题目描述：

LeetCode 813. Largest Sum of Averages

We partition a row of numbers A into at most K adjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve?

Note that our partition must use every number in A, and that scores are not necessarily integers.

Example:
Input: 
A = [9,1,2,3,9]
K = 3
Output: 20
Explanation: 
The best choice is to 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], for example.
That partition would lead to a score of 5 + 2 + 6 = 13, which is worse.

Note:

• 1 <= A.length <= 100.
• 1 <= A[i] <= 10000.
• 1 <= K <= A.length.
• Answers within 10^-6 of the correct answer will be accepted as correct.

题目大意：

给定整数数组，至多将其分成K个连续的子数组，求各子数组平均值之和的最大值。

解题思路：

动态规划（Dynamic Programming）

状态转移方程：

dp[x][y] = max(dp[x][y], dp[x - 1][z] + avg(z..y))    0 <= x < K,  x <= y < N, x <= z < y

上式中，dp[x][y]表示将数组的前y个元素至多分成x个子数组的最优解

Python代码：

class Solution(object):
    def largestSumOfAverages(self, A, K):
        """
        :type A: List[int]
        :type K: int
        :rtype: float
        """
        N = len(A)
        S = [0] * (N + 1)
        for x, a in enumerate(A):
            S[x + 1] += S[x] + a

        dp = [1.0 * S[x] / x for x in range(1, N + 1)]
        for x in range(K - 1):
            dp0 = [0] * N
            for y in range(x, N):
                for z in range(x, y):
                    dp0[y] = max(dp0[y], dp[z] + 1.0 * (S[y + 1] - S[z + 1]) / (y - z))
            dp = dp0
        return dp[-1]

 

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