Given an positive large integer n. Count the number of positive divisors of n2 which are not divisible by any divisor of n(1 <= n <= 1012).
Input: 6 Output: 5 Explanation Total divisors of 62 are 9 i.e., 1, 2, 3, 4, 6, 9, 12, 18, 36 Total divisors of '6' are 4, 1, 2, 3, 6 Total divisor of '36' which are not divisible by divisors of '6' are '5' i.e., 4, 9, 12, 18, 36 Input: 8 Output: 3
Simple approach is to traverse for every divisor of n2 and count only those divisors which are not divisor of ‘n’. Time complexity of this approach is O(n).
Efficient approach is to use prime factorization to count total divisors of n2. A number ‘n’ can be represented as product of primes 
. Refer this to understand more.
Letfor some primes p1 and p2.Squaring both the sides
Total factors of n2 will be,
Total factors of 'n' will be,
Difference between the two gives the requiredanswer
for some primes p1 and p2.Squaring both the sides
Total factors of n2 will be,
Total factors of 'n' will be,
Difference between the two gives the requiredanswer
C++
// C++ program to count number of// divisors of n^2 which are not// divisible by divisor of n#include <bits/stdc++.h>using namespace std;// Function to count divisors of n^2// having no factors of 'n'int factors(long long n){ unordered_map<int, int> prime; for (int i = 2; i <= sqrt(n); ++i) { while (n % i == 0) { // Increment count of i-th prime divisor ++prime[i]; // Find next prime divisor n = n / i; } } // Increment count if divisor still remains if (n > 2) ++prime[n]; // Initialize variable for counting the factors // of n^2 and n as ans1 and ans2 respectively int ans1 = 1, ans2 = 1; // Range based for-loop for (auto it : prime) { // Use formula as discussed in above ans1 *= 2 * it.second + 1; ans2 *= it.second + 1; } // return the difference of answers return ans1 - ans2;}// Driver codeint main(){ long long n = 5; cout << factors(n) << endl; n = 8; cout << factors(n); return 0;} |
Java
// Java program to count number of// divisors of n^2 which are not// divisible by divisor of nimport java.util.*;class GFG {// Function to count divisors of n^2// having no factors of 'n'static int factors(int n){ HashMap<Integer, Integer>prime = new HashMap<Integer, Integer>(); for (int i = 2; i <= Math.sqrt(n); ++i) { while (n % i == 0) { // Increment count of i-th prime divisor if (prime.containsKey(i)) { prime.put(i, prime.get(i) + 1); } else { prime.put(i, 1); } // Find next prime divisor n = n / i; } } // Increment count if divisor still remains if (n > 2) { if(prime.containsKey(n)) { prime.put(n, prime.get(n) + 1); } else { prime.put(n, 1); } } // Initialize variable for counting the factors // of n^2 and n as ans1 and ans2 respectively int ans1 = 1, ans2 = 1; // Range based for-loop for (Map.Entry<Integer, Integer> it : prime.entrySet()) { // Use formula as discussed in above ans1 *= 2 * it.getValue() + 1; ans2 *= it.getValue() + 1; } // return the difference of answers return ans1 - ans2;}// Driver codepublic static void main(String[] args) { int n = 5; System.out.println(factors(n)); n = 8; System.out.println(factors(n));}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 program to count number of# divisors of n^2 which are not# divisible by divisor of nimport math as mt# Function to count divisors of n^2# having no factors of 'n'def factors(n): prime = dict() for i in range(2, mt.ceil(mt.sqrt(n + 1))): while (n % i == 0): # Increment count of i-th # prime divisor if i in prime.keys(): prime[i] += 1 else: prime[i] = 1 # Find next prime divisor n = n // i # Increment count if divisor # still remains if (n > 2): if n in prime.keys(): prime[n] += 1 else: prime[n] = 1 # Initialize variable for counting # the factors of n^2 and n as ans1 # and ans2 respectively ans1 = 1 ans2 = 1 # Range based for-loop for it in prime: # Use formula as discussed in above ans1 *= 2 * prime[it] + 1 ans2 *= prime[it] + 1 # return the difference of answers return ans1 - ans2# Driver coden = 5print(factors(n))n = 8print(factors(n))# This code is contributed by # Mohit kumar 29 |
C#
// C# program to count number of// divisors of n^2 which are not// divisible by divisor of nusing System;using System.Collections.Generic; class GFG {// Function to count divisors of n^2// having no factors of 'n'static int factors(int n){ Dictionary<int, int> prime = new Dictionary<int, int>(); for (int i = 2; i <= Math.Sqrt(n); ++i) { while (n % i == 0) { // Increment count of i-th prime divisor if (prime.ContainsKey(i)) { prime[i] = prime[i] + 1; } else { prime.Add(i, 1); } // Find next prime divisor n = n / i; } } // Increment count if divisor still remains if (n > 2) { if(prime.ContainsKey(n)) { prime[n] = prime[n] + 1; } else { prime.Add(n, 1); } } // Initialize variable for counting the factors // of n^2 and n as ans1 and ans2 respectively int ans1 = 1, ans2 = 1; // Range based for-loop foreach(KeyValuePair<int, int> it in prime) { // Use formula as discussed in above ans1 *= 2 * it.Value + 1; ans2 *= it.Value + 1; } // return the difference of answers return ans1 - ans2;}// Driver codepublic static void Main(String[] args) { int n = 5; Console.WriteLine(factors(n)); n = 8; Console.WriteLine(factors(n));}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program to count number of// divisors of n^2 which are not// divisible by divisor of n// Function to count divisors of n^2// having no factors of 'n'function factors(n){ let prime = new Map(); for (let i = 2; i <= Math.sqrt(n); ++i) { while (n % i == 0) { // Increment count of i-th prime divisor if (prime.has(i)) { prime.set(i, prime.get(i) + 1); } else { prime.set(i, 1); } // Find next prime divisor n = Math.floor(n / i); } } // Increment count if divisor still remains if (n > 2) { if(prime.has(n)) { prime.set(n, prime.get(n) + 1); } else { prime.set(n, 1); } } // Initialize variable for counting the factors // of n^2 and n as ans1 and ans2 respectively let ans1 = 1, ans2 = 1; // Range based for-loop for (let [key, value] of prime.entries()) { // Use formula as discussed in above ans1 *= 2 * value + 1; ans2 *= value + 1; } // return the difference of answers return ans1 - ans2;}// Driver codelet n = 5;document.write(factors(n)+"<br>");n = 8;document.write(factors(n)+"<br>"); // This code is contributed by unknown2108</script> |
Output:
1 3
Time complexity: O(log(n))
Auxiliary space: O(n) as using unordered_map
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