Given an integer N, the task is to count all pairs of divisors of N such that the sum of each pair is coprime with N.
Examples:
Input: N = 24
Output: 2
Explanation:
There are 2 pairs (1, 24) and (2, 3) whose sum is coprime with 24Input: 105
Output: 4
Explanation:
There are 4 pairs (1, 105), (3, 35), (5, 21) and (7, 15) whose sum is coprime with 105
Approach:
To solve the problem mentioned above we can easily calculate the result by finding all divisors in ?N complexity, and check for each pair, whether its sum is coprime with N or not.’
Below is the implementation of the above approach:
C++
// C++ program to count all pairs// of divisors such that their sum// is coprime with N#include <bits/stdc++.h>using namespace std;// Function to calculate GCDint gcd(int a, int b){ if (b == 0) return a; return (gcd(b, a % b));}// Function to count all valid pairsint CountPairs(int n){ // Initialize count int cnt = 0; for (int i = 1; i * i <= n; i++) { if (n % i == 0) { int div1 = i; int div2 = n / i; int sum = div1 + div2; // Check if sum of pair // and n are coprime if (gcd(sum, n) == 1) cnt += 1; } } // Return the result return cnt;}// Driver codeint main(){ int n = 24; cout << CountPairs(n) << endl; return 0;} |
Java
// Java program to count all pairs// of divisors such that their sum// is coprime with Nimport java.util.*;class GFG{// Function to calculate GCDpublic static int gcd(int a, int b){ if (b == 0) return a; return (gcd(b, a % b));}// Function to count all valid pairspublic static int CountPairs(int n){ // Initialize count int cnt = 0; for(int i = 1; i * i <= n; i++) { if (n % i == 0) { int div1 = i; int div2 = n / i; int sum = div1 + div2; // Check if sum of pair // and n are coprime if (gcd(sum, n) == 1) cnt += 1; } } // Return the result return cnt;}// Driver codepublic static void main(String[] args){ int n = 24; System.out.println(CountPairs(n));}}// This code is contributed by equbalzeeshan |
Python3
# Python3 program to count all pairs# of divisors such that their sum# is coprime with Nimport math as m# Function to count all valid pairsdef CountPairs(n): # initialize count cnt = 0 i = 1 while i * i <= n : if(n % i == 0): div1 = i div2 = n//i sum = div1 + div2; # Check if sum of pair # and n are coprime if( m.gcd(sum, n) == 1): cnt += 1 i += 1 # Return the result return cnt # Driver code n = 24print(CountPairs(n)) |
C#
// C# program to count all pairs of// divisors such that their sum// is coprime with Nusing System;class GFG { // Function to find gcd of a and b static int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a % b); } // Function to count all valid pairs static int CountPairs(int n) { // Initialize count int cnt = 0; for (int i = 1; i * i <= n; i++) { if (n % i == 0) { int div1 = i; int div2 = n / i; int sum = div1 + div2; // Check if sum of pair // and n are coprime if (gcd(sum, n) == 1) cnt += 1; } } // Return the result return cnt; } // Driver method public static void Main() { int n = 24; Console.WriteLine(CountPairs(n)); }} |
Javascript
<script>// JavaScript program to count all pairs // of divisors such that their sum // is coprime with N // Function to calculate GCD function gcd(a, b) { if (b == 0) return a; return (gcd(b, a % b)); } // Function to count all valid pairs function CountPairs(n) { // Initialize count let cnt = 0; for (let i = 1; i * i <= n; i++) { if (n % i == 0) { let div1 = i; let div2 = Math.floor(n / i); let sum = div1 + div2; // Check if sum of pair // and n are coprime if (gcd(sum, n) == 1) cnt += 1; } } // Return the result return cnt; } // Driver code let n = 24; document.write(CountPairs(n) + "<br>"); // This code is contributed by Surbhi Tyagi.</script> |
Output:
2
Time Complexity: O(?N * log(N))
Auxiliary Space: O(log(min(a,b)) for call stack
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