Mayor potencia de 2 que divide el MCM de los primeros N números naturales.

Dado un número N , la tarea es encontrar la mayor potencia de 2 que divide a MCM de los primeros N números naturales.

Ejemplos:

Entrada: N = 5
Salida: 2
Explicación:
MCM de {1, 2, 3, 4, 5} = 60
60 es divisible por 2 ²

Entrada: N = 15
Salida: 3
Explicación:
MCM de {1, 2, 3…..14, 15} = 360360
360360 es divisible por 2 ³

Enfoque ingenuo: la idea es encontrar el mínimo común múltiplo de los primeros N números naturales. Luego, itere un ciclo desde i = 1 y verifique si 2 ⁱ divide el LCM o no y mantenga la pista del máximo i que divide el LCM.

A continuación se muestra la implementación del enfoque anterior:

C++

// C++ implementation of the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find LCM of
// first N natural numbers
int findlcm(int n)
{
    // Initialize result
    int ans = 1;
 
    // Ans contains LCM of 1, 2, 3, ..i
    // after i'th iteration
    for (int i = 1; i <= n; i++)
        ans = (((i * ans)) / (__gcd(i, ans)));
    return ans;
}
 
// Function to find the
// highest power of 2
// which divides LCM of
// first n natural numbers
int highestPower(int n)
{
    // Find lcm of first
    // N natural numbers
    int lcm = findlcm(n);
 
    // To store the highest
    // required power of 2
    int ans = 0;
 
    // Counting number of consecutive zeros
    // from the end in the given binary string
    for (int i = 1;; i++) {
        int x = pow(2, i);
        if (lcm % x == 0) {
            ans = i;
        }
        if (x > n)
            break;
    }
    return ans;
}
 
// Driver code
int main()
{
    int n = 15;
    cout << highestPower(n);
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG{
 
// Function to find LCM of
// first N natural numbers
static int findlcm(int n)
{
     
    // Initialize result
    int ans = 1;
 
    // Ans contains LCM of 1, 2, 3, ..i
    // after i'th iteration
    for(int i = 1; i <= n; i++)
        ans = (((i * ans)) / (__gcd(i, ans)));
         
    return ans;
}
 
// Function to find the
// highest power of 2
// which divides LCM of
// first n natural numbers
static int highestPower(int n)
{
     
    // Find lcm of first
    // N natural numbers
    int lcm = findlcm(n);
 
    // To store the highest
    // required power of 2
    int ans = 0;
 
    // Counting number of consecutive zeros
    // from the end in the given binary String
    for(int i = 1;; i++)
    {
        int x = (int) Math.pow(2, i);
        if (lcm % x == 0)
        {
            ans = i;
        }
        if (x > n)
            break;
    }
    return ans;
}
 
static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);    
}
 
// Driver code
public static void main(String[] args)
{
    int n = 15;
     
    System.out.print(highestPower(n));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
 
# Function to find LCM of
# first N natural numbers
def findlcm(n):
     
    # Initialize result
    ans = 1;
 
    # Ans contains LCM of 1, 2, 3, ..i
    # after i'th iteration
    for i in range(1, n + 1):
        ans = (((i * ans)) //
          (__gcd(i, ans)));
 
    return ans;
 
# Function to find the highest power
# of 2 which divides LCM of first n
# natural numbers
def highestPower(n):
     
    # Find lcm of first
    # N natural numbers
    lcm = findlcm(n);
 
    # To store the highest
    # required power of 2
    ans = 0;
 
    # Counting number of consecutive zeros
    # from the end in the given binary String
    for i in range(1, n):
        x = int(pow(2, i));
         
        if (lcm % x == 0):
            ans = i;
        if (x > n):
            break;
 
    return ans;
 
def __gcd(a, b):
     
    if (b == 0):
        return a;
    else:
        return __gcd(b, a % b);
 
# Driver code
if __name__ == '__main__':
     
    n = 15;
 
    print(highestPower(n));
 
# This code is contributed by 29AjayKumar

C#

// C# implementation of the approach
using System;
class GFG{
 
// Function to find LCM of
// first N natural numbers
static int findlcm(int n)
{   
    // Initialize result
    int ans = 1;
 
    // Ans contains LCM of 1, 2, 3, ..i
    // after i'th iteration
    for(int i = 1; i <= n; i++)
        ans = (((i * ans)) /
               (__gcd(i, ans)));
         
    return ans;
}
 
// Function to find the
// highest power of 2
// which divides LCM of
// first n natural numbers
static int highestPower(int n)
{   
    // Find lcm of first
    // N natural numbers
    int lcm = findlcm(n);
 
    // To store the highest
    // required power of 2
    int ans = 0;
 
    // Counting number of consecutive zeros
    // from the end in the given binary String
    for(int i = 1;; i++)
    {
        int x = (int) Math.Pow(2, i);
        if (lcm % x == 0)
        {
            ans = i;
        }
        if (x > n)
            break;
    }
    return ans;
}
 
static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);    
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 15;   
    Console.Write(highestPower(n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
// JavaScript program for the
// above approach
 
// Function to find LCM of
// first N natural numbers
function findlcm(n)
{
      
    // Initialize result
    let ans = 1;
  
    // Ans contains LCM of 1, 2, 3, ..i
    // after i'th iteration
    for(let i = 1; i <= n; i++)
        ans = (((i * ans)) / (__gcd(i, ans)));
          
    return ans;
}
  
// Function to find the
// highest power of 2
// which divides LCM of
// first n natural numbers
function highestPower(n)
{
      
    // Find lcm of first
    // N natural numbers
    let lcm = findlcm(n);
  
    // To store the highest
    // required power of 2
    let ans = 0;
  
    // Counting number of consecutive zeros
    // from the end in the given binary String
    for(let i = 1;; i++)
    {
        let x =  Math.pow(2, i);
        if (lcm % x == 0)
        {
            ans = i;
        }
        if (x > n)
            break;
    }
    return ans;
}
  
function __gcd(a, b)
{
    return b == 0 ? a : __gcd(b, a % b);   
}
 
// Driver Code
 
    let n = 15;
      
    document.write(highestPower(n));
 
</script>

Producción

Complejidad de tiempo: O(N)

Espacio Auxiliar: O(1)

Enfoque eficiente: El MCM de los primeros N números naturales siempre es divisible por una potencia de 2 y dado que el MCM de los primeros N números naturales contiene el producto 2 * 4 * 8 * 16 ……N. Por lo tanto, la mayor potencia de 2 que divide a MCM de los primeros N números naturales siempre será
$\lfloor \log_2 N \rfloor$

A continuación se muestra la implementación del enfoque anterior:

C++

// C++ implementation of the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the
// highest power of 2
// which divides LCM of
// first n natural numbers
int highestPower(int n)
{
    return log(n) / log(2);
}
 
// Driver code
int main()
{
    int n = 15;
    cout << highestPower(n);
    return 0;
}

Java

// Java implementation of the approach
class GFG{
     
// Function to find the highest
// power of 2 which divides LCM of
// first n natural numbers
static int highestPower(int n)
{
    return (int)(Math.log(n) / Math.log(2));
}
 
// Driver code
public static void main(String[] args)
{
    int n = 15;
    System.out.println(highestPower(n));
}
}
 
// This code is contributed by dewantipandeydp

Python3

# Python3 implementation of the approach
import math
 
# Function to find the highest
# power of 2 which divides LCM of
# first n natural numbers
def highestPower(n):
     
    return int((math.log(n) // math.log(2)));
 
# Driver code
if __name__ == '__main__':
     
    n = 15;
    print(highestPower(n));
 
# This code is contributed by Rajput-Ji

C#

// C# implementation of the approach
using System;
 
class GFG{
     
// Function to find the highest
// power of 2 which divides LCM of
// first n natural numbers
static int highestPower(int n)
{
    return (int)(Math.Log(n) / Math.Log(2));
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 15;
     
    Console.WriteLine(highestPower(n));
}
}
 
// This code is contributed by sapnasingh4991

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function to find the
// highest power of 2
// which divides LCM of
// first n natural numbers
function highestPower(n)
{
    return parseInt(Math.log(n) / Math.log(2));
}
 
// Driver code
var n = 15;
document.write( highestPower(n));
 
 
</script>

Producción

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)

Publicación traducida automáticamente

Artículo escrito por spp____ y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

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