重新排列数组以获得前缀 GCD 串联的最大可能值

给定的阵列ARR []由N个正整数，任务是重新排列的数组元素，使得通过连接所述阵列ARR的元素的GCD形成数[]从索引0至i为每个索引i是最大可能的.

例子：

Input: arr[] = {4, 2, 5}
Output: 5 4 2
Explanation:
X = 511 is the maximum value of X that can be obtained among all the rearrangement of arr[].
Possible arrangements of arr[] are:
arr[] = [2, 4, 5] → X = 221
arr[] = [2, 5, 4] → X = 211
arr[] = [4, 2, 5] → X = 421
arr[] = [4, 5, 2] → X = 411
arr[] = [5, 4, 2] → X = 511
arr[] = [5, 2, 4] → X = 511

Input: arr[] = {2, 4, 6, 8}
Output: 8 4 6 2
Explanation: 
X = 842 is the maximum value of X that can be obtained among all the rearrangement of arr[].
Possible arrangements of arr[] are:
arr[] = [4, 6, 8] → X = 422
arr[] = [4, 8, 6] → X = 442
arr[] = [6, 4, 8] → X = 622
arr[] = [6, 8, 4] → X = 622
arr[] = [8, 4, 6] → X = 842
arr[] = [8, 6, 4] → X = 822

方法：数字的 GCD 本身就是数字本身，因此X的第一个数字，即X[0]将始终等于arr[0] 。因此，为了确保X在所有可获得的数字中最大， arr[0]需要最大。然后继续跟踪已排列的arr[]最长前缀的 GCD，并找到要放置在此前缀之后的连续元素的值。请按照以下步骤解决上述问题：

1. 数组中最大的元素被设置为第一个元素，因此第一个前缀正确排列在数组arr[]中。
2. 现在找到与前缀的最后一个元素连续的元素，即arr[1] 。
3. 这里的 最长前缀(说G)的GCD是等于给Arr [0]，从而遍历其余阵列找到给出其中G最大的GCD的元素。
4. 现在，交换元件ARR [1]给出最大GCD与值G的元素，更新G的值到该最大GCD获得即，G = GCD(G，ARR [1])。
5. 现在最长的固定前缀变成了arr[0] , arr[1] ，继续这个过程寻找arr[2] , arr[3] , …, arr[N – 1] ，得到所需的数组。
6. 在上述步骤后打印重新排列的数组。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the maximum number
// obtainable from prefix GCDs
void prefixGCD(int arr[], int N)
{
    // Stores the GCD of the
    // longest prefix
    int gcc;
 
    // Sort the array
    sort(arr, arr + N);
 
    // Reverse the array
    reverse(arr, arr + N);
 
    // GCD of a[0] is a[0]
    gcc = arr[0];
    int start = 0;
 
    // Iterate to place the arr[start + 1]
    // element at it's correct position
    while (start < N - 1) {
 
        int g = 0, s1;
 
        for (int i = start + 1; i < N; i++) {
 
            // Find the element with
            // maximum GCD
            int gc = __gcd(gcc, arr[i]);
 
            // Update the value of g
            if (gc > g) {
                g = gc;
                s1 = i;
            }
        }
 
        // Update GCD of prefix
        gcc = g;
 
        // Place arr[s1] to it's
        // correct position
        swap(arr[s1], arr[start + 1]);
 
        // Increment start for the
        // remaining elements
        start++;
    }
 
    // Print the rearranged array
    for (int i = 0; i < N; i++) {
        cout << arr[i] << " ";
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 2, 3, 4 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    prefixGCD(arr, N);
 
    return 0;
}

Java

//Java program for
// the above approach
import java.util.*;
class GFG{
 
//Function to find the maximum number
//obtainable from prefix GCDs
static void prefixGCD(int arr[], int N)
{
  // Stores the GCD of the
  // longest prefix
  int gcc;
 
  // Sort the array
  Arrays.sort(arr);
 
  // Reverse the array
  arr = reverse(arr);
 
  // GCD of a[0] is a[0]
  gcc = arr[0];
  int start = 0;
 
  // Iterate to place
  // the arr[start + 1]
  // element at it's
  // correct position
  while (start < N - 1)
  {
    int g = 0, s1 = 0;
 
    for (int i = start + 1; i < N; i++)
    {
      // Find the element with
      // maximum GCD
      int gc = __gcd(gcc, arr[i]);
 
      // Update the value of g
      if (gc > g)
      {
        g = gc;
        s1 = i;
      }
    }
 
    // Update GCD of prefix
    gcc = g;
 
    // Place arr[s1] to it's
    // correct position
    arr = swap(arr, s1, start + 1);
 
    // Increment start for the
    // remaining elements
    start++;
  }
 
  // Print the rearranged array
  for (int i = 0; i < N; i++)
  {
    System.out.print(arr[i] + " ");
  }
}
   
static int __gcd(int a, int b) 
{ 
  return b == 0 ? a : __gcd(b, a % b);    
}
 
static int[] reverse(int a[])
{
  int i, n = a.length, t;
  for (i = 0; i < n / 2; i++)
  {
    t = a[i];
    a[i] = a[n - i - 1];
    a[n - i - 1] = t;
  }
  return a;
}
 
static int[] swap(int []arr,
                  int i, int j)
{
  int temp = arr[i];
  arr[i] = arr[j];
  arr[j] = temp;
  return arr;
}
     
