GEMEINSAME TEILER ZWEIER ZAHLEN

Bei zwei gegebenen ganzen Zahlen besteht die Aufgabe darin, die Anzahl aller gemeinsamen Teiler gegebener Zahlen zu ermitteln.

Beispiele:

Input : a = 12 b = 24 Output: 6 // all common divisors are 1 2 3 // 4 6 and 12 Input : a = 3 b = 17 Output: 1 // all common divisors are 1 Input : a = 20 b = 36 Output: 3 // all common divisors are 1 2 4

Recommended Practice Gemeinsame Teiler Probieren Sie es aus!

Es wird empfohlen, darauf zu verweisen alle Teiler einer gegebenen Zahl als Voraussetzung für diesen Artikel.

Naive Lösung
Eine einfache Lösung besteht darin, zunächst alle Teiler der ersten Zahl zu finden und sie in einem Array oder Hash zu speichern. Finden Sie dann die gemeinsamen Teiler der zweiten Zahl und speichern Sie sie. Zum Schluss werden gemeinsame Elemente zweier gespeicherter Arrays oder Hashs ausgegeben. Der Schlüssel liegt darin, dass die Größe der Potenzen der Primfaktoren eines Teilers gleich der minimalen Potenz zweier Primfaktoren von a und b sein sollte.

Finden Sie die Primfaktoren einer Verwendung Primfaktorzerlegung .
Finden Sie die Anzahl jedes Primfaktors von A und speichern Sie es in einer Hashmap.
Primfaktorisieren B unter Verwendung unterschiedlicher Primfaktoren von A .
Dann wäre die Gesamtzahl der Teiler gleich dem Produkt von (Anzahl + 1)
jedes Faktors.

Dies ergibt die Anzahl aller Teiler von A Und B .

C++

// C++ implementation of program  #include    using namespace std; // Map to store the count of each // prime factor of a  map<int int> ma; // Function that calculate the count of  // each prime factor of a number  void primeFactorize(int a)  {   for(int i = 2; i * i <= a; i += 2)   {   int cnt = 0;   while (a % i == 0)   {   cnt++;   a /= i;   }   ma[i] = cnt;   }   if (a > 1)  {  ma[a] = 1;  } }  // Function to calculate all common // divisors of two given numbers  // a b --> input integer numbers  int commDiv(int a int b)  {     // Find count of each prime factor of a   primeFactorize(a);   // stores number of common divisors   int res = 1;   // Find the count of prime factors   // of b using distinct prime factors of a   for(auto m = ma.begin();  m != ma.end(); m++)  {  int cnt = 0;   int key = m->first;   int value = m->second;   while (b % key == 0)   {   b /= key;   cnt++;   }   // Prime factor of common divisor   // has minimum cnt of both a and b   res *= (min(cnt value) + 1);   }   return res;  }  // Driver code  int main() {  int a = 12 b = 24;     cout << commDiv(a b) << endl;     return 0; } // This code is contributed by divyeshrabadiya07

Java

// Java implementation of program import java.util.*; import java.io.*; class GFG {  // map to store the count of each prime factor of a  static HashMap<Integer Integer> ma = new HashMap<>();  // method that calculate the count of  // each prime factor of a number  static void primeFactorize(int a)  {  for (int i = 2; i * i <= a; i += 2) {  int cnt = 0;  while (a % i == 0) {  cnt++;  a /= i;  }  ma.put(i cnt);  }  if (a > 1)  ma.put(a 1);  }  // method to calculate all common divisors  // of two given numbers  // a b --> input integer numbers  static int commDiv(int a int b)  {  // Find count of each prime factor of a  primeFactorize(a);  // stores number of common divisors  int res = 1;  // Find the count of prime factors of b using  // distinct prime factors of a  for (Map.Entry<Integer Integer> m : ma.entrySet()) {  int cnt = 0;  int key = m.getKey();  int value = m.getValue();  while (b % key == 0) {  b /= key;  cnt++;  }  // prime factor of common divisor  // has minimum cnt of both a and b  res *= (Math.min(cnt value) + 1);  }  return res;  }  // Driver method  public static void main(String args[])  {  int a = 12 b = 24;  System.out.println(commDiv(a b));  } }

