最少插入以制作一个 Co-prime 数组

给定一个包含 N 个元素的数组，找到将给定数组转换为互素数组的最小插入次数。也打印结果数组。
互素数组：每对相邻元素互素的数组。 IE， $gcd(a, b) = 1$ .

例子：

Input : A[] = {2, 7, 28}
Output : 1
Explanation : 
Here, 1st pair = {2, 7} are co-primes( gcd(2, 7) = 1).
2nd pair = {7, 28} are not co-primes, insert 9
between them. gcd(7, 9) = 1 and gcd(9, 28) = 1.


Input : A[] = {5, 10, 20}
Output : 2
Explanation : 
Here, there is no pair which are co-primes. 
Insert 7 between (5, 10) and 1 between (10, 20).

请注意，要使一对成为互质，我们必须插入一个数字，使新形成的对成为互质。因此，我们必须检查每个相邻对的共素性，并在需要时插入一个数字。现在，应该插入的数字是多少？让我们取两个数字 a 和 b。如果 a 或 b 中的任何一个为 1，则 GCD(a, b) = 1。因此，我们可以在每一对插入 ONE(1)。为了提高效率，我们使用欧拉的 gcd函数。

下面是上述方法的实现：

C++

// CPP program for minimum insertions to
// make a Co-prime Array.
#include 
using namespace std;
 
void printResult(int arr[], int n)
{
    // Counting adjacent pairs that are not
    // co-prime.
    int count = 0;
    for (int i = 1; i < n; i++)    
        if (__gcd(arr[i], arr[i - 1]) != 1)
            count++;
 
    cout << count << endl; // No.of insertions
    cout << arr[0] << " ";
    for (int i = 1; i < n; i++)
    {
        // If pair is not a co-prime insert 1.
        if (__gcd(arr[i], arr[i - 1]) != 1)
            cout << 1 << " ";
        cout << arr[i] << " ";
    }
}
 
// Driver Function
int main()
{
    int A[] = { 5, 10, 20 };
    int n = sizeof(A) / sizeof(A[0]);
    printResult(A, n);
    return 0;
}

Java

//Java program for minimum insertions
// to make a Co-prime Array.
import java.io.*;
 
class GFG {
     
    // Recursive function to return
    // gcd of a and b
    static int gcd(int a, int b)
    {
        // Everything divides 0
        if (a == 0 || b == 0)
        return 0;
     
        // base case
        if (a == b)
            return a;
     
        // a is greater
        if (a > b)
            return gcd(a-b, b);
 
        return gcd(a, b-a);
    }
     
    static void printResult(int arr[], int n)
    {
         
        // Counting adjacent pairs that are not
        // co-prime.
        int count = 0;
 
        for (int i = 1; i < n; i++)    
            if (gcd(arr[i], arr[i - 1]) != 1)
                count++;
     
        // No.of insertions
        System.out.println(count );
        System.out.print (arr[0] + " ");
 
        for (int i = 1; i < n; i++)
        {
             
            // If pair is not a co-prime insert 1.
            if (gcd(arr[i], arr[i - 1]) != 1)
                System.out.print( 1 + " ");
            System.out.print(arr[i] + " ");
        }
    }
     
    // Driver Function
    public static void main(String args[])
    {
        int A[] = { 5, 10, 20 };
        int n = A.length;
        printResult(A, n);
    }
}
 
/*This code is contributed by Nikita Tiwari.*/

Python3

# Python3 code for minimum insertions
# to make a Co-prime Array.
from fractions import gcd
 
def printResult(arr, n):
 
    # Counting adjacent pairs that
    # are not co-prime.
    count = 0
    for i in range(1,n):
        if (gcd(arr[i], arr[i - 1]) != 1):
            count+=1
     
    print(count)     # No.of insertions
    print( arr[0], end = " ")
    for i in range(1,n):
         
        # If pair is not a co-prime insert 1.
        if (gcd(arr[i], arr[i - 1]) != 1):
            print(1, end = " ")
        print(arr[i] , end = " ")
         
# Driver Code
A = [ 5, 10, 20 ]
n = len(A)
printResult(A, n)
 
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program for minimum insertions
// to make a Co-prime Array.
using System;
 
class GFG {
 
    // Recursive function to return
    // gcd of a and b
    static int gcd(int a, int b)
    {
        // Everything divides 0
        if (a == 0 || b == 0)
            return 0;
 
        // base case
        if (a == b)
            return a;
 
        // a is greater
        if (a > b)
            return gcd(a - b, b);
 
        return gcd(a, b - a);
    }
 
    static void printResult(int[] arr, int n)
    {
        // Counting adjacent pairs that
        // are not co-prime.
        int count = 0;
 
        for (int i = 1; i < n; i++)
            if (gcd(arr[i], arr[i - 1]) != 1)
                count++;
 
        // No.of insertions
        Console.WriteLine(count);
        Console.Write(arr[0] + " ");
 
        for (int i = 1; i < n; i++) {
 
            // If pair is not a co-prime insert 1.
            if (gcd(arr[i], arr[i - 1]) != 1)
                Console.Write(1 + " ");
            Console.Write(arr[i] + " ");
        }
    }
 
    // Driver Function
    public static void Main()
    {
        int[] A = { 5, 10, 20 };
        int n = A.Length;
        printResult(A, n);
    }
}
 
/*This code is contributed by vt_m.*/

PHP

 $b)
        return gcd($a - $b, $b);
 
    return gcd($a, $b - $a);
}
 
function printResult($arr, $n)
{
    // Counting adjacent pairs
    // that are not co-prime.
    $count = 0;
 
    for ($i = 1; $i < $n; $i++)
        if (gcd($arr[$i],
                $arr[$i - 1]) != 1)
            $count++;
 
    // No.of insertions
    echo $count, "\n";
    echo $arr[0] , " ";
 
    for ($i = 1; $i < $n; $i++)
    {
 
        // If pair is not a
        // co-prime insert 1.
        if (gcd($arr[$i],
                $arr[$i - 1]) != 1)
            echo 1 , " ";
        echo $arr[$i] , " ";
    }
}
 
// Driver Code
$A = array(5, 10, 20);
$n = sizeof($A);
printResult($A, $n);
 
// This code is contributed
// by ajit
?>

Javascript

输出：

2
5 1 10 1 20

时间复杂度： O(n)。