Given an integer N, the task is to find the sum of exponents of prime factors of numbers 1 to N.
Examples:
Input: N = 4
Output: 4
Explanation:
Numbers up to 4 are 1, 2, 3, 4 where
The exponent of 1 in the prime factorization of 1 is 0 (20),
For 2 it is 1 (21),
For 3 it is 1 (31), and
For 4 it is 2 (22).
The sum of the exponent of prime factors of each number up to 4 is 0 + 1 + 1 + 2 = 4.Input: N = 10
Output: 15
Explanation:
sum of the exponent of prime factors of each number up to 10 is 15.
Approach: The idea is to use the concept of Prime factors and their powers. Below are the steps:
- Iterate for each number from 2 to N and for each number do the following:
- find the power of prime factors for each number N.
- Find the summation of each power in the above steps
- Print the summation of all the powers of prime factors of N and print the sum.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to implement sieve of// eratosthenesvoid sieveOfEratosthenes(int N, int s[]){ // Create a boolean array and // initialize all entries as false vector<bool> prime(N + 1, false); // Initializing smallest // factor equal to 2 // for all the even numbers for (int i = 2; i <= N; i += 2) s[i] = 2; // Iterate for odd numbers // less than equal to n for (int i = 3; i <= N; i += 2) { if (prime[i] == false) { // s(i) for a prime is // the number itself s[i] = i; // For all multiples of // current prime number for (int j = i; j * i <= N; j += 2) { if (prime[i * j] == false) { prime[i * j] = true; // i is the smallest // prime factor for // number "i*j" s[i * j] = i; } } } }}// Function to generate prime// factors and its powerint generatePrimeFactors(int N){ // s[i] is going to store // smallest prime factor of i int s[N + 1]; int sum = 0; sieveOfEratosthenes(N, s); // Current prime factor of N int curr = s[N]; // Power of current prime factor int cnt = 1; // Calculating prime factors // and their powers sum while (N > 1) { N /= s[N]; if (curr == s[N]) { // Increment the count and // continue the process cnt++; continue; } // Add count to the sum sum = sum + cnt; curr = s[N]; // Reinitialize count cnt = 1; } // Return the result return sum;}// Function to find the sum of all the// power of prime factors of Nvoid findSum(int N){ int sum = 0; // Iterate for in [2, N] for (int i = 2; i <= N; i++) { sum += generatePrimeFactors(i); } cout << sum << endl;}// Driver Codeint main(){ // Given Number N int N = 4; // Function Call findSum(N); return 0;} |
Java
// Java program for the above approach import java.io.*;public class GFG{ // Function to implement sieve of // eratosthenes static void sieveOfEratosthenes(int N, int s[]) { // Create a boolean array and // initialize all entries as false boolean []prime = new boolean[N + 1]; // Initializing smallest // factor equal to 2 // for all the even numbers for(int i = 2; i <= N; i += 2) s[i] = 2; // Iterate for odd numbers // less than equal to n for(int i = 3; i <= N; i += 2) { if (prime[i] == false) { // s(i) for a prime is // the number itself s[i] = i; // For all multiples of // current prime number for(int j = i; j * i <= N; j += 2) { if (prime[i * j] == false) { prime[i * j] = true; // i is the smallest // prime factor for // number "i*j" s[i * j] = i; } } } } } // Function to generate prime // factors and its power static int generatePrimeFactors(int N) { // s[i] is going to store // smallest prime factor of i int []s = new int[N + 1]; int sum = 0; sieveOfEratosthenes(N, s); // Current prime factor of N int curr = s[N]; // Power of current prime factor int cnt = 1; // Calculating prime factors // and their powers sum while (N > 1) { N /= s[N]; if (curr == s[N]) { // Increment the count and // continue the process cnt++; continue; } // Add count to the sum sum = sum + cnt; curr = s[N]; // Reinitialize count cnt = 1; } // Return the result return sum; } // Function to find the sum of all the // power of prime factors of N static void findSum(int N) { int sum = 0; // Iterate for in [2, N] for(int i = 2; i <= N; i++) { sum += generatePrimeFactors(i); } System.out.print(sum + "\n"); } // Driver Code public static void main(String[] args) { // Given Number N int N = 4; // Function Call findSum(N); }}// This code is contributed by Ajay Kumar |
Python3
