求 K 的最小值，使指数中的元素之和最大化，指数是 K 的倍数

原文:https://www . geeksforgeeks . org/find-最小 k 值-最大化 k 倍数指数上的元素总和/

给定一个由 N 个整数组成的数组 arr[] ，任务是找到 K 的最小值，使得指数上的元素之和是 K 的倍数是最大可能的。

示例:

输入: arr[] = {-3，4} 输出: 2 解释:对于给定的数组，它取 K = 1 的值，那么 K 的倍数为{1，2}，其和为 arr[1] + arr[2] = -3 + 4 = 1。对于 K = 2，K 的有效倍数为 2，因此总和为 arr[2] = 4，这是最大可能值。因此，K = 2 是一个有效答案。

输入: arr[] = {-1，-2，-3 } T3】输出: 2

方法:给定的问题可以用类似于厄拉多塞的筛的方法来解决。其思想是通过迭代 K 的每个倍数来计算【1，N】范围内 K 的所有可能值的总和，就像在筛子中标记非素元素时所做的那样。给出最大和的 K 的值就是需要的答案。

下面是上述方法的实现:

C++

// C++ Program of the above approach
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)

// 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.

java 描述语言

<script>
      // 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
  </script>

Output

2

时间复杂度: O(Nlog N)* 辅助空间: O(1)

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