FÆLLES DIVISORER AF TO TAL

Givet to heltal er opgaven at finde antallet af alle fælles divisorer af givne tal?

Eksempler:

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 Fælles Divisorer Prøv det!

Det anbefales at henvise alle divisorer af et givet tal som en forudsætning for denne artikel.

Naiv løsning
En simpel løsning er først at finde alle divisorer af første tal og gemme dem i et array eller hash. Find derefter fælles divisorer for andet tal og gem dem. Udskriv endelig fælles elementer i to lagrede arrays eller hash. Nøglen er, at størrelsen af potenser af primfaktorer af en divisor skal være lig med minimumspotensen af to primfaktorer af a og b.

Find de primære faktorer ved en brug primfaktorisering .
Find antallet af hver primfaktor på -en og gem det i en Hashmap.
Prime faktorisere b ved hjælp af forskellige primfaktorer -en .
Så ville det samlede antal divisorer være lig med produktet af (tæl + 1)
af hver faktor.

-en

Dette giver optællingen af alle divisorer af -en og 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>

Produktion:

Tidskompleksitet : O(?n log n)
Hjælpeplads: På)

Effektiv løsning -
En bedre løsning er at beregne største fælles divisor (gcd) af givet to tal og tæl så divisorer af den 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>

Output:

Tidskompleksitet: På^1/2) hvor n er gcd af to tal.
Hjælpeplads: O(1)

En anden tilgang:

1. Definer en funktion 'gcd', der tager to heltal 'a' og 'b' og returnerer deres største fælles divisor (GCD) ved hjælp af den euklidiske algoritme.
2. Definer en funktion 'count_common_divisors', der tager to heltal 'a' og 'b' og tæller antallet af fælles divisorer af 'a' og 'b' ved hjælp af deres GCD.
3. Beregn GCD for 'a' og 'b' ved hjælp af 'gcd'-funktionen.
4. Initialiser en tæller 'tæller' til 0.
5. Loop gennem alle mulige divisorer af GCD'en for 'a' og 'b' fra 1 til kvadratroden af GCD'en.
6. Hvis den aktuelle divisor dividerer GCD ligeligt, øges tælleren med 2 (fordi både 'a' og 'b' er delelige med divisoren).
7. Hvis kvadratet af den aktuelle divisor er lig med GCD, dekrementeres tælleren med 1 (fordi vi allerede har talt denne divisor én gang).
8. Returner den endelige optælling af fælles divisorer.
9. Definer to heltal 'a' og 'b' i hovedfunktionen, og kald funktionen 'count_common_divisors' med disse heltal.
10. Udskriv antallet af fælles divisorer for 'a' og 'b' ved hjælp af printf-funktionen.

#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}.`);

Produktion

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

Tidskompleksiteten af gcd()-funktionen er O(log(min(a b))), da den bruger Euklids algoritme, som tager logaritmisk tid i forhold til det mindste af de to tal.

Tidskompleksiteten af funktionen count_common_divisors() er O(sqrt(gcd(a b))), da den itererer op til kvadratroden af gcd af de to tal.

Pladskompleksiteten af begge funktioner er O(1), da de kun bruger en konstant mængde hukommelse uanset inputstørrelsen.

TechCodeview

Fælles divisorer af to tal

En anden tilgang: