重新排列数组以最大化数组元素及其索引的GCD之和

给定一个由前N个自然数组成的数组arr [] ，任务是通过重新排列给定的数组元素来找到ΣGCD(arr [i]，i) (基于1的索引)的最大可能值。

例子：

Input: arr[] = { 2, 1} 
Output: 6 
Explanation: 
Rearranging the given array to { 2, 1}. 
ΣGCD(arr[i], i) = GCD(arr[1], 1) + GCD(arr[2], 2) = GCD(2, 1) + GCD(1, 2)= 2 
Rearranging the given array to { 1, 2 }. 
ΣGCD(arr[i], i) = GCD(arr[1], 1) + GCD(arr[2], 2) = GCD(1, 1) + GCD(2, 2) = 3 
Therefore, the required output is 3

Input: arr[] = { 4, 5, 3, 2, 1 } 
Output: 15

天真的方法：解决问题的最简单方法是遍历数组并生成给定数组的所有可能排列，对于每个排列，找到ΣGCD(arr [i]，i)的值。最后，从每个排列中打印ΣGCD(arr [i]，i)的最大值。

下面是上述方法的实现：

C++

// C++ program to impement
// the above approach
 
#include 
using namespace std;
 
// Function to find the maximum sum of
// GCD(arr[i], i) by rearranging the array
int findMaxValByRearrArr(int arr[], int N)
{
 
    // Sort the array in
    // ascending order
    sort(arr, arr + N);
 
    // Stores maximum sum of GCD(arr[i], i)
    // by rearranging the array elements
    int res = 0;
 
    // Generate all possible
    // permutations of the array
    do {
 
        // Stores sum of GCD(arr[i], i)
        int sum = 0;
 
        // Traverse the array
        for (int i = 0; i < N; i++) {
 
            // Update sum
            sum += __gcd(i + 1, arr[i]);
        }
 
        // Update res
        res = max(res, sum);
 
    } while (next_permutation(arr, arr + N));
 
    return res;
}
 
// Driver Code
int main()
{
 
    int arr[] = { 3, 2, 1 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << findMaxValByRearrArr(arr, N);
    return 0;
}

Java

// Java program to impement
// the above approach
import java.util.*;
class GFG{
 
  // Function to find the maximum sum of
  // GCD(arr[i], i) by rearranging the array
  static int findMaxValByRearrArr(int arr[], int N)
  {
 
    // Sort the array in
    // ascending order
    Arrays.sort(arr);
 
    // Stores maximum sum of GCD(arr[i], i)
    // by rearranging the array elements
    int res = 0;
 
    // Generate all possible
    // permutations of the array
    do {
 
      // Stores sum of GCD(arr[i], i)
      int sum = 0;
 
      // Traverse the array
      for (int i = 0; i < N; i++) {
 
        // Update sum
        sum += __gcd(i + 1, arr[i]);
      }
 
      // Update res
      res = Math.max(res, sum);
 
    } while (next_permutation(arr));
 
    return res;
  }
  static int __gcd(int a, int b) 
  { 
    return b == 0? a:__gcd(b, a % b);    
  }
  static boolean next_permutation(int[] p)
  {
    for (int a = p.length - 2; a >= 0; --a)
      if (p[a] < p[a + 1])
        for (int b = p.length - 1;; --b)
          if (p[b] > p[a])
          {
            int t = p[a];
            p[a] = p[b];
            p[b] = t;
            for (++a, b = p.length - 1; a < b; ++a, --b)
            {
              t = p[a];
              p[a] = p[b];
              p[b] = t;
            }
            return true;
          }
    return false;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    int arr[] = { 3, 2, 1 };
    int N = arr.length;
    System.out.print(findMaxValByRearrArr(arr, N));
  }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to implement
# the above approach
 
# Function to find the maximum sum of
# GCD(arr[i], i) by rearranging the array
def findMaxValByRearrArr(arr, N):
     
    # Sort the array in
    # ascending order
    arr.sort()
     
    # Stores maximum sum of GCD(arr[i], i)
    # by rearranging the array elements
    res = 0
     
    # Generate all possible
    # permutations of the array
    while (True):
         
        # Stores sum of GCD(arr[i], i)
        Sum = 0
         
        # Traverse the array
        for i in range(N):
             
            # Update sum
            Sum += __gcd(i + 1, arr[i])
         
        # Update res
        res = max(res, Sum)
         
        if (not next_permutation(arr)):
            break
         
    return res
     
def __gcd(a, b): 
     
    if b == 0:
        return a
    else:
        return __gcd(b, a % b)
 
def next_permutation(p):
     
    for a in range(len(p) - 2, -1, -1):
        if (p[a] < p[a + 1]):
            b = len(p) - 1
             
            while True:
                if (p[b] > p[a]):
                    t = p[a]
                    p[a] = p[b]
                    p[b] = t
                    a += 1
                    b = len(p) - 1
                     
                    while a < b:
                        t = p[a]
                        p[a] = p[b]
                        p[b] = t
                        a += 1
                        b -= 1
                         
                    return True
                     
                b -= 1
                 
    return False
 
# Driver code   
arr = [ 3, 2, 1 ]
N = len(arr)
 
print(findMaxValByRearrArr(arr, N))
 
