数组的最大平均和分区

给定一个数组，我们将一行数字 A 划分为至多 K 个相邻(非空)组，那么分数是每个组的平均值之和。最高可以打多少分?
例子：

Input : A = { 9, 1, 2, 3, 9 } 
K = 3 
Output : 20 
Explanation : We can partition A into [9], [1, 2, 3], [9]. The answer is 9 + (1 + 2 + 3) / 3 + 9 = 20. 
We could have also partitioned A into [9, 1], [2], [3, 9]. That partition would lead to a score of 5 + 2 + 6 = 13, which is worse.
Input : A[] = { 1, 2, 3, 4, 5, 6, 7 } 
K = 4 
Output : 20.5 
Explanation : We can partition A into [1, 2, 3, 4], [5], [6], [7]. The answer is 2.5 + 5 + 6 + 7 = 20.5.

一个简单的解决方案是使用递归。一个有效的解决方案是记忆，我们将最大的分数保持在 k 以内，即 1、2、3……直到 k；
让 memo[i][k] 成为将 A[i..n-1] 分成最多 K 个部分的最佳分数。在第一组中，我们将 A[i..n-1] 划分为 A[i..j-1] 和 A[j..n-1]，那么我们的候选划分的分数为 average(i, j) + score(j, k-1))，其中 average(i, j) = (A[i] + A[i+1] + … + A[j-1]) / (j – i)。我们取其中的最高分。
总的来说，我们在一般情况下的递归是：
memo[n][k] = max(memo[n][k], score(memo, i, A, k-1) + average(i, j))
对于所有 i 从 n-1 到 1 。

C++

// CPP program for maximum average sum partition
#include 
using namespace std;
 
#define MAX 1000
 
double memo[MAX][MAX];
 
// bottom up approach to calculate score
double score(int n, vector& A, int k)
{
    if (memo[n][k] > 0)
        return memo[n][k];
    double sum = 0;
    for (int i = n - 1; i > 0; i--) {
        sum += A[i];
        memo[n][k] = max(memo[n][k], score(i, A, k - 1) +
                                          sum / (n - i));
    }
    return memo[n][k];
}
 
double largestSumOfAverages(vector& A, int K)
{
    int n = A.size();
    double sum = 0;
    memset(memo, 0.0, sizeof(memo));
    for (int i = 0; i < n; i++) {
        sum += A[i];
 
        // storing averages from starting to each i ;
        memo[i + 1][1] = sum / (i + 1);
    }
    return score(n, A, K);
}
 
int main()
{
    vector A = { 9, 1, 2, 3, 9 };
    int K = 3; // atmost partioning size
    cout << largestSumOfAverages(A, K) << endl;
    return 0;
}

Java

// Java program for maximum average sum partition
import java.util.Arrays;
import java.util.Vector;
 
class GFG
{
 
    static int MAX = 1000;
    static double[][] memo = new double[MAX][MAX];
 
    // bottom up approach to calculate score
    public static double score(int n, Vector A, int k)
    {
        if (memo[n][k] > 0)
            return memo[n][k];
 
        double sum = 0;
        for (int i = n - 1; i > 0; i--)
        {
            sum += A.elementAt(i);
 
            memo[n][k] = Math.max(memo[n][k],
                                  score(i, A, k - 1) +
                                  sum / (n - i));
        }
        return memo[n][k];
    }
 
    public static double largestSumOfAverages(Vector A, int K)
    {
        int n = A.size();
        double sum = 0;
 
        for (int i = 0; i < memo.length; i++)
        {
            for (int j = 0; j < memo[i].length; j++)
                memo[i][j] = 0.0;
        }
 
        for (int i = 0; i < n; i++)
        {
            sum += A.elementAt(i);
 
            // storing averages from starting to each i ;
            memo[i + 1][1] = sum / (i + 1);
        }
 
        return score(n, A, K);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        Vector A = new Vector<>(Arrays.asList(9, 1, 2, 3, 9));
        int K = 3;
        System.out.println(largestSumOfAverages(A, K));
 
    }
}
 
// This code is contributed by sanjeev2552

Python3

# Python3 program for maximum average sum partition
MAX = 1000
 
memo = [[0.0 for i in range(MAX)]
             for i in range(MAX)]
 
# bottom up approach to calculate score
def score(n, A, k):
    if (memo[n][k] > 0):
        return memo[n][k]
    sum = 0
    i = n - 1
    while(i > 0):
        sum += A[i]
        memo[n][k] = max(memo[n][k], score(i, A, k - 1) +
                                       int(sum / (n - i)))
 
        i -= 1
     
    return memo[n][k]
 
def largestSumOfAverages(A, K):
    n = len(A)
    sum = 0
    for i in range(n):
        sum += A[i]
 
        # storing averages from starting to each i ;
        memo[i + 1][1] = int(sum / (i + 1))
     
    return score(n, A, K)
 
# Driver Code
if __name__ == '__main__':
    A = [9, 1, 2, 3, 9]
    K = 3 # atmost partioning size
    print(largestSumOfAverages(A, K))
     
# This code is contributed by
# Surendra_Gangwar

C#

// C# program for maximum average sum partition
using System;
using System.Collections.Generic;
 
class GFG
{
    static int MAX = 1000;
    static double[,] memo = new double[MAX, MAX];
 
    // bottom up approach to calculate score
    public static double score(int n,
                               List A, int k)
    {
        if (memo[n, k] > 0)
            return memo[n, k];
 
        double sum = 0;
        for (int i = n - 1; i > 0; i--)
        {
            sum += A[i];
 
            memo[n, k] = Math.Max(memo[n, k],
                                  score(i, A, k - 1) +
                                  sum / (n - i));
        }
        return memo[n, k];
    }
 
    public static double largestSumOfAverages(List A,
                                                   int K)
    {
        int n = A.Count;
        double sum = 0;
 
        for (int i = 0;
                 i < memo.GetLength(0); i++)
        {
            for (int j = 0;
                     j < memo.GetLength(1); j++)
                memo[i, j] = 0.0;
        }
 
        for (int i = 0; i < n; i++)
        {
            sum += A[i];
 
            // storing averages from
            // starting to each i;
            memo[i + 1, 1] = sum / (i + 1);
        }
 
        return score(n, A, K);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int [] arr = {9, 1, 2, 3, 9};
        List A = new List(arr);
        int K = 3;
        Console.WriteLine(largestSumOfAverages(A, K));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// JavaScript program for maximum average sum partition
 
 
let MAX = 1000;
 
let memo = new Array(MAX).fill(0).map(() => new Array(MAX).fill(0));
 
// bottom up approach to calculate score
function score(n, A, k) {
    if (memo[n][k] > 0)
        return memo[n][k];
    let sum = 0;
    for (let i = n - 1; i > 0; i--) {
        sum += A[i];
        memo[n][k] = Math.max(memo[n][k], score(i, A, k - 1) +
            sum / (n - i));
    }
    return memo[n][k];
}
 
function largestSumOfAverages(A, K) {
    let n = A.length;
    let sum = 0;
 
    for (let i = 0; i < n; i++) {
        sum += A[i];
 
        // storing averages from starting to each i ;
        memo[i + 1][1] = sum / (i + 1);
    }
    return score(n, A, K);
}
 
 
let A = [9, 1, 2, 3, 9];
let K = 3; // atmost partioning size
document.write(largestSumOfAverages(A, K) + "<br>");
 

输出：

20

上述问题现在可以很容易地理解为动态规划。
让 dp(i, k) 是将 A[i:j] 分成最多 K 个部分的最佳分数。如果我们将 A[i:j] 划分成的第一组在 j 之前结束，那么我们的候选划分的分数为 average(i, j) + dp(j, k-1))。一般情况下的递归是 dp(i, k) = max(average(i, N), (average(i, j) + dp(j, k-1)))。我们可以预先计算前缀和以快速执行输出代码。

C++

// CPP program for maximum average sum partition
#include 
using namespace std;
 
double largestSumOfAverages(vector& A, int K)
{
    int n = A.size();
 
    // storing prefix sums
    double pre_sum[n+1];
    pre_sum[0] = 0;
    for (int i = 0; i < n; i++)
        pre_sum[i + 1] = pre_sum[i] + A[i];
 
    // for each i to n storing averages
    double dp[n] = {0};
    double sum = 0;
    for (int i = 0; i < n; i++)
        dp[i] = (pre_sum[n] - pre_sum[i]) / (n - i);
     
    for (int k = 0; k < K - 1; k++)
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++)
                dp[i] = max(dp[i], (pre_sum[j] -
                         pre_sum[i]) / (j - i) + dp[j]);
     
    return dp[0];
}
 
// Driver code
int main()
{
    vector A = { 9, 1, 2, 3, 9 };
    int K = 3; // atmost partioning size
    cout << largestSumOfAverages(A, K) << endl;
    return 0;
}

Java

// Java program for maximum average sum partition
import java.util.*;
 
class GFG
{
 
static double largestSumOfAverages(int[] A, int K)
{
    int n = A.length;
 
    // storing prefix sums
    double []pre_sum = new double[n + 1];
    pre_sum[0] = 0;
    for (int i = 0; i < n; i++)
        pre_sum[i + 1] = pre_sum[i] + A[i];
 
    // for each i to n storing averages
    double []dp = new double[n];
    double sum = 0;
    for (int i = 0; i < n; i++)
        dp[i] = (pre_sum[n] - pre_sum[i]) / (n - i);
     
    for (int k = 0; k < K - 1; k++)
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++)
                dp[i] = Math.max(dp[i], (pre_sum[j] -
                        pre_sum[i]) / (j - i) + dp[j]);
     
    return dp[0];
}
 
// Driver code
public static void main(String[] args)
{
    int []A = { 9, 1, 2, 3, 9 };
    int K = 3; // atmost partioning size
    System.out.println(largestSumOfAverages(A, K));
}
}
 
// This code is contributed by PrinciRaj1992

蟒蛇3

# Python3 program for maximum average
# sum partition
def largestSumOfAverages(A, K):
 
    n = len(A);
 
    # storing prefix sums
    pre_sum = [0] * (n + 1);
    pre_sum[0] = 0;
    for i in range(n):
        pre_sum[i + 1] = pre_sum[i] + A[i];
 
    # for each i to n storing averages
    dp = [0] * n;
    sum = 0;
    for i in range(n):
        dp[i] = (pre_sum[n] -
                 pre_sum[i]) / (n - i);
     
    for k in range(K - 1):
        for i in range(n):
            for j in range(i + 1, n):
                dp[i] = max(dp[i], (pre_sum[j] -
                                    pre_sum[i]) /
                                    (j - i) + dp[j]);
     
    return int(dp[0]);
 
# Driver code
A = [ 9, 1, 2, 3, 9 ];
K = 3; # atmost partioning size
print(largestSumOfAverages(A, K));
 
# This code is contributed by Rajput-Ji

C＃

// C# program for maximum average sum partition
using System;
using System.Collections.Generic;
     
class GFG
{
static double largestSumOfAverages(int[] A,
                                   int K)
{
    int n = A.Length;
 
    // storing prefix sums
    double []pre_sum = new double[n + 1];
    pre_sum[0] = 0;
    for (int i = 0; i < n; i++)
        pre_sum[i + 1] = pre_sum[i] + A[i];
 
    // for each i to n storing averages
    double []dp = new double[n];
    for (int i = 0; i < n; i++)
        dp[i] = (pre_sum[n] - pre_sum[i]) / (n - i);
     
    for (int k = 0; k < K - 1; k++)
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++)
                dp[i] = Math.Max(dp[i], (pre_sum[j] -
                        pre_sum[i]) / (j - i) + dp[j]);
     
    return dp[0];
}
 
// Driver code
public static void Main(String[] args)
{
    int []A = { 9, 1, 2, 3, 9 };
    int K = 3; // atmost partioning size
    Console.WriteLine(largestSumOfAverages(A, K));
}
}
 
// This code is contributed by PrinciRaj1992

输出：

20

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