Python3

# Python3 implementation of program  import math # Map to store the count of each # prime factor of a  ma = {} # Function that calculate the count of  # each prime factor of a number  def primeFactorize(a): sqt = int(math.sqrt(a)) for i in range(2 sqt 2): cnt = 0 while (a % i == 0): cnt += 1 a /= i ma[i] = cnt if (a > 1): ma[a] = 1 # Function to calculate all common # divisors of two given numbers  # a b --> input integer numbers  def commDiv(a b): # Find count of each prime factor of a  primeFactorize(a) # stores number of common divisors  res = 1 # Find the count of prime factors  # of b using distinct prime factors of a  for key value in ma.items(): cnt = 0 while (b % key == 0): b /= key cnt += 1 # Prime factor of common divisor  # has minimum cnt of both a and b  res *= (min(cnt value) + 1) return res # Driver code  a = 12 b = 24 print(commDiv(a b)) # This code is contributed by Stream_Cipher

// C# implementation of program using System; using System.Collections.Generic;  class GFG{   // Map to store the count of each  // prime factor of a static Dictionary<int  int> ma = new Dictionary<int  int>(); // Function that calculate the count of // each prime factor of a number static void primeFactorize(int a) {  for(int i = 2; i * i <= a; i += 2)  {  int cnt = 0;  while (a % i == 0)  {  cnt++;  a /= i;  }  ma.Add(i cnt);  }    if (a > 1)  ma.Add(a 1); } // Function to calculate all common  // divisors of two given numbers // a b --> input integer numbers static int commDiv(int a int b) {    // Find count of each prime factor of a  primeFactorize(a);    // Stores number of common divisors  int res = 1;    // Find the count of prime factors  // of b using distinct prime factors of a  foreach(KeyValuePair<int int> m in ma)  {  int cnt = 0;  int key = m.Key;  int value = m.Value;    while (b % key == 0)  {  b /= key;  cnt++;  }  // Prime factor of common divisor  // has minimum cnt of both a and b  res *= (Math.Min(cnt value) + 1);  }  return res; } // Driver code  static void Main() {  int a = 12 b = 24;    Console.WriteLine(commDiv(a b)); } } // This code is contributed by divyesh072019

JavaScript

<script>   // JavaScript implementation of program  // Map to store the count of each  // prime factor of a  let ma = new Map();  // Function that calculate the count of  // each prime factor of a number  function primeFactorize(a)  {  for(let i = 2; i * i <= a; i += 2)  {  let cnt = 0;  while (a % i == 0)  {  cnt++;  a = parseInt(a / i 10);  }  ma.set(i cnt);  }  if (a > 1)  {  ma.set(a 1);  }  }  // Function to calculate all common  // divisors of two given numbers  // a b --> input integer numbers  function commDiv(ab)  {  // Find count of each prime factor of a  primeFactorize(a);  // stores number of common divisors  let res = 1;  // Find the count of prime factors  // of b using distinct prime factors of a  ma.forEach((valueskeys)=>{  let cnt = 0;  let key = keys;  let value = values;  while (b % key == 0)  {  b = parseInt(b / key 10);  cnt++;  }  // Prime factor of common divisor  // has minimum cnt of both a and b  res *= (Math.min(cnt value) + 1);  })  return res;  }  // Driver code  let a = 12 b = 24;    document.write(commDiv(a b));   </script>

Ausgabe:

Zeitkomplexität : O(?n log n)
Hilfsraum: An)

Effiziente Lösung –
Eine bessere Lösung besteht darin, die zu berechnen größter gemeinsamer Teiler (ggT) der gegebenen zwei Zahlen und zählen Sie dann die Teiler dieser gcd.

C++

// C++ implementation of program #include    using namespace std; // Function to calculate gcd of two numbers int gcd(int a int b) {  if (a == 0)  return b;  return gcd(b % a a); } // Function to calculate all common divisors // of two given numbers // a b --> input integer numbers int commDiv(int a int b) {  // find gcd of a b  int n = gcd(a b);  // Count divisors of n.  int result = 0;  for (int i = 1; i <= sqrt(n); i++) {  // if 'i' is factor of n  if (n % i == 0) {  // check if divisors are equal  if (n / i == i)  result += 1;  else  result += 2;  }  }  return result; } // Driver program to run the case int main() {  int a = 12 b = 24;  cout << commDiv(a b);  return 0; }

Java

// Java implementation of program class Test {  // method to calculate gcd of two numbers  static int gcd(int a int b)  {  if (a == 0)  return b;  return gcd(b % a a);  }  // method to calculate all common divisors  // of two given numbers  // a b --> input integer numbers  static int commDiv(int a int b)  {  // find gcd of a b  int n = gcd(a b);  // Count divisors of n.  int result = 0;  for (int i = 1; i <= Math.sqrt(n); i++) {  // if 'i' is factor of n  if (n % i == 0) {  // check if divisors are equal  if (n / i == i)  result += 1;  else  result += 2;  }  }  return result;  }  // Driver method  public static void main(String args[])  {  int a = 12 b = 24;  System.out.println(commDiv(a b));  } }