//Driver Code
public static void main(String[] args)
{
  // Given array arr[]
  int arr[] = {1, 2, 3, 4};
 
  int N = arr.length;
 
  // Function Call
  prefixGCD(arr, N);
}
}
 
//This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
from math import gcd
 
# Function to find the maximum number
# obtainable from prefix GCDs
def prefixGCD(arr, N):
     
    # Stores the GCD of the
    # longest prefix
    gcc = 0
 
    # Sort the array
    arr = sorted(arr)
 
    # Reverse the array
    arr = arr[::-1]
 
    # GCD of a[0] is a[0]
    gcc = arr[0]
    start = 0
 
    # Iterate to place the arr[start + 1]
    # element at it's correct position
    while (start < N - 1):
        g = 0
        s1 = 0
 
        for i in range(start + 1, N):
 
            # Find the element with
            # maximum GCD
            gc = gcd(gcc, arr[i])
 
            # Update the value of g
            if (gc > g):
                g = gc
                s1 = i
 
        # Update GCD of prefix
        gcc = g
 
        # Place arr[s1] to it's
        # correct position
        arr[s1], arr[start + 1] = arr[start + 1], arr[s1]
 
        # Increment start for the
        # remaining elements
        start += 1
 
    # Print the rearranged array
    for i in range(N):
        print(arr[i], end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    # Given array arr[]
    arr = [ 1, 2, 3, 4 ]
 
    N = len(arr)
 
    # Function Call
    prefixGCD(arr, N)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach 
using System;
class GFG{
 
// Function to find the maximum number
// obtainable from prefix GCDs
static void prefixGCD(int[] arr, int N)
{
     
  // Stores the GCD of the
  // longest prefix
  int gcc;
 
  // Sort the array
  Array.Sort(arr);
 
  // Reverse the array
  arr = reverse(arr);
 
  // GCD of a[0] is a[0]
  gcc = arr[0];
  int start = 0;
 
  // Iterate to place the
  // arr[start + 1] element
  // at it's correct position
  while (start < N - 1)
  {
    int g = 0, s1 = 0;
 
    for(int i = start + 1; i < N; i++)
    {
         
      // Find the element with
      // maximum GCD
      int gc = __gcd(gcc, arr[i]);
 
      // Update the value of g
      if (gc > g)
      {
        g = gc;
        s1 = i;
      }
    }
 
    // Update GCD of prefix
    gcc = g;
 
    // Place arr[s1] to it's
    // correct position
    arr = swap(arr, s1, start + 1);
 
    // Increment start for the
    // remaining elements
    start++;
  }
 
  // Print the rearranged array
  for(int i = 0; i < N; i++)
  {
    Console.Write(arr[i] + " ");
  }
}
   
static int __gcd(int a, int b) 
{ 
  return b == 0 ? a : __gcd(b, a % b);    
}
 
static int[] reverse(int[] a)
{
  int i, n = a.Length, t;
   
  for(i = 0; i < n / 2; i++)
  {
    t = a[i];
    a[i] = a[n - i - 1];
    a[n - i - 1] = t;
  }
  return a;
}
 
static int[] swap(int []arr, int i,
                             int j)
{
  int temp = arr[i];
  arr[i] = arr[j];
  arr[j] = temp;
  return arr;
}
     
//Driver Code
public static void Main()
{
     
  // Given array arr[]
  int[] arr = { 1, 2, 3, 4 };
 
  int N = arr.Length;
 
  // Function call
  prefixGCD(arr, N);
}
}
 
// This code is contributed by sanjoy_62

Javascript

 
// Javascript program to implement
// the above approach
 
//Function to find the maximum number
//obtainable from prefix GCDs
function prefixGCD(arr, N)
{
  // Stores the GCD of the
  // longest prefix
  let gcc;
 
  // Sort the array
  arr.sort();
 
  // Reverse the array
  arr = reverse(arr);
 
  // GCD of a[0] is a[0]
  gcc = arr[0];
  let start = 0;
 
  // Iterate to place
  // the arr[start + 1]
  // element at it's
  // correct position
  while (start < N - 1)
  {
    let g = 0, s1 = 0;
 
    for (let i = start + 1; i < N; i++)
    {
      // Find the element with
      // maximum GCD
      let gc = __gcd(gcc, arr[i]);
 
      // Update the value of g
      if (gc > g)
      {
        g = gc;
        s1 = i;
      }
    }
 
    // Update GCD of prefix
    gcc = g;
 
    // Place arr[s1] to it's
    // correct position
    arr = swap(arr, s1, start + 1);
 
    // Increment start for the
    // remaining elements
    start++;
  }
 
  // Prlet the rearranged array
  for (let i = 0; i < N; i++)
  {
    document.write(arr[i] + " ");
  }
}
   
function __gcd(a, b) 
{ 
  return b == 0 ? a : __gcd(b, a % b);    
}
 
function reverse(a)
{
  let i, n = a.length, t;
  for (i = 0; i < n / 2; i++)
  {
    t = a[i];
    a[i] = a[n - i - 1];
    a[n - i - 1] = t;
  }
  return a;
}
 
function swap(arr, i, j)
{
  let temp = arr[i];
  arr[i] = arr[j];
  arr[j] = temp;
  return arr;
}
 
// Driver Code
 
    // Given array arr[]
  let arr = [1, 2, 3, 4];
 
  let N = arr.length;
 
  // Function Call
  prefixGCD(arr, N);
 

输出：

4 2 3 1

时间复杂度： O(N ² )
辅助空间： O(1)

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