<p>You are given an integer array <code>nums</code> and an integer <code>k</code>. You can partition the array into <strong>at most</strong> <code>k</code> non-empty adjacent subarrays. The <strong>score</strong> of a partition is the sum of the averages of each subarray.</p>

<p>Note that the partition must use every integer in <code>nums</code>, and that the score is not necessarily an integer.</p>

<p>Return <em>the maximum <strong>score</strong> you can achieve of all the possible partitions</em>. Answers within <code>10<sup>-6</sup></code> of the actual answer will be accepted.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [9,1,2,3,9], k = 3
<strong>Output:</strong> 20.00000
<strong>Explanation:</strong> 
The best choice is to partition nums into [9], [1, 2, 3], [9]. The answer is 9 + (1 + 2 + 3) / 3 + 9 = 20.
We could have also partitioned nums into [9, 1], [2], [3, 9], for example.
That partition would lead to a score of 5 + 2 + 6 = 13, which is worse.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 4
<strong>Output:</strong> 20.50000
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>

class Solution {
public:
    double largestSumOfAverages(vector<int>& nums, int k) {
        vector<int> sum;
        sum.resize(nums.size() + 1, 0);
        for(int i = 0; i < nums.size(); ++i){
            sum[i + 1] = sum[i] + nums[i];
        }

        vector<double> dp;
        dp.resize(nums.size(), 0.0);
        for(int i = 0; i < nums.size(); ++i){
            dp[i] = double(sum[nums.size()] - sum[i]) / (nums.size() - i);
        }

        for(int i = 0; i < k - 1; ++i){
            for(int l = 0; l < nums.size() - 1; ++l){
                for(int r = l + 1; r < nums.size(); ++r){
                    dp[l] = std::max(dp[l], dp[r] + double(sum[r] - sum[l]) / (r - l));
                }
            }
        }

        return dp[0];
    }
};
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