Máximo divisor que divide todos los números naturales en el rango [L, R]

Dados dos enteros L y R , la tarea es encontrar el mayor divisor que divida a todos los números naturales en el rango [L, R] .
Ejemplos: 
 

Entrada: L = 3, R = 12 
Salida: 1
Entrada: L = 24, R = 24 
Salida: 24 
 

Enfoque: para un rango de elementos enteros consecutivos, hay dos casos: 
 

• Si L = R entonces la respuesta será L .
• Si L < R , todos los números naturales consecutivos en este rango son coprimos. Entonces, 1 es el único número que podrá dividir todos los elementos del rango.

A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the greatest divisor that
// divides all the natural numbers in the range [l, r]
int find_greatest_divisor(int l, int r)
{
    if (l == r)
        return l;
 
    return 1;
}
 
// Driver Code
int main()
{
    int l = 2, r = 12;
 
    cout << find_greatest_divisor(l, r);
}

// C implementation of the approach
#include <stdio.h>
 
// Function to return the greatest divisor that
// divides all the natural numbers in the range [l, r]
int find_greatest_divisor(int l, int r)
{
    if (l == r)
        return l;
 
    return 1;
}
 
// Driver Code
int main()
{
    int l = 2, r = 12;
    printf("%d",find_greatest_divisor(l, r));
}
 
// This code is contributed by kothvvsaakash.

// Java implementation of the approach
 
class GFG {
 
// Function to return the greatest divisor that
// divides all the natural numbers in the range [l, r]
    static int find_greatest_divisor(int l, int r) {
        if (l == r) {
            return l;
        }
 
        return 1;
    }
 
// Driver Code
    public static void main(String[] args) {
        int l = 2, r = 12;
 
        System.out.println(find_greatest_divisor(l, r));
    }
}
// This code is contributed by PrinciRaj1992

# Python3 implementation of the approach
 
# Function to return the greatest divisor that
# divides all the natural numbers in the range [l, r]
def find_greatest_divisor(l, r):
 
    if (l == r):
        return l;
 
    return 1;
 
 
# Driver Code
 
l = 2;
r = 12;
 
print(find_greatest_divisor(l, r));
 
#This code is contributed by Shivi_Aggarwal

// C# implementation of the approach
using System;
 
class GFG {
 
    // Function to return the greatest divisor that
    // divides all the natural numbers in the range [l, r]
    static int find_greatest_divisor(int l, int r) {
        if (l == r) {
            return l;
        }
 
        return 1;
    }
 
    // Driver Code
    public static void Main() {
        int l = 2,  r = 12 ;
 
        Console.WriteLine(find_greatest_divisor(l, r));
    }
    // This code is contributed by Ryuga
 
}

<?php
// PHP implementation of the approach
 
// Function to return the greatest
// divisor that divides all the natural
// numbers in the range [l, r]
function find_greatest_divisor($l, $r)
{
    if ($l == $r)
        return $l;
 
    return 1;
}
 
// Driver Code
$l = 2;
$r = 12;
 
echo find_greatest_divisor($l, $r);
 
// This code is contributed by jit_t
?>

<script>
 
// Javascript implementation of the approach
 
// Function to return the greatest
// divisor that divides all the natural
// numbers in the range [l, r]
function find_greatest_divisor(l, r)
{
    if (l == r)
        return l;
 
    return 1;
}
 
// Driver Code
let l = 2;
let r = 12;
 
document.write( find_greatest_divisor(l, r));
 
// This code is contributed
// by bobby
 
</script>

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Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)