找到 K 的最小值以最大化 K 倍数的索引上的元素总和

给定一个包含N个整数的数组arr[] ，任务是找到K的最小值，使得索引上的元素总和是K的倍数是可能的最大值。

例子：

Input: arr[] = {-3, 4}
Output: 2
Explanation: For the given array, it the value of K = 1, then the multiples of K are {1, 2} which sums up to arr[1] + arr[2] = -3 + 4 = 1. For K = 2, the valid multiple of K is 2 and hence the sum is arr[2] = 4, which is the maximum possible. Hence, K = 2 is a valid answer. 

Input: arr[] = {-1, -2, -3}
Output: 2

方法：给定的问题可以通过类似于埃拉托色尼筛法的方法来解决。这个想法是通过迭代K的每个倍数来计算范围[1, N]内所有可能值的总和，同时在筛子中标记非素数元素。给出最大和的K值是所需的答案。

下面是上述方法的实现：

C++

// C++ Program of the above approach
#include 
using namespace std;
 
// Function to find minimum K such that
// the sum of elements on indices that
// are multiples of K is maximum possible
int maxSum(int arr[], int N)
{
    // Stores the maximum sum and
    // respective K value
    int maxSum = INT_MIN, ans = -1;
 
    // Loop to iterate over all
    // value of K in range [1, N]
    for (int K = 1; K <= N; K++) {
        int sum = 0;
 
        // Iterating over all
        // multiples of K
        for (int i = K; i <= N; i += K) {
            sum += arr[i - 1];
        }
 
        // Update Maximum Sum
        if (sum > maxSum) {
            maxSum = sum;
            ans = K;
        }
    }
 
    // Return Answer
    return ans;
}
 
// Driver Code
int main()
{
    int arr[] = { -1, -2, -3 };
    int N = sizeof(arr) / sizeof(int);
 
    cout << maxSum(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to find minimum K such that
// the sum of elements on indices that
// are multiples of K is maximum possible
static int maxSum(int arr[], int N)
{
     
    // Stores the maximum sum and
    // respective K value
    int maxSum = Integer.MIN_VALUE, ans = -1;
 
    // Loop to iterate over all
    // value of K in range [1, N]
    for(int K = 1; K <= N; K++)
    {
        int sum = 0;
 
        // Iterating over all
        // multiples of K
        for(int i = K; i <= N; i += K)
        {
            sum += arr[i - 1];
        }
 
        // Update Maximum Sum
        if (sum > maxSum)
        {
            maxSum = sum;
            ans = K;
        }
    }
 
    // Return Answer
    return ans;
}
 
// Driver Code
public static void main(String args[])
{
    int arr[] = { -1, -2, -3 };
    int N = arr.length;
 
    System.out.println(maxSum(arr, N));
}
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python program for the above approach
import sys
 
# Function to find minimum K such that
# the sum of elements on indices that
# are multiples of K is maximum possible
def maxSum(arr, N):
 
    # Stores the maximum sum and
    # respective K value
    maxSum = -sys.maxsize;
    ans = -1;
 
    # Loop to iterate over all
    # value of K in range [1, N]
    for K in range(1,N+1):
        sum = 0;
 
        # Iterating over all
        # multiples of K
        for i in range(K, N + 1,K):
            sum += arr[i - 1];
         
        # Update Maximum Sum
        if (sum > maxSum):
            maxSum = sum;
            ans = K;
         
    # Return Answer
    return ans;
 
# Driver Code
if __name__ == '__main__':
    arr = [ -1, -2, -3 ];
    N = len(arr);
 
    print(maxSum(arr, N));
 
# This code is contributed by gauravrajput1

C#

// C# program for the above approach
using System;
 
public class GFG{
 
  // Function to find minimum K such that
  // the sum of elements on indices that
  // are multiples of K is maximum possible
  static int maxSum(int []arr, int N)
  {
 
    // Stores the maximum sum and
    // respective K value
    int maxSum = int.MinValue, ans = -1;
 
    // Loop to iterate over all
    // value of K in range [1, N]
    for(int K = 1; K <= N; K++)
    {
      int sum = 0;
 
      // Iterating over all
      // multiples of K
      for(int i = K; i <= N; i += K)
      {
        sum += arr[i - 1];
      }
 
      // Update Maximum Sum
      if (sum > maxSum)
      {
        maxSum = sum;
        ans = K;
      }
    }
 
    // Return Answer
    return ans;
  }
 
  // Driver Code
  public static void Main(String []args)
  {
    int []arr = { -1, -2, -3 };
    int N = arr.Length;
 
    Console.WriteLine(maxSum(arr, N));
  }
}
 
// This code is contributed by 29AjayKumar

Javascript

      // JavaScript code for the above approach
 
 
      // Function to find minimum K such that
      // the sum of elements on indices that
      // are multiples of K is maximum possible
      function maxSum(arr, N) {
          // Stores the maximum sum and
          // respective K value
          let maxSum = -999999;
          let ans = -1;
 
          // Loop to iterate over all
          // value of K in range [1, N]
          for (let K = 1; K <= N; K++) {
              let sum = 0;
 
              // Iterating over all
              // multiples of K
              for (let i = K; i <= N; i += K) {
                  sum = sum + arr[i - 1];
              }
 
              // Update Maximum Sum
              if (sum > maxSum) {
                  maxSum = sum;
                  ans = K;
              }
          }
 
          // Return Answer
          return ans;
      }
 
      // Driver Code
 
      let arr = [-1, -2, -3];
      let N = arr.length;
 
      document.write(maxSum(arr, N));
 
 
// This code is contributed by Potta Lokesh
  

输出

2

时间复杂度： O(N*log N)
辅助空间： O(1)