给定一个由N 个整数和两个整数X和Y组成的数组A[] ，任务是找到通过将数组拆分为长度在[X, Y]范围内的子序列而获得的每个子序列的平均值的最大和。
注意： X和Y的值总是可以组成这样的组。
例子：
Input: A[] = {4, 10, 6, 5}, X = 2, Y = 3
Output: 12.50
Explanation:
Divide the given array into two groups as {4, 10}, {6, 5} such that their sizes lies over the range [2, 3].
The average of the first group = (4 + 10) / 2 = 7.
The average of the second group = (6 + 5) / 2 = 5.5.
Therefore, the sum of average = 7 + 5.5 = 12.5, which is minimum among all possible groups.
Input: A[] = {3, 3, 1}
Output: 3.00
方法：可以使用贪心方法解决给定的问题。这个想法是按升序对数组进行排序并选择大小为X的组，因为平均值与元素数量成反比。最后，将剩余元素添加到最后一组。请按照以下步骤解决问题：
- 按升序对给定的数组arr[]进行排序。
- 将变量sum、res和count初始化为0，以存储数组元素的总和、结果和当前组中元素的计数。
- 使用变量i在范围[0, N] 上迭代并执行以下步骤：
- 将arr[i]的值添加到变量sum并将count的值增加1 。
- 如果count的值等于X ，并且剩余的数组元素不能组成一个组，则将它们添加到当前组并在变量res 中添加平均值。
- 否则，将迄今为止形成的组的平均值添加到变量res 中。
- 完成以上步骤后，打印res的值作为答案。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the maximum sum
// of average of groups
void maxAverage(int A[], int N, int X,
int Y)
{
// Sort the given array
sort(A, A + N);
int sum = 0;
// Stores the sum of averages
double res = 0;
// Stores count of array element
int count = 0;
for (int i = 0; i < N; i++) {
// Add the current value to
// the variable sum
sum += A[i];
// Increment the count by 1
count++;
// If the current size is X
if (count == X) {
// If the remaining elements
// can't become a group
if (N - i - 1 < X) {
i++;
int cnt = 0;
// Iterate until i is
// less than N
while (i < N) {
cnt++;
sum += A[i];
i++;
}
// Update the value of X
X = X + cnt;
// Update the average
res += (double)sum / double(X);
break;
}
// Find the average
res += (double)sum / double(X);
// Reset the sum and count
sum = 0;
count = 0;
}
}
// Print maximum sum of averages
cout << fixed << setprecision(2)
<< res << "\n";
}
// Driver Code
int main()
{
int A[] = { 4, 10, 6, 5 };
int N = sizeof(A) / sizeof(A[0]);
int X = 2, Y = 3;
maxAverage(A, N, X, Y);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the maximum sum
// of average of groups
static void maxAverage(int A[], int N, int X,
int Y)
{
// Sort the given array
Arrays.sort(A);
int sum = 0;
// Stores the sum of averages
double res = 0;
// Stores count of array element
int count = 0;
for(int i = 0; i < N; i++)
{
// Add the current value to
// the variable sum
sum += A[i];
// Increment the count by 1
count++;
// If the current size is X
if (count == X)
{
// If the remaining elements
// can't become a group
if (N - i - 1 < X)
{
i++;
int cnt = 0;
// Iterate until i is
// less than N
while (i < N)
{
cnt++;
sum += A[i];
i++;
}
// Update the value of X
X = X + cnt;
// Update the average
res += (double)sum / (double)(X);
break;
}
// Find the average
res += (double)sum / (double)(X);
// Reset the sum and count
sum = 0;
count = 0;
}
}
// Print maximum sum of averages
System.out.println(res);
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 4, 10, 6, 5 };
int N = A.length;
int X = 2, Y = 3;
maxAverage(A, N, X, Y);
}
}
// This code is contributed by gauravrajput1Python3
# Python 3 program for the above approach
# Function to find the maximum sum
# of average of groups
def maxAverage(A,N,X,Y):
# Sort the given array
A.sort()
sum = 0
# Stores the sum of averages
res = 0
# Stores count of array element
count = 0
for i in range(N):
# Add the current value to
# the variable sum
sum += A[i]
# Increment the count by 1
count += 1
# If the current size is X
if (count == X):
# If the remaining elements
# can't become a group
if (N - i - 1 < X):
i += 1
cnt = 0
# Iterate until i is
# less than N
while (i < N):
cnt += 1
sum += A[i]
i += 1
# Update the value of X
X = X + cnt
# Update the average
res += sum / X
break
# Find the average
res += sum / X
# Reset the sum and count
sum = 0
count = 0
# Print maximum sum of averages
print(res)
# Driver Code
if __name__ == '__main__':
A = [4, 10, 6, 5]
N = len(A)
X = 2
Y = 3
maxAverage(A, N, X, Y)
# This code is contributed by ipg2016107.C#
// C# program for the above approach
using System;
public class GFG{
// Function to find the maximum sum
// of average of groups
static void maxAverage(int []A, int N, int X,
int Y)
{
// Sort the given array
Array.Sort(A);
int sum = 0;
// Stores the sum of averages
double res = 0;
// Stores count of array element
int count = 0;
for(int i = 0; i < N; i++)
{
// Add the current value to
// the variable sum
sum += A[i];
// Increment the count by 1
count++;
// If the current size is X
if (count == X)
{
// If the remaining elements
// can't become a group
if (N - i - 1 < X)
{
i++;
int cnt = 0;
// Iterate until i is
// less than N
while (i < N)
{
cnt++;
sum += A[i];
i++;
}
// Update the value of X
X = X + cnt;
// Update the average
res += (double)sum / (double)(X);
break;
}
// Find the average
res += (double)sum / (double)(X);
// Reset the sum and count
sum = 0;
count = 0;
}
}
// Print maximum sum of averages
Console.WriteLine(res);
}
// Driver Code
public static void Main(String[] args)
{
int []A = { 4, 10, 6, 5 };
int N = A.Length;
int X = 2, Y = 3;
maxAverage(A, N, X, Y);
}
}
// This code is contributed by 29AjayKumarJavascript
输出：
12.50时间复杂度： O(N * log N)
辅助空间： O(1)
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