# Python3 program for the above approach # Function to implement sieve of # eratosthenes def sieveOfEratosthenes(N, s): # Create a boolean array and # initialize all entries as false prime = [False] * (N + 1) # Initializing smallest # factor equal to 2 # for all the even numbers for i in range(2, N + 1, 2): s[i] = 2 # Iterate for odd numbers # less than equal to n for i in range(3, N + 1, 2): if (prime[i] == False): # s(i) for a prime is # the number itself s[i] = i # For all multiples of # current prime number j = i while (j * i <= N): if (prime[i * j] == False): prime[i * j] = True # i is the smallest # prime factor for # number "i*j" s[i * j] = i j += 2# Function to generate prime # factors and its power def generatePrimeFactors(N): # s[i] is going to store # smallest prime factor of i s = [0] * (N + 1) sum = 0 sieveOfEratosthenes(N, s) # Current prime factor of N curr = s[N] # Power of current prime factor cnt = 1 # Calculating prime factors # and their powers sum while (N > 1): N //= s[N] if (curr == s[N]): # Increment the count and # continue the process cnt += 1 continue # Add count to the sum sum = sum + cnt curr = s[N] # Reinitialize count cnt = 1 # Return the result return sum# Function to find the sum of all the # power of prime factors of N def findSum (N): sum = 0 # Iterate for in [2, N] for i in range(2, N + 1): sum += generatePrimeFactors(i) print(sum)# Driver Codeif __name__ == '__main__': # Given number N N = 4 # Function call findSum(N)# This code is contributed by himanshu77 |
C#
// C# program for the above approach using System;class GFG{ // Function to implement sieve of // eratosthenes static void sieveOfEratosthenes(int N, int []s) { // Create a bool array and // initialize all entries as false bool []prime = new bool[N + 1]; // Initializing smallest // factor equal to 2 // for all the even numbers for(int i = 2; i <= N; i += 2) s[i] = 2; // Iterate for odd numbers // less than equal to n for(int i = 3; i <= N; i += 2) { if (prime[i] == false) { // s(i) for a prime is // the number itself s[i] = i; // For all multiples of // current prime number for(int j = i; j * i <= N; j += 2) { if (prime[i * j] == false) { prime[i * j] = true; // i is the smallest // prime factor for // number "i*j" s[i * j] = i; } } } } } // Function to generate prime // factors and its power static int generatePrimeFactors(int N) { // s[i] is going to store // smallest prime factor of i int []s = new int[N + 1]; int sum = 0; sieveOfEratosthenes(N, s); // Current prime factor of N int curr = s[N]; // Power of current prime factor int cnt = 1; // Calculating prime factors // and their powers sum while (N > 1) { N /= s[N]; if (curr == s[N]) { // Increment the count and // continue the process cnt++; continue; } // Add count to the sum sum = sum + cnt; curr = s[N]; // Reinitialize count cnt = 1; } // Return the result return sum; } // Function to find the sum of all the // power of prime factors of N static void findSum(int N) { int sum = 0; // Iterate for in [2, N] for(int i = 2; i <= N; i++) { sum += generatePrimeFactors(i); } Console.Write(sum + "\n"); } // Driver Code public static void Main(String[] args) { // Given number N int N = 4; // Function call findSum(N); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program for the above approach // Function to implement sieve of // eratosthenes function sieveOfEratosthenes(N, s) { // Create a boolean array and // initialize all entries as false let prime = Array.from({length: N+1}, (_, i) => 0); // Initializing smallest // factor equal to 2 // for all the even numbers for(let i = 2; i <= N; i += 2) s[i] = 2; // Iterate for odd numbers // less than equal to n for(let i = 3; i <= N; i += 2) { if (prime[i] == false) { // s(i) for a prime is // the number itself s[i] = i; // For all multiples of // current prime number for(let j = i; j * i <= N; j += 2) { if (prime[i * j] == false) { prime[i * j] = true; // i is the smallest // prime factor for // number "i*j" s[i * j] = i; } } } } } // Function to generate prime // factors and its power function generatePrimeFactors(N) { // s[i] is going to store // smallest prime factor of i let s = Array.from({length: N+1}, (_, i) => 0); let sum = 0; sieveOfEratosthenes(N, s); // Current prime factor of N let curr = s[N]; // Power of current prime factor let cnt = 1; // Calculating prime factors // and their powers sum while (N > 1) { N /= s[N]; if (curr == s[N]) { // Increment the count and // continue the process cnt++; continue; } // Add count to the sum sum = sum + cnt; curr = s[N]; // Reinitialize count cnt = 1; } // Return the result return sum; } // Function to find the sum of all the // power of prime factors of N function findSum(N) { let sum = 0; // Iterate for in [2, N] for(let i = 2; i <= N; i++) { sum += generatePrimeFactors(i); } document.write(sum + "\n"); } // Driver Code // Given Number N let N = 4; // Function Call findSum(N); </script> |
Output:
4
Time Complexity: O(N*log2N)
Auxiliary Space: O(N)
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