在采用长度为 K 的非重叠子数组后最大化数组总和

给定一个长度为N的整数数组arr[]和一个整数K ，任务是选择一些不重叠的子数组，使得每个子数组的长度正好为K ，没有两个子数组相邻并且所有子数组的总和所选子数组的元素最大。
例子：

Input: arr[] = {1, 2, 3, 4, 5}, K = 2 
Output: 12 
Sub-arrays that maximizes sum will be {{1, 2}, {4, 5}}. 
Thus, the answer will be 12.
Input: arr[] = {1, 1, 1, 1, 1}, K = 1 
Output: 3 
 

方法：这个问题可以用动态规划解决。假设我们处于索引i 。令dp[i]定义为满足上述条件的子数组arr[i…n-1]的所有可能子集的元素的最大和。
我们将有两种可能的选择，即选择子数组arr[i…i+k-1]并求解dp[i + k + 1]或拒绝它并求解dp[i + 1] 。
因此，递推关系将是

dp[i] = max(dp[i + 1], arr[i] + arr[i + 1] + arr[i + 2] + … + arr[i + k – 1] + dp[i + k + 1]) 
 

由于K的值可能很大，我们将使用前缀和数组在 O(1) 中找到子数组arr[i…i + k – 1]的所有元素的总和。
总的来说，算法的时间复杂度为O(N) 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
#define maxLen 10
using namespace std;
 
// To store the states of dp
int dp[maxLen];
 
// To check if a given state
// has been solved
bool v[maxLen];
 
// To store the prefix-sum
int prefix_sum[maxLen];
 
// Function to fill the prefix_sum[] with
// the prefix sum of the given array
void findPrefixSum(int arr[], int n)
{
    prefix_sum[0] = arr[0];
    for (int i = 1; i < n; i++)
        prefix_sum[i] = arr[i] + prefix_sum[i - 1];
}
 
// Function to find the maximum sum subsequence
// such that no two elements are adjacent
int maxSum(int arr[], int i, int n, int k)
{
    // Base case
    if (i + k > n)
        return 0;
 
    // To check if a state has
    // been solved
    if (v[i])
        return dp[i];
    v[i] = 1;
 
    int x;
 
    if (i == 0)
        x = prefix_sum[k - 1];
    else
        x = prefix_sum[i + k - 1] - prefix_sum[i - 1];
 
    // Required recurrence relation
    dp[i] = max(maxSum(arr, i + 1, n, k),
                x + maxSum(arr, i + k + 1, n, k));
 
    // Returning the value
    return dp[i];
}
 
// Driver code
int main()
{
    int arr[] = { 1, 3, 7, 6 };
    int n = sizeof(arr) / sizeof(int);
    int k = 1;
 
    // Finding prefix-sum
    findPrefixSum(arr, n);
 
    // Finding the maximum possible sum
    cout << maxSum(arr, 0, n, k);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
    static int maxLen = 10;
 
    // To store the states of dp
    static int[] dp = new int[maxLen];
 
    // To check if a given state
    // has been solved
    static boolean[] v = new boolean[maxLen];
 
    // To store the prefix-sum
    static int[] prefix_sum = new int[maxLen];
 
    // Function to fill the prefix_sum[] with
    // the prefix sum of the given array
    static void findPrefixSum(int arr[], int n)
    {
        prefix_sum[0] = arr[0];
        for (int i = 1; i < n; i++)
        {
            prefix_sum[i] = arr[i] + prefix_sum[i - 1];
        }
    }
 
    // Function to find the maximum sum subsequence
    // such that no two elements are adjacent
    static int maxSum(int arr[], int i, int n, int k)
    {
        // Base case
        if (i + k > n)
        {
            return 0;
        }
 
        // To check if a state has
        // been solved
        if (v[i])
        {
            return dp[i];
        }
        v[i] = true;
 
        int x;
 
        if (i == 0)
        {
            x = prefix_sum[k - 1];
        }
        else
        {
            x = prefix_sum[i + k - 1] - prefix_sum[i - 1];
        }
 
        // Required recurrence relation
        dp[i] = Math.max(maxSum(arr, i + 1, n, k),
                x + maxSum(arr, i + k + 1, n, k));
 
        // Returning the value
        return dp[i];
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {1, 3, 7, 6};
        int n = arr.length;
        int k = 1;
 
        // Finding prefix-sum
        findPrefixSum(arr, n);
 
        // Finding the maximum possible sum
        System.out.println(maxSum(arr, 0, n, k));
    }
}
 
// This code contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
 
maxLen = 10
 
# To store the states of dp
dp = [0]*maxLen;
 
# To check if a given state
# has been solved
v = [0]*maxLen;
 
# To store the prefix-sum
prefix_sum = [0]*maxLen;
 
# Function to fill the prefix_sum[] with
# the prefix sum of the given array
def findPrefixSum(arr, n) :
 
    prefix_sum[0] = arr[0];
    for i in range(n) :
        prefix_sum[i] = arr[i] + prefix_sum[i - 1];
 
 
# Function to find the maximum sum subsequence
# such that no two elements are adjacent
def maxSum(arr, i, n, k) :
 
    # Base case
    if (i + k > n) :
        return 0;
 
    # To check if a state has
    # been solved
    if (v[i]) :
        return dp[i];
         
    v[i] = 1;
 
    if (i == 0) :
        x = prefix_sum[k - 1];
    else :
        x = prefix_sum[i + k - 1] - prefix_sum[i - 1];
 
    # Required recurrence relation
    dp[i] = max(maxSum(arr, i + 1, n, k),
                x + maxSum(arr, i + k + 1, n, k));
 
    # Returning the value
    return dp[i];
 
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 1, 3, 7, 6 ];
     
    n = len(arr);
    k = 1;
 
    # Finding prefix-sum
    findPrefixSum(arr, n);
 
    # Finding the maximum possible sum
    print(maxSum(arr, 0, n, k));
     
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
    static int maxLen = 10;
 
    // To store the states of dp
    static int[] dp = new int[maxLen];
 
    // To check if a given state
    // has been solved
    static bool[] v = new bool[maxLen];
 
    // To store the prefix-sum
    static int[] prefix_sum = new int[maxLen];
 
    // Function to fill the prefix_sum[] with
    // the prefix sum of the given array
    static void findPrefixSum(int []arr, int n)
    {
        prefix_sum[0] = arr[0];
        for (int i = 1; i < n; i++)
        {
            prefix_sum[i] = arr[i] + prefix_sum[i - 1];
        }
    }
 
    // Function to find the maximum sum subsequence
    // such that no two elements are adjacent
    static int maxSum(int []arr, int i, int n, int k)
    {
        // Base case
        if (i + k > n)
        {
            return 0;
        }
 
        // To check if a state has
        // been solved
        if (v[i])
        {
            return dp[i];
        }
        v[i] = true;
 
        int x;
 
        if (i == 0)
        {
            x = prefix_sum[k - 1];
        }
        else
        {
            x = prefix_sum[i + k - 1] - prefix_sum[i - 1];
        }
 
        // Required recurrence relation
        dp[i] = Math.Max(maxSum(arr, i + 1, n, k),
                x + maxSum(arr, i + k + 1, n, k));
 
        // Returning the value
        return dp[i];
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {1, 3, 7, 6};
        int n = arr.Length;
        int k = 1;
 
        // Finding prefix-sum
        findPrefixSum(arr, n);
 
        // Finding the maximum possible sum
        Console.Write(maxSum(arr, 0, n, k));
    }
}
 
// This code is contributed by Princi Singh

Javascript

 
// Javascript implementation of the approach
var maxLen = 10
 
// To store the states of dp
var dp = Array(maxLen);
 
// To check if a given state
// has been solved
var v = Array(maxLen);
 
// To store the prefix-sum
var prefix_sum = Array(maxLen);;
 
// Function to fill the prefix_sum[] with
// the prefix sum of the given array
function findPrefixSum(arr, n)
{
    prefix_sum[0] = arr[0];
    for (var i = 1; i < n; i++)
        prefix_sum[i] = arr[i] + prefix_sum[i - 1];
}
 
// Function to find the maximum sum subsequence
// such that no two elements are adjacent
function maxSum(arr, i, n, k)
{
    // Base case
    if (i + k > n)
        return 0;
 
    // To check if a state has
    // been solved
    if (v[i])
        return dp[i];
    v[i] = 1;
 
    var x;
 
    if (i == 0)
        x = prefix_sum[k - 1];
    else
        x = prefix_sum[i + k - 1] - prefix_sum[i - 1];
 
    // Required recurrence relation
    dp[i] = Math.max(maxSum(arr, i + 1, n, k),
                x + maxSum(arr, i + k + 1, n, k));
 
    // Returning the value
    return dp[i];
}
 
// Driver code
var arr = [1, 3, 7, 6];
var n = arr.length;
var k = 1;
// Finding prefix-sum
findPrefixSum(arr, n);
// Finding the maximum possible sum
document.write( maxSum(arr, 0, n, k));
 

输出：

9

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