Número de subarreglos con GCD igual a 1

Dada una array arr[] , la tarea es encontrar el número de sub-arrays con un valor GCD igual a 1 .

Ejemplos:

Entrada: arr[] = {1, 1, 1} 
Salida: 6 
Todos los subarreglos de la array dada 
tendrán GCD igual a 1.
Entrada: arr[] = {2, 2, 2} 
Salida: 0  

Enfoque: La observación clave es que si se conoce el GCD de todos los elementos del subconjunto arr[l…r ] , entonces se puede obtener el GCD de todos los elementos del subconjunto arr[l…r+1]. simplemente tomando el GCD del subarreglo anterior con arr[r + 1] . 
Por lo tanto, para cada índice i , siga iterando hacia adelante y calcule el GCD del índice i al j y verifique si es igual a 1 .

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 required count
int cntSubArr(int* arr, int n)
{
    // To store the final answer
    int ans = 0;
 
    for (int i = 0; i < n; i++) {
 
        // To store the GCD starting from
        // index 'i'
        int curr_gcd = 0;
 
        // Loop to find the gcd of each subarray
        // from arr[i] to arr[i...n-1]
        for (int j = i; j < n; j++) {
            curr_gcd = __gcd(curr_gcd, arr[j]);
 
            // Increment the count if curr_gcd = 1
            ans += (curr_gcd == 1);
        }
    }
 
    // Return the final answer
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 1, 1 };
    int n = sizeof(arr) / sizeof(int);
 
    cout << cntSubArr(arr, n);
 
    return 0;
}

// Java implementation of the approach
class GFG
{
 
// Function to return the required count
static int cntSubArr(int []arr, int n)
{
    // To store the final answer
    int ans = 0;
 
    for (int i = 0; i < n; i++)
    {
 
        // To store the GCD starting from
        // index 'i'
        int curr_gcd = 0;
 
        // Loop to find the gcd of each subarray
        // from arr[i] to arr[i...n-1]
        for (int j = i; j < n; j++)
        {
            curr_gcd = __gcd(curr_gcd, arr[j]);
 
            // Increment the count if curr_gcd = 1
            ans += (curr_gcd == 1) ? 1 : 0;
        }
    }
 
    // Return the final answer
    return ans;
}
 
static int __gcd(int a, int b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);    
}
 
// Driver code
public static void main(String []args)
{
    int arr[] = { 1, 1, 1 };
    int n = arr.length;
 
    System.out.println(cntSubArr(arr, n));
}
}
 
// This code is contributed by Rajput-Ji

# Python3 implementation of the approach
from math import gcd
 
# Function to return the required count
def cntSubArr(arr, n) :
 
    # To store the final answer
    ans = 0;
 
    for i in range(n) :
 
        # To store the GCD starting from
        # index 'i'
        curr_gcd = 0;
 
        # Loop to find the gcd of each subarray
        # from arr[i] to arr[i...n-1]
        for j in range(i, n) :
            curr_gcd = gcd(curr_gcd, arr[j]);
 
            # Increment the count if curr_gcd = 1
            ans += (curr_gcd == 1);
 
    # Return the final answer
    return ans;
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 1, 1, 1 ];
    n = len(arr);
 
    print(cntSubArr(arr, n));
 
# This code is contributed by AnkitRai01

// C# implementation of the approach
using System;
                     
class GFG
{
 
// Function to return the required count
static int cntSubArr(int []arr, int n)
{
    // To store the final answer
    int ans = 0;
 
    for (int i = 0; i < n; i++)
    {
 
        // To store the GCD starting from
        // index 'i'
        int curr_gcd = 0;
 
        // Loop to find the gcd of each subarray
        // from arr[i] to arr[i...n-1]
        for (int j = i; j < n; j++)
        {
            curr_gcd = __gcd(curr_gcd, arr[j]);
 
            // Increment the count if curr_gcd = 1
            ans += (curr_gcd == 1) ? 1 : 0;
        }
    }
 
    // Return the final answer
    return ans;
}
 
static int __gcd(int a, int b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);    
}
 
// Driver code
public static void Main(String []args)
{
    int []arr = { 1, 1, 1 };
    int n = arr.Length;
 
    Console.WriteLine(cntSubArr(arr, n));
}
}
 
// This code is contributed by Rajput-Ji

<script>
 
// Javascript implementation of the approach
 
function __gcd(a, b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);    
}
 
 
// Function to return the required count
function cntSubArr(arr, n)
{
    // To store the final answer
    var ans = 0;
 
    for (var i = 0; i < n; i++) {
 
        // To store the GCD starting from
        // index 'i'
        var curr_gcd = 0;
 
        // Loop to find the gcd of each subarray
        // from arr[i] to arr[i...n-1]
        for (var j = i; j < n; j++) {
            curr_gcd = __gcd(curr_gcd, arr[j]);
 
            // Increment the count if curr_gcd = 1
            ans += (curr_gcd == 1);
        }
    }
 
    // Return the final answer
    return ans;
}
 
// Driver code
var arr = [1, 1, 1];
var n = arr.length;
document.write( cntSubArr(arr, n));
 
</script>

6

Complejidad de tiempo: O(N ² log(max(arr[])))

Espacio Auxiliar: O(1)