maximum sqrt(r) time each to find all the factors and check if the factors are prime.
Auxiliary Space: O(1)