# This code is contributed by divyesh072019

C#

// C# program to impement
// the above approach
using System;
class GFG
{
 
  // Function to find the maximum sum of
  // GCD(arr[i], i) by rearranging the array
  static int findMaxValByRearrArr(int []arr, int N)
  {
 
    // Sort the array in
    // ascending order
    Array.Sort(arr);
 
    // Stores maximum sum of GCD(arr[i], i)
    // by rearranging the array elements
    int res = 0;
 
    // Generate all possible
    // permutations of the array
    do
    {
 
      // Stores sum of GCD(arr[i], i)
      int sum = 0;
 
      // Traverse the array
      for (int i = 0; i < N; i++)
      {
 
        // Update sum
        sum += __gcd(i + 1, arr[i]);
      }
 
      // Update res
      res = Math.Max(res, sum);
 
    } while (next_permutation(arr));
 
    return res;
  }
  static int __gcd(int a, int b) 
  { 
    return b == 0? a:__gcd(b, a % b);    
  }
  static bool next_permutation(int[] p)
  {
    for (int a = p.Length - 2; a >= 0; --a)
      if (p[a] < p[a + 1])
        for (int b = p.Length - 1;; --b)
          if (p[b] > p[a])
          {
            int t = p[a];
            p[a] = p[b];
            p[b] = t;
            for (++a, b = p.Length - 1; a < b; ++a, --b)
            {
              t = p[a];
              p[a] = p[b];
              p[b] = t;
            }
            return true;
          }
    return false;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int []arr = { 3, 2, 1 };
    int N = arr.Length;
    Console.Write(findMaxValByRearrArr(arr, N));
  }
}
 
// This code is contributed by 29AjayKumar

输出：

6

时间复杂度： O(N！)
辅助空间： O(1)

高效的方法：可以基于以下观察来优化上述方法：

Maximum possible value of GCD(X, Y) = min(X, Y) 
Therefore, the maximum possible value of GCD(i, arr[i]) = min(i, arr[i]) 
If array is sorted then i = arr[i] and the value of GCD(i, arr[i]) = i 
ΣGCD(arr[i], i) = Σi = N * (N + 1) / 2 
 

请按照以下步骤解决问题：

• 初始化一个变量，例如res ，以存储ΣGCD(arr [i]，i)的最大可能和。
• 更新res =(N *(N +1)/ 2) 。
• 最后，打印res的值。

下面是上述方法的实现：

C++

// C++ program to impement
// the above approach
 
#include 
using namespace std;
 
// Function to find the maximum sum of
// GCD(arr[i], i) by rearranging the array
int findMaxValByRearrArr(int arr[], int N)
{
 
    // Stores maximum sum of GCD(arr[i], i)
    // by rearranging the array elements
    int res = 0;
 
    // Update res
    res = (N * (N + 1)) / 2;
 
    return res;
}
 
// Driver Code
int main()
{
 
    int arr[] = { 3, 2, 1 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << findMaxValByRearrArr(arr, N);
    return 0;
}

Java

// Java program to impement
// the above approach
import java.util.*;
class GFG
{
 
// Function to find the maximum sum of
// GCD(arr[i], i) by rearranging the array
static int findMaxValByRearrArr(int arr[], int N)
{
 
    // Stores maximum sum of GCD(arr[i], i)
    // by rearranging the array elements
    int res = 0;
 
    // Update res
    res = (N * (N + 1)) / 2;
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 2, 1 };
    int N = arr.length;
    System.out.print(findMaxValByRearrArr(arr, N));
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python program to impement
# the above approach
 
# Function to find the maximum sum of
# GCD(arr[i], i) by rearranging the array
def findMaxValByRearrArr(arr, N):
 
    # Stores maximum sum of GCD(arr[i], i)
    # by rearranging the array elements
    res = 0;
 
    # Update res
    res = (N * (N + 1)) // 2;
    return res;
 
# Driver Code
if __name__ == '__main__':
    arr = [ 3, 2, 1 ];
    N = len(arr);
    print(findMaxValByRearrArr(arr, N));
 
# This code contributed by shikhasingrajput

C＃

// C# program to impement
// the above approach
using System;
 
class GFG{
     
// Function to find the maximum sum of
// GCD(arr[i], i) by rearranging the array
static int findMaxValByRearrArr(int []arr, int N)
{
     
    // Stores maximum sum of GCD(arr[i], i)
    // by rearranging the array elements
    int res = 0;
     
    // Update res
    res = (N * (N + 1)) / 2;
    return res;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 3, 2, 1 };
    int N = arr.Length;
     
    Console.Write(findMaxValByRearrArr(arr, N));
}
}
 
// This code is contributed by shikhasingrajput

Java脚本

 
    // Javascript program to impement
    // the above approach
     
    // Function to find the maximum sum of
    // GCD(arr[i], i) by rearranging the array
    function findMaxValByRearrArr(arr, N)
    {
 
        // Stores maximum sum of GCD(arr[i], i)
        // by rearranging the array elements
        let res = 0;
 
        // Update res
        res = parseInt((N * (N + 1)) / 2, 10);
        return res;
    }
     
    let arr = [ 3, 2, 1 ];
    let N = arr.length;
      
    document.write(findMaxValByRearrArr(arr, N));
 

输出：

6

时间复杂度： O(1)
辅助空间： O(1)