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MCM de los primeros n números naturales

Dado un número n tal que 1<= N <= 10^6 the Task is to Find the LCM of First n Natural Numbers. 

Ejemplos:  

Input : n = 5 Output : 60 Input : n = 6 Output : 60 Input : n = 7 Output : 420 

Le recomendamos encarecidamente que haga clic aquí y lo practique antes de pasar a la solución.

Hemos discutido una solución simple en el siguiente artículo. 
Número más pequeño divisible por los primeros n números
La solución anterior funciona bien para una sola entrada. Pero si tenemos múltiples entradas es una buena idea usar Tamiz de Eratóstenes para almacenar todos los factores primos. Como sabemos, si MCM (a b) = X, cualquier factor primo de aob también será el factor primo de 'X'.  



  1. Inicializar la variable mcm con 1
  2. Genere un tamiz de Eratóstenes (el vector bool es primo) de longitud 10 ^ 6 (idealmente debe ser igual al número de dígitos en factorial)
  3. Ahora, para cada número en el vector bool isPrime, si el número es primo (isPrime[i] es verdadero), encuentre el número máximo que sea menor que el número dado e igual a la potencia del primo.
  4. Luego multiplique este número con la variable mcm.
  5. Repita los pasos 3 y 4 hasta que el número primo sea menor que el número dado.

Ilustración:  

For example if n = 10 8 will be the first number which is equal to 2^3 then 9 which is equal to 3^2 then 5 which is equal to 5^1 then 7 which is equal to 7^1 Finally we multiply those numbers 8*9*5*7 = 2520

A continuación se muestra la implementación de la idea anterior.  

C++
// C++ program to find LCM of First N Natural Numbers. #include    #define MAX 100000 using namespace std; vector<bool> isPrime (MAX true); // utility function for sieve of sieve of Eratosthenes void sieve() {    for (int i = 2; i * i <= MAX; i++)  {  if (isPrime[i] == true)  for (int j = i*i; j<= MAX; j+=i)  isPrime[j] = false;  } } // Function to find LCM of first n Natural Numbers long long LCM(int n) {  long long lcm = 1;  int i=2;   while(i<=n) {  if(isPrime[i]){  int pp = i;  while (pp * i <= n)  pp = pp * i;  lcm *= pp;  }  i++;  }  return lcm; } // Driver code int main() {  // build sieve  sieve();  int N = 7;  // Function call  cout << LCM(N);  return 0; } 
Java
// Java program to find LCM of First N Natural Numbers. import java.util.*; class GFG  {  static int MAX = 100000;  // array to store all prime less than and equal to 10^6  static ArrayList<Integer> primes  = new ArrayList<Integer>();  // utility function for sieve of sieve of Eratosthenes  static void sieve()  {  boolean[] isComposite = new boolean[MAX + 1];  for (int i = 2; i * i <= MAX; i++)   {  if (isComposite[i] == false)  for (int j = 2; j * i <= MAX; j++)  isComposite[i * j] = true;  }  // Store all prime numbers in vector primes[]  for (int i = 2; i <= MAX; i++)  if (isComposite[i] == false)  primes.add(i);  }  // Function to find LCM of first n Natural Numbers  static long LCM(int n)  {  long lcm = 1;  for (int i = 0;  i < primes.size() && primes.get(i) <= n;   i++)   {  // Find the highest power of prime primes[i]  // that is less than or equal to n  int pp = primes.get(i);  while (pp * primes.get(i) <= n)  pp = pp * primes.get(i);  // multiply lcm with highest power of prime[i]  lcm *= pp;  lcm %= 1000000007;  }  return lcm;  }  // Driver code  public static void main(String[] args)  {  sieve();  int N = 7;    // Function call  System.out.println(LCM(N));  } } // This code is contributed by mits 
Python3
# Python3 program to find LCM of # First N Natural Numbers. MAX = 100000 # array to store all prime less # than and equal to 10^6 primes = [] # utility function for # sieve of Eratosthenes def sieve(): isComposite = [False]*(MAX+1) i = 2 while (i * i <= MAX): if (isComposite[i] == False): j = 2 while (j * i <= MAX): isComposite[i * j] = True j += 1 i += 1 # Store all prime numbers in # vector primes[] for i in range(2 MAX+1): if (isComposite[i] == False): primes.append(i) # Function to find LCM of # first n Natural Numbers def LCM(n): lcm = 1 i = 0 while (i < len(primes) and primes[i] <= n): # Find the highest power of prime # primes[i] that is less than or # equal to n pp = primes[i] while (pp * primes[i] <= n): pp = pp * primes[i] # multiply lcm with highest # power of prime[i] lcm *= pp lcm %= 1000000007 i += 1 return lcm # Driver code sieve() N = 7 # Function call print(LCM(N)) # This code is contributed by mits 
C#
// C# program to find LCM of First N // Natural Numbers. using System.Collections; using System; class GFG {  static int MAX = 100000;  // array to store all prime less than  // and equal to 10^6  static ArrayList primes = new ArrayList();  // utility function for sieve of  // sieve of Eratosthenes  static void sieve()  {  bool[] isComposite = new bool[MAX + 1];  for (int i = 2; i * i <= MAX; i++)   {  if (isComposite[i] == false)  for (int j = 2; j * i <= MAX; j++)  isComposite[i * j] = true;  }  // Store all prime numbers in vector primes[]  for (int i = 2; i <= MAX; i++)  if (isComposite[i] == false)  primes.Add(i);  }  // Function to find LCM of first  // n Natural Numbers  static long LCM(int n)  {  long lcm = 1;  for (int i = 0;  i < primes.Count && (int)primes[i] <= n;   i++)   {  // Find the highest power of prime primes[i]  // that is less than or equal to n  int pp = (int)primes[i];  while (pp * (int)primes[i] <= n)  pp = pp * (int)primes[i];  // multiply lcm with highest power of prime[i]  lcm *= pp;  lcm %= 1000000007;  }  return lcm;  }  // Driver code  public static void Main()  {  sieve();  int N = 7;    // Function call  Console.WriteLine(LCM(N));  } } // This code is contributed by mits 
JavaScript
<script>  // Javascript program to find LCM of First N  // Natural Numbers.    let MAX = 100000;    // array to store all prime less than  // and equal to 10^6  let primes = [];    // utility function for sieve of  // sieve of Eratosthenes  function sieve()  {  let isComposite = new Array(MAX + 1);  isComposite.fill(false);  for (let i = 2; i * i <= MAX; i++)  {  if (isComposite[i] == false)  for (let j = 2; j * i <= MAX; j++)  isComposite[i * j] = true;  }    // Store all prime numbers in vector primes[]  for (let i = 2; i <= MAX; i++)  if (isComposite[i] == false)  primes.push(i);  }    // Function to find LCM of first  // n Natural Numbers  function LCM(n)  {  let lcm = 1;  for (let i = 0;  i < primes.length && primes[i] <= n;  i++)  {  // Find the highest power of prime primes[i]  // that is less than or equal to n  let pp = primes[i];  while (pp * primes[i] <= n)  pp = pp * primes[i];    // multiply lcm with highest power of prime[i]  lcm *= pp;  lcm %= 1000000007;  }  return lcm;  }    sieve();  let N = 7;  // Function call  document.write(LCM(N)); // This code is contributed by decode2207. </script> 
PHP
 // PHP program to find LCM of // First N Natural Numbers. $MAX = 100000; // array to store all prime less // than and equal to 10^6 $primes = array(); // utility function for // sieve of Eratosthenes function sieve() { global $MAX $primes; $isComposite = array_fill(0 $MAX false); for ($i = 2; $i * $i <= $MAX; $i++) { if ($isComposite[$i] == false) for ($j = 2; $j * $i <= $MAX; $j++) $isComposite[$i * $j] = true; } // Store all prime numbers in // vector primes[] for ($i = 2; $i <= $MAX; $i++) if ($isComposite[$i] == false) array_push($primes $i); } // Function to find LCM of  // first n Natural Numbers function LCM($n) { global $MAX $primes; $lcm = 1; for ($i = 0; $i < count($primes) && $primes[$i] <= $n; $i++) { // Find the highest power of prime  // primes[i] that is less than or  // equal to n $pp = $primes[$i]; while ($pp * $primes[$i] <= $n) $pp = $pp * $primes[$i]; // multiply lcm with highest // power of prime[i] $lcm *= $pp; $lcm %= 1000000007; } return $lcm; } // Driver code sieve(); $N = 7; // Function call echo LCM($N); // This code is contributed by mits ?> 

Producción
420

Complejidad del tiempo : En2
Espacio Auxiliar: En)

Otro enfoque:

 La idea es que si el número es menor que 3, entonces se devuelve el número. Si el número es mayor que 2, encuentre el MCM de nn-1

  • Digamos x=MCM(nn-1)
  • nuevamente x=MCM(xn-2)
  • de nuevo x=MCM(xn-3) ...
  • .
  • .
  • de nuevo x=MCM(x1) ...

ahora el resultado es x.