Python3

# Python implementation of program from math import sqrt # Function to calculate gcd of two numbers def gcd(a b): if a == 0: return b return gcd(b % a a) # Function to calculate all common divisors  # of two given numbers  # a b --> input integer numbers  def commDiv(a b): # find GCD of a b n = gcd(a b) # Count divisors of n result = 0 for i in range(1int(sqrt(n))+1): # if i is a factor of n if n % i == 0: # check if divisors are equal if n/i == i: result += 1 else: result += 2 return result # Driver program to run the case  if __name__ == '__main__': a = 12 b = 24; print(commDiv(a b))

// C# implementation of program using System; class GFG {  // method to calculate gcd  // of two numbers  static int gcd(int a int b)  {  if (a == 0)  return b;  return gcd(b % a a);  }  // method to calculate all  // common divisors of two  // given numbers a b -->  // input integer numbers  static int commDiv(int a int b)  {  // find gcd of a b  int n = gcd(a b);  // Count divisors of n.  int result = 0;  for (int i = 1; i <= Math.Sqrt(n); i++) {  // if 'i' is factor of n  if (n % i == 0) {  // check if divisors are equal  if (n / i == i)  result += 1;  else  result += 2;  }  }  return result;  }  // Driver method  public static void Main(String[] args)  {  int a = 12 b = 24;  Console.Write(commDiv(a b));  } } // This code contributed by parashar.

PHP

 // PHP implementation of program // Function to calculate  // gcd of two numbers function gcd($a $b) { if ($a == 0) return $b; return gcd($b % $a $a); } // Function to calculate all common  // divisors of two given numbers // a b --> input integer numbers function commDiv($a $b) { // find gcd of a b $n = gcd($a $b); // Count divisors of n. $result = 0; for ($i = 1; $i <= sqrt($n); $i++) { // if 'i' is factor of n if ($n % $i == 0) { // check if divisors  // are equal if ($n / $i == $i) $result += 1; else $result += 2; } } return $result; } // Driver Code $a = 12; $b = 24; echo(commDiv($a $b)); // This code is contributed by Ajit. ?>

JavaScript

<script>  // Javascript implementation of program    // Function to calculate gcd of two numbers  function gcd(a b)  {  if (a == 0)  return b;  return gcd(b % a a);  }  // Function to calculate all common divisors  // of two given numbers  // a b --> input integer numbers  function commDiv(a b)  {  // find gcd of a b  let n = gcd(a b);  // Count divisors of n.  let result = 0;  for (let i = 1; i <= Math.sqrt(n); i++) {  // if 'i' is factor of n  if (n % i == 0) {  // check if divisors are equal  if (n / i == i)  result += 1;  else  result += 2;  }  }  return result;  }  let a = 12 b = 24;  document.write(commDiv(a b));   </script>

Ausgabe :

Zeitkomplexität: An^1/2), wobei n der ggT zweier Zahlen ist.
Hilfsraum: O(1)

Ein anderer Ansatz:

1. Definieren Sie eine Funktion „gcd“, die zwei ganze Zahlen „a“ und „b“ akzeptiert und mithilfe des euklidischen Algorithmus ihren größten gemeinsamen Teiler (GCD) zurückgibt.
2. Definieren Sie eine Funktion „count_common_divisors“, die zwei Ganzzahlen „a“ und „b“ akzeptiert und die Anzahl der gemeinsamen Teiler von „a“ und „b“ anhand ihres GCD zählt.
3. Berechnen Sie den GCD von „a“ und „b“ mit der Funktion „gcd“.
4. Initialisieren Sie einen Zähler „count“ auf 0.
5. Durchlaufen Sie alle möglichen Teiler des GCD von „a“ und „b“ von 1 bis zur Quadratwurzel des GCD.
6. Wenn der aktuelle Divisor den GCD teilt, erhöhen Sie den Zähler gleichmäßig um 2 (da sowohl „a“ als auch „b“ durch den Divisor teilbar sind).
7. Wenn das Quadrat des aktuellen Divisors gleich dem GCD ist, dekrementieren Sie den Zähler um 1 (da wir diesen Divisor bereits einmal gezählt haben).
8. Geben Sie die endgültige Anzahl der gemeinsamen Teiler zurück.
9. Definieren Sie in der Hauptfunktion zwei Ganzzahlen „a“ und „b“ und rufen Sie die Funktion „count_common_divisors“ mit diesen Ganzzahlen auf.
10. Drucken Sie die Anzahl der gemeinsamen Teiler von „a“ und „b“ mit der printf-Funktion aus.

