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Count all possible values of K less than Y such that GCD(X, Y) = GCD(X+K, Y)
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Generate an array having sum of Euler Totient Function of all elements equal to N

Last Updated : 17 Jan, 2022
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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 3

Input: 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 N void 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 Code int main() {     int N = 12;      // Function Call     constructArray(N);      return 0; } 
Java
// Java program for the above approach import java.util.*;  class GFG{      // Function to construct the array such // the sum of values of Euler Totient // functions of all array elements is N static 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 Code public static void main(String[] args) {     int N = 12;      // Function Call     constructArray(N); } }  // This code is contributed by splevel62 
Python3
# Python3 program for the above approach from math import sqrt  # Function to construct the array such # the sum of values of Euler Totient # functions of all array elements is N def 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 Code if __name__ == '__main__':          N = 12          # Function Call     constructArray(N)  # This code is contributed by ipg2016107 
C#
// C# program for the above approach using 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 N static 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 Code public 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 N function 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 Code var N = 12;  // Function Call constructArray(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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Count all possible values of K less than Y such that GCD(X, Y) = GCD(X+K, Y)

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Article Tags :
  • Algorithms
  • Mathematical
  • Technical Scripter
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  • Technical Scripter 2020
  • divisors
  • euler-totient
Practice Tags :
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