Given a positive integer N, the task is to generate an array such that the sum of the Euler Totient Function of each element is equal to N.
Examples:
Input: N = 6
Output: 1 6 2 3Input: N = 12
Output: 1 12 2 6 3 4
Approach: The given problem can be solved based on the divisor sum property of the Euler Totient Function, i.e.,
- The Euler Totient Function of a number N< is the number of integers from 1 to N that gives GCD(i, N) as 1 and a number N can be represented as the summation of the Euler Totient Function values of all the divisors of N.
- Therefore, the idea is to find the divisors of the given number N as the resultant array.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to construct the array such// the sum of values of Euler Totient// functions of all array elements is Nvoid constructArray(int N){ // Stores the resultant array vector<int> ans; // Find divisors in sqrt(N) for (int i = 1; i * i <= N; i++) { // If N is divisible by i if (N % i == 0) { // Push the current divisor ans.push_back(i); // If N is not a // perfect square if (N != (i * i)) { // Push the second divisor ans.push_back(N / i); } } } // Print the resultant array for (auto it : ans) { cout << it << " "; }}// Driver Codeint main(){ int N = 12; // Function Call constructArray(N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{ // Function to construct the array such// the sum of values of Euler Totient// functions of all array elements is Nstatic void constructArray(int N){ // Stores the resultant array ArrayList<Integer> ans = new ArrayList<Integer>(); // Find divisors in sqrt(N) for(int i = 1; i * i <= N; i++) { // If N is divisible by i if (N % i == 0) { // Push the current divisor ans.add(i); // If N is not a // perfect square if (N != (i * i)) { // Push the second divisor ans.add(N / i); } } } // Print the resultant array for(int it : ans) { System.out.print(it + " "); }}// Driver Codepublic static void main(String[] args){ int N = 12; // Function Call constructArray(N);}}// This code is contributed by splevel62 |
Python3
# Python3 program for the above approachfrom math import sqrt# Function to construct the array such# the sum of values of Euler Totient# functions of all array elements is Ndef constructArray(N): # Stores the resultant array ans = [] # Find divisors in sqrt(N) for i in range(1, int(sqrt(N)) + 1, 1): # If N is divisible by i if (N % i == 0): # Push the current divisor ans.append(i) # If N is not a # perfect square if (N != (i * i)): # Push the second divisor ans.append(N / i) # Print the resultant array for it in ans: print(int(it), end = " ")# Driver Codeif __name__ == '__main__': N = 12 # Function Call constructArray(N)# This code is contributed by ipg2016107 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ // Function to construct the array such// the sum of values of Euler Totient// functions of all array elements is Nstatic void constructArray(int N){ // Stores the resultant array List<int> ans = new List<int>(); // Find divisors in sqrt(N) for(int i = 1; i * i <= N; i++) { // If N is divisible by i if (N % i == 0) { // Push the current divisor ans.Add(i); // If N is not a // perfect square if (N != (i * i)) { // Push the second divisor ans.Add(N / i); } } } // Print the resultant array foreach(int it in ans) { Console.Write(it + " "); }}// Driver Codepublic static void Main(){ int N = 12; // Function Call constructArray(N);}}// This code is contributed by ukasp |
Javascript
<script>// javascript program for the above approach // Function to construct the array such// the sum of values of Euler Totient// functions of all array elements is Nfunction constructArray(N){ // Stores the resultant array var ans = []; // Find divisors in sqrt(N) for(var i = 1; i * i <= N; i++) { // If N is divisible by i if (N % i == 0) { // Push the current divisor ans.push(i); // If N is not a // perfect square if (N != (i * i)) { // Push the second divisor ans.push(N / i); } } } // Print the resultant array document.write(ans);}// Driver Codevar N = 12;// Function CallconstructArray(N);// This code contributed by shikhasingrajput </script> |
Output:
1 12 2 6 3 4
Time Complexity: O(√N)
Auxiliary Space: O(N)
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