#include  int gcd(int a int b) {  if(b == 0) {  return a;  }  return gcd(b a % b); } int count_common_divisors(int a int b) {  int gcd_ab = gcd(a b);  int count = 0;  for(int i = 1; i * i <= gcd_ab; i++) {  if(gcd_ab % i == 0) {  count += 2;  if(i * i == gcd_ab) {  count--;  }  }  }  return count; } int main() {  int a = 12;  int b = 18;  int common_divisors = count_common_divisors(a b);  printf('The number of common divisors of %d and %d is %d.n' a b common_divisors);  return 0; }

C++

#include    using namespace std; int gcd(int a int b) {  if(b == 0) {  return a;  }  return gcd(b a % b); } int count_common_divisors(int a int b) {  int gcd_ab = gcd(a b);  int count = 0;  for(int i = 1; i * i <= gcd_ab; i++) {  if(gcd_ab % i == 0) {  count += 2;  if(i * i == gcd_ab) {  count--;  }  }  }  return count; } int main() {  int a = 12;  int b = 18;  int common_divisors = count_common_divisors(a b);  cout<<'The number of common divisors of '<<a<<' and '<<b<<' is '<<common_divisors<<'.'<<endl;  return 0; }

Java

import java.util.*; public class Main {  public static int gcd(int a int b) {  if(b == 0) {  return a;  }  return gcd(b a % b);  }  public static int countCommonDivisors(int a int b) {  int gcd_ab = gcd(a b);  int count = 0;  for(int i = 1; i * i <= gcd_ab; i++) {  if(gcd_ab % i == 0) {  count += 2;  if(i * i == gcd_ab) {  count--;  }  }  }  return count;  }  public static void main(String[] args) {  int a = 12;  int b = 18;  int commonDivisors = countCommonDivisors(a b);  System.out.println('The number of common divisors of ' + a + ' and ' + b + ' is ' + commonDivisors + '.');  } }

Python3

import math def gcd(a b): if b == 0: return a return gcd(b a % b) def count_common_divisors(a b): gcd_ab = gcd(a b) count = 0 for i in range(1 int(math.sqrt(gcd_ab)) + 1): if gcd_ab % i == 0: count += 2 if i * i == gcd_ab: count -= 1 return count a = 12 b = 18 common_divisors = count_common_divisors(a b) print('The number of common divisors of' a 'and' b 'is' common_divisors '.') # This code is contributed by Prajwal Kandekar

using System; public class MainClass {  public static int GCD(int a int b)  {  if (b == 0)  {  return a;  }  return GCD(b a % b);  }  public static int CountCommonDivisors(int a int b)  {  int gcd_ab = GCD(a b);  int count = 0;  for (int i = 1; i * i <= gcd_ab; i++)  {  if (gcd_ab % i == 0)  {  count += 2;  if (i * i == gcd_ab)  {  count--;  }  }  }  return count;  }  public static void Main()  {  int a = 12;  int b = 18;  int commonDivisors = CountCommonDivisors(a b);  Console.WriteLine('The number of common divisors of {0} and {1} is {2}.' a b commonDivisors);  } }

JavaScript

// Function to calculate the greatest common divisor of  // two integers a and b using the Euclidean algorithm function gcd(a b) {  if(b === 0) {  return a;  }  return gcd(b a % b); } // Function to count the number of common divisors of two integers a and b function count_common_divisors(a b) {  let gcd_ab = gcd(a b);  let count = 0;  for(let i = 1; i * i <= gcd_ab; i++) {  if(gcd_ab % i === 0) {  count += 2;  if(i * i === gcd_ab) {  count--;  }  }  }  return count; } let a = 12; let b = 18; let common_divisors = count_common_divisors(a b); console.log(`The number of common divisors of ${a} and ${b} is ${common_divisors}.`);

Ausgabe

The number of common divisors of 12 and 18 is 4.

Die Zeitkomplexität der gcd()-Funktion beträgt O(log(min(a b))), da sie den Euklid-Algorithmus verwendet, der logarithmische Zeit in Bezug auf die kleinere der beiden Zahlen benötigt.

Die zeitliche Komplexität der Funktion count_common_divisors() beträgt O(sqrt(gcd(a b))), da sie bis zur Quadratwurzel des gcd der beiden Zahlen iteriert.

Die räumliche Komplexität beider Funktionen beträgt O(1), da sie unabhängig von der Eingabegröße nur eine konstante Menge an Speicher belegen.

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Gemeinsame Teiler zweier Zahlen

Ein anderer Ansatz: