给定一个数组以及两个数字M和K。我们需要找到数组中大小为K(不重叠)的最大M个子数组的总和。 (数组的顺序保持不变)。 K是子数组的大小，M是子数组的计数。可以假设数组的大小大于m * k。如果总数组大小不是k的倍数，那么我们可以取部分最后的数组。
例子 ：
Input: N = 7, M = 3, K = 1
arr[] = {2, 10, 7, 18, 5, 33, 0};
Output: 61
Explanation: subsets are: 33, 18, 10 (3 subsets of size 1)
Input: N = 4, M = 2, K = 2
arr[] = {3, 2, 100, 1};
Output: 106
Explanation: subsets are: (3, 2), (100, 1) 2 subsets of size 2
在这里我们可以看到我们需要找到每个大小为K的M个子数组，因此，
1.我们创建一个假定数组，该数组在每个索引中包含给定数组中从’ index ‘到‘index + K’的所有元素的和。和数组的大小将为n + 1-k 。
2.现在，如果我们包含大小为k的子数组，那么我们将无法再将该子数组的任何元素包含在任何其他子数组中，因为它将创建重叠的子数组。因此，我们通过排除包含的子数组的k个元素进行递归调用。
3.如果我们排除一个子数组，那么我们可以在其他子数组中使用该子数组的下一个k-1元素，因此我们将通过排除该子数组的第一个元素来进行递归调用。
4.最后返回最大值(包括和，不包括和)。
C++
// C++ program to find Max sum of M non-overlapping
// subarray of size K in an Array
#include
using namespace std;
// calculating presum of array. presum[i]
// is going to store prefix sum of subarray
// of size k beginning with arr[i]
void calculatePresumArray(int presum[],
int arr[], int n, int k)
{
for (int i=0; i size - 1)
return 0;
int mx = 0;
// if including subarray of size k
int includeMax = presum[start] +
maxSumMnonOverlappingSubarray(presum,
m - 1, size, k, start + k);
// if excluding element and searching
// in all next possible subarrays
int excludeMax =
maxSumMnonOverlappingSubarray(presum,
m, size, k, start + 1);
// return max
return max(includeMax, excludeMax);
}
// Driver code
int main()
{
int arr[] = { 2, 10, 7, 18, 5, 33, 0 };
int n = sizeof(arr)/sizeof(arr[0]);
int m = 3, k = 1;
int presum[n + 1 - k] = { 0 };
calculatePresumArray(presum, arr, n, k);
// resulting presum array will have a size = n+1-k
cout << maxSumMnonOverlappingSubarray(presum,
m, n + 1 - k, k, 0);
return 0;
} Java
// Java program to find Max sum
// of M non-overlapping subarray
// of size K in an Array
import java.io.*;
class GFG
{
// calculating presum of array.
// presum[i] is going to store
// prefix sum of subarray of
// size k beginning with arr[i]
static void calculatePresumArray(int presum[],
int arr[],
int n, int k)
{
for (int i = 0; i < k; i++)
presum[0] += arr[i];
// store sum of array index i to i+k
// in presum array at index i of it.
for (int i = 1; i <= n - k; i++)
presum[i] += presum[i - 1] + arr[i + k - 1] -
arr[i - 1];
}
// calculating maximum sum of
// m non overlapping array
static int maxSumMnonOverlappingSubarray(int presum[],
int m, int size,
int k, int start)
{
// if m is zero then no need
// any array of any size so
// return 0.
if (m == 0)
return 0;
// if start is greater then the
// size of presum array return 0.
if (start > size - 1)
return 0;
int mx = 0;
// if including subarray of size k
int includeMax = presum[start] +
maxSumMnonOverlappingSubarray(presum,
m - 1, size, k, start + k);
// if excluding element and searching
// in all next possible subarrays
int excludeMax =
maxSumMnonOverlappingSubarray(presum,
m, size, k, start + 1);
// return max
return Math.max(includeMax, excludeMax);
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 2, 10, 7, 18, 5, 33, 0 };
int n = arr.length;
int m = 3, k = 1;
int presum[] = new int[n + 1 - k] ;
calculatePresumArray(presum, arr, n, k);
// resulting presum array
// will have a size = n+1-k
System.out.println(maxSumMnonOverlappingSubarray(presum,
m, n + 1 - k, k, 0));
}
}
// This code is contributed by anuj_67.Python
# Python3 program to find Max sum of M non-overlapping
# subarray of size K in an Array
# calculating presum of array. presum[i]
# is going to store prefix sum of subarray
# of size k beginning with arr[i]
def calculatePresumArray(presum,arr, n, k):
for i in range(k):
presum[0] += arr[i]
# store sum of array index i to i+k
# in presum array at index i of it.
for i in range(1,n - k + 1):
presum[i] += presum[i - 1] + arr[i + k - 1] - arr[i - 1]
# calculating maximum sum of m non overlapping array
def maxSumMnonOverlappingSubarray(presum, m, size, k, start):
# if m is zero then no need any array
# of any size so return 0.
if (m == 0):
return 0
# if start is greater then the size
# of presum array return 0.
if (start > size - 1):
return 0
mx = 0
# if including subarray of size k
includeMax = presum[start] + maxSumMnonOverlappingSubarray(presum,m - 1, size, k, start + k)
# if excluding element and searching
# in all next possible subarrays
excludeMax = maxSumMnonOverlappingSubarray(presum,m, size, k, start + 1)
# return max
return max(includeMax, excludeMax)
# Driver code
arr = [2, 10, 7, 18, 5, 33, 0]
n = len(arr)
m, k = 3, 1
presum = [0 for i in range(n + 1 - k)]
calculatePresumArray(presum, arr, n, k)
# resulting presum array will have a size = n+1-k
print(maxSumMnonOverlappingSubarray(presum, m, n + 1 - k, k, 0))
# This code is contributed by mohit kumarC#
// C# program to find Max sum of M
// non-overlapping subarray of size
// K in an Array
using System;
class GFG {
// calculating presum of array.
// presum[i] is going to store
// prefix sum of subarray of
// size k beginning with arr[i]
static void calculatePresumArray(int []presum,
int []arr, int n, int k)
{
for (int i = 0; i < k; i++)
presum[0] += arr[i];
// store sum of array index i to i+k
// in presum array at index i of it.
for (int i = 1; i <= n - k; i++)
presum[i] += presum[i - 1] + arr[i + k - 1]
- arr[i - 1];
}
// calculating maximum sum of
// m non overlapping array
static int maxSumMnonOverlappingSubarray(
int []presum, int m, int size,
int k, int start)
{
// if m is zero then no need
// any array of any size so
// return 0.
if (m == 0)
return 0;
// if start is greater then the
// size of presum array return 0.
if (start > size - 1)
return 0;
//int mx = 0;
// if including subarray of size k
int includeMax = presum[start] +
maxSumMnonOverlappingSubarray(presum,
m - 1, size, k, start + k);
// if excluding element and searching
// in all next possible subarrays
int excludeMax =
maxSumMnonOverlappingSubarray(presum,
m, size, k, start + 1);
// return max
return Math.Max(includeMax, excludeMax);
}
// Driver code
public static void Main ()
{
int []arr = { 2, 10, 7, 18, 5, 33, 0 };
int n = arr.Length;
int m = 3, k = 1;
int []presum = new int[n + 1 - k] ;
calculatePresumArray(presum, arr, n, k);
// resulting presum array
// will have a size = n+1-k
Console.WriteLine(
maxSumMnonOverlappingSubarray(presum,
m, n + 1 - k, k, 0));
}
}
// This code is contributed by anuj_67.输出 ：
61