Para encontrar MCM(ab) usamos una función hcf(ab) que devolverá HCF de (ab)

sabemos que LCM(ab)= (a*b)/HCF(ab)

Ilustración: 

For example if n = 7 function call lcm(76) now lets say a=7  b=6 Now  b!= 1 Hence a=lcm(76) = 42 and b=6-1=5 function call lcm(425) a=lcm(425) = 210 and b=5-1=4 function call lcm(2104) a=lcm(2104) = 420 and b=4-1=3 function call lcm(4203) a=lcm(4203) = 420 and b=3-1=2 function call lcm(4202) a=lcm(4202) = 420 and b=2-1=1 Now b=1 Hence return a=420

A continuación se muestra la implementación del enfoque anterior.

C++
// C++ program to find LCM of First N Natural Numbers. #include    using namespace std; // to calculate hcf int hcf(int a int b) {  if (b == 0)  return a;  return hcf(b a % b); } int findlcm(int aint b) {  if (b == 1)    // lcm(ab)=(a*b)/hcf(ab)  return a;     // assign a=lcm of nn-1  a = (a * b) / hcf(a b);     // b=b-1  b -= 1;   return findlcm(a b); } // Driver code int main() {  int n = 7;  if (n < 3)  cout << n; // base case  else    // Function call  // pass nn-1 in function to find LCM of first n natural  // number  cout << findlcm(n n - 1);    return 0; } // contributed by ajaykr00kj 
Java
// Java program to find LCM of First N Natural Numbers public class Main {  // to calculate hcf  static int hcf(int a int b)  {  if (b == 0)  return a;  return hcf(b a % b);  }  static int findlcm(int aint b)  {  if (b == 1)  // lcm(ab)=(a*b)/hcf(ab)  return a;   // assign a=lcm of nn-1  a = (a * b) / hcf(a b);   // b=b-1  b -= 1;   return findlcm(a b);  }  // Driver code.  public static void main(String[] args)   {  int n = 7;  if (n < 3)  System.out.print(n); // base case  else  // Function call  // pass nn-1 in function to find LCM of first n natural  // number  System.out.print(findlcm(n n - 1));  } } // This code is contributed by divyeshrabadiya07. 
Python3
# Python3 program to find LCM  # of First N Natural Numbers. # To calculate hcf def hcf(a b): if (b == 0): return a return hcf(b a % b) def findlcm(a b): if (b == 1): # lcm(ab)=(a*b)//hcf(ab) return a # Assign a=lcm of nn-1 a = (a * b) // hcf(a b) # b=b-1 b -= 1 return findlcm(a b) # Driver code n = 7 if (n < 3): print(n) else: # Function call # pass nn-1 in function # to find LCM of first n  # natural number print(findlcm(n n - 1)) # This code is contributed by Shubham_Singh 
C#
// C# program to find LCM of First N Natural Numbers. using System; class GFG {  // to calculate hcf  static int hcf(int a int b)  {  if (b == 0)  return a;  return hcf(b a % b);  }  static int findlcm(int aint b)  {  if (b == 1)  // lcm(ab)=(a*b)/hcf(ab)  return a;   // assign a=lcm of nn-1  a = (a * b) / hcf(a b);   // b=b-1  b -= 1;   return findlcm(a b);  }  // Driver code  static void Main() {  int n = 7;  if (n < 3)  Console.Write(n); // base case  else  // Function call  // pass nn-1 in function to find LCM of first n natural  // number  Console.Write(findlcm(n n - 1));  } } // This code is contributed by divyesh072019. 
JavaScript
<script>  // Javascript program to find LCM of First N Natural Numbers.    // to calculate hcf  function hcf(a b)  {  if (b == 0)  return a;  return hcf(b a % b);  }  function findlcm(ab)  {  if (b == 1)  // lcm(ab)=(a*b)/hcf(ab)  return a;  // assign a=lcm of nn-1  a = (a * b) / hcf(a b);  // b=b-1  b -= 1;  return findlcm(a b);  }    let n = 7;  if (n < 3)  document.write(n); // base case  else    // Function call  // pass nn-1 in function to find LCM of first n natural  // number  document.write(findlcm(n n - 1));   </script> 

Producción
420

Complejidad del tiempo: O(n iniciar sesión n)
Espacio Auxiliar: O(1)