Dado un entero grande positivo n. Cuente el número de divisores positivos de n 2 que no son divisibles por ningún divisor de n(1 <= n <= 10 12 ).
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
El enfoque simple es recorrer para cada divisor de n 2 y contar solo aquellos divisores que no son divisores de ‘n’. La complejidad temporal de este enfoque es O(n).
El enfoque eficiente es utilizar la descomposición en factores primos para contar los divisores totales de n 2 . Un número ‘n’ se puede representar como producto de números primos . Consulte esto para comprender más.
Let 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 required answer
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 code int 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 n import 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 code public 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 n import 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 code n = 5 print(factors(n)) n = 8 print(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 n using 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 code public 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 code let n = 5; document.write(factors(n)+"<br>"); n = 8; document.write(factors(n)+"<br>"); // This code is contributed by unknown2108 </script>
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Complejidad temporal: O(sqrt(n))
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por Shubham Bansal 13 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA