# # 【LeetCode】813. Largest Sum of Averages 解题报告（Python）

## # 题目描述：

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. 1 <= A.length <= 100.
2. 1 <= A[i] <= 10000.
3. 1 <= K <= A.length.
4. Answers within 10^-6 of the correct answer will be accepted as correct.

## # 解题方法

``````class Solution:
def largestSumOfAverages(self, A, K):
"""
:type A: List[int]
:type K: int
:rtype: float
"""
n = len(A)
m_ = [[0 for i in range(n + 1)] for j in range(K + 1)]
sum_ = [0] * (n + 1)
for i in range(1, n + 1):
sum_[i] = sum_[i - 1] + A[i - 1]
return self.LSA(A, sum_, m_, n, K)

# Largest sum of averages for first n elements in A partioned into K groups
def LSA(self, A, sum_, m_, n, k):
if m_[k][n] > 0: return m_[k][n]
if k == 1: return sum_[n] / n
for i in range(k - 1, n):
m_[k][n] = max(m_[k][n], self.LSA(A, sum_, m_, i, k - 1) + (sum_[n] - sum_[i]) / (n - i))
return m_[k][n]
``````