Количество делителей произведения N чисел
Учитывая массив целых чисел arr [] , задача состоит в том, чтобы подсчитать количество делителей произведения всех элементов данного массива.
Примеры:
Input: arr[] = {3, 5, 7}
Output: 8
3 * 5 * 7 = 105.
Factors of 105 are 1, 3, 5, 7, 15, 21, 35 and 105.
Input: arr[] = {5, 5}
Output: 3
5 * 5 = 25.
Factors of 25 are 1, 5 and 25.
Простое решение - умножить все N целых чисел и посчитать количество делителей произведения. Однако, если произведение превышает 10 7, мы не можем использовать этот подход, потому что числа больше 10 ^ 7 не могут быть эффективно разложены на простые множители с использованием подхода сита.
Эффективное решение не предполагает вычисления произведения всех чисел. Мы уже знаем, что когда мы умножаем 2 числа, степени складываются. Например,
A = 27, B = 23
A * B = 210
Therefore, we need to maintain the count of every power in the product of numbers which can be done by adding counts of powers from every element.
Следовательно, при вычислении числа делителей основное внимание уделяется подсчету встречающихся простых чисел. Поэтому мы будем делать упор только на простых элементах, встречающихся в продукте, не беспокоясь о самом продукте. Обходя массив, мы ведем подсчет всех встреченных простых чисел.
Number of divisors = (p1 + 1) * (p2 + 1) * (p3 + 1) * … * (pn + 1)
where p1, p2, p3, …, pn are the primes encountered in the prime factorization of all the elements.
Ниже представлена реализация описанного выше подхода:
C ++
// C++ implementation of the approach #include <bits/stdc++.h> #define MAX 10000002 using namespace std; int prime[MAX]; // Array to store count of primes int prime_count[MAX]; // Function to store smallest prime factor // of every number till MAX void sieve() { memset (prime, 0, sizeof (prime)); prime[0] = prime[1] = 1; for ( int i = 2; i * i < MAX; i++) { if (prime[i] == 0) { for ( int j = i * 2; j < MAX; j += i) { if (prime[j] == 0) prime[j] = i; } } } for ( int i = 2; i < MAX; i++) { // If the number is prime then it's // smallest prime factor is the number // itself if (prime[i] == 0) prime[i] = i; } } // Function to return the count of the divisors for // the product of all the numbers from the array long long numberOfDivisorsOfProduct( const int * arr, int n) { memset (prime_count, 0, sizeof (prime_count)); for ( int i = 0; i < n; i++) { int temp = arr[i]; while (temp != 1) { // Increase the count of prime // encountered prime_count[prime[temp]]++; temp = temp / prime[temp]; } } long long ans = 1; // Multiplying the count of primes // encountered for ( int i = 2; i < MAX; i++) { ans = ans * (prime_count[i] + 1); } return ans; } // Driver code int main() { sieve(); int arr[] = { 2, 4, 6 }; int n = sizeof (arr) / sizeof (arr[0]); cout << numberOfDivisorsOfProduct(arr, n); return 0; } |
Джава
// Java implementation of the approach import java.util.Arrays; // Java implementation of the approach class GFG { final static int MAX = 10000002 ; static int prime[] = new int [MAX]; // Array to store count of primes static int prime_count[] = new int [MAX]; // Function to store smallest prime factor // of every number till MAX static void sieve() { Arrays.fill(prime, 0 , MAX, 0 ); prime[ 0 ] = prime[ 1 ] = 1 ; for ( int i = 2 ; i * i < MAX; i++) { if (prime[i] == 0 ) { for ( int j = i * 2 ; j < MAX; j += i) { if (prime[j] == 0 ) { prime[j] = i; } } } } for ( int i = 2 ; i < MAX; i++) { // If the number is prime then it's // smallest prime factor is the number // itself if (prime[i] == 0 ) { prime[i] = i; } } } // Function to return the count of the divisors for // the product of all the numbers from the array static long numberOfDivisorsOfProduct( int [] arr, int n) { Arrays.fill(prime_count, 0 , MAX, 0 ); for ( int i = 0 ; i < n; i++) { int temp = arr[i]; while (temp != 1 ) { // Increase the count of prime // encountered prime_count[prime[temp]]++; temp = temp / prime[temp]; } } long ans = 1 ; // Multiplying the count of primes // encountered for ( int i = 2 ; i < MAX; i++) { ans = ans * (prime_count[i] + 1 ); } return ans; } // Driver code public static void main(String[] args) { sieve(); int arr[] = { 2 , 4 , 6 }; int n = arr.length; System.out.println(numberOfDivisorsOfProduct(arr, n)); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach MAX = 10000002 prime = [ 0 ] * ( MAX ) MAX_sqrt = int ( MAX * * ( 0.5 )) # Array to store count of primes prime_count = [ 0 ] * ( MAX ) # Function to store smallest prime # factor in prime[] def sieve(): prime[ 0 ], prime[ 1 ] = 1 , 1 for i in range ( 2 , MAX_sqrt): if prime[i] = = 0 : for j in range (i * 2 , MAX , i): if prime[j] = = 0 : prime[j] = i for i in range ( 2 , MAX ): # If the number is prime then it's # the smallest prime factor is the # number itself if prime[i] = = 0 : prime[i] = i # Function to return the count of the divisors for # the product of all the numbers from the array def numberOfDivisorsOfProduct(arr, n): for i in range ( 0 , n): temp = arr[i] while temp ! = 1 : # Increase the count of prime # encountered prime_count[prime[temp]] + = 1 temp = temp / / prime[temp] ans = 1 # Multiplying the count of primes # encountered for i in range ( 2 , len (prime_count)): ans = ans * (prime_count[i] + 1 ) return ans # Driver code if __name__ = = "__main__" : sieve() arr = [ 2 , 4 , 6 ] n = len (arr) print (numberOfDivisorsOfProduct(arr, n)) # This code is contributed by Rituraj Jain |
C #
// C# implementation of the approach using System; public class GFG { static int MAX = 1000000; static int []prime = new int [MAX]; // Array to store count of primes static int []prime_count = new int [MAX]; // Function to store smallest prime factor // of every number till MAX static void sieve() { for ( int i =0;i<MAX;i++) prime[i]=0; prime[0] = prime[1] = 1; for ( int i = 2; i * i < MAX; i++) { if (prime[i] == 0) { for ( int j = i * 2; j < MAX; j += i) { if (prime[j] == 0) { prime[j] = i; } } } } for ( int i = 2; i < MAX; i++) { // If the number is prime then it's // smallest prime factor is the number // itself if (prime[i] == 0) { prime[i] = i; } } } // Function to return the count of the divisors for // the product of all the numbers from the array static long numberOfDivisorsOfProduct( int [] arr, int n) { for ( int i =0;i<MAX;i++) prime_count[i]=0; for ( int i = 0; i < n; i++) { int temp = arr[i]; while (temp != 1) { // Increase the count of prime // encountered prime_count[prime[temp]]++; temp = temp / prime[temp]; } } long ans = 1; // Multiplying the count of primes // encountered for ( int i = 2; i < MAX; i++) { ans = ans * (prime_count[i] + 1); } return ans; } // Driver code public static void Main() { sieve(); int []arr = {2, 4, 6}; int n = arr.Length; Console.Write(numberOfDivisorsOfProduct(arr, n)); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // Javascript implementation of the approach let MAX = 10000002 let prime = new Array(MAX); // Array to store count of primes let prime_count = new Array(MAX); // Function to store smallest prime factor // of every number till MAX function sieve() { prime.fill(0) prime[0] = prime[1] = 1; for (let i = 2; i * i < MAX; i++) { if (prime[i] == 0) { for (let j = i * 2; j < MAX; j += i) { if (prime[j] == 0) prime[j] = i; } } } for (let i = 2; i < MAX; i++) { // If the number is prime then it's // smallest prime factor is the number // itself if (prime[i] == 0) prime[i] = i; } } // Function to return the count of the divisors for // the product of all the numbers from the array function numberOfDivisorsOfProduct(arr, n) { prime_count.fill(0) for (let i = 0; i < n; i++) { let temp = arr[i]; while (temp != 1) { // Increase the count of prime // encountered prime_count[prime[temp]]++; temp = temp / prime[temp]; } } let ans = 1; // Multiplying the count of primes // encountered for (let i = 2; i < MAX; i++) { ans = ans * (prime_count[i] + 1); } return ans; } // Driver code sieve(); let arr = [2, 4, 6]; let n = arr.length; document.write(numberOfDivisorsOfProduct(arr, n)); // This code is contributed by gfgking </script> |
10
В подходе с эффективным использованием памяти массив может быть заменен неупорядоченной картой для хранения количества только тех простых чисел, которые были встречены.
Ниже представлена реализация подхода с эффективным использованием памяти:
C ++
// C++ implementation of the approach #include <bits/stdc++.h> #define MAX 10000002 using namespace std; int prime[MAX]; // Map to store count of primes unordered_map< int , int > prime_count; // Function to store smallest prime factor // in prime[] void sieve() { memset (prime, 0, sizeof (prime)); prime[0] = prime[1] = 1; for ( int i = 2; i * i < MAX; i++) { if (prime[i] == 0) { for ( int j = i * 2; j < MAX; j += i) { if (prime[j] == 0) prime[j] = i; } } } for ( int i = 2; i < MAX; i++) { // If the number is prime then // it's the smallest prime factor // is the number itself if (prime[i] == 0) prime[i] = i; } } // Function to return the count of the divisors for // the product of all the numbers from the array long long numberOfDivisorsOfProduct( const int * arr, int n) { for ( int i = 0; i < n; i++) { int temp = arr[i]; while (temp != 1) { // Increase the count of prime // encountered prime_count[prime[temp]]++; temp = temp / prime[temp]; } } long long ans = 1; // Multiplying the count of primes // encountered unordered_map< int , int >::iterator it; for (it = prime_count.begin(); it != prime_count.end(); it++) { ans = ans * (it->second + 1); } return ans; } // Driver code int main() { sieve(); int arr[] = { 3, 5, 7 }; int n = sizeof (arr) / sizeof (arr[0]); cout << numberOfDivisorsOfProduct(arr, n); return 0; } |
Джава
// Java implementation of the approach import java.util.*; class GFG { static int MAX = 10000002 ; static int []prime = new int [MAX]; // Map to store count of primes<
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