<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:
    vector<vector<vector<double>>> dp;

    double f(vector<int>& nums, int i, double c, int ele, int k) {
        if (i < 0 && k == 0) {
            return c / ele;
        }
        if (i < 0 || k < 0) {
            return 0;
        }
        if (dp[i][ele][k] != -1) {
            return dp[i][ele][k];
        }

        double nt = f(nums, i - 1, c + nums[i], ele + 1, k);
        double t = 0;
        if (ele > 0 && k > 0) {
            t = (c / ele) + f(nums, i - 1, nums[i], 1, k - 1);
        }

        return dp[i][ele][k] = max(nt, t);
    }

    double largestSumOfAverages(vector<int>& nums, int k) {
        int n = nums.size();
        dp = vector<vector<vector<double>>>(n, vector<vector<double>>(n + 1, vector<double>(k, -1)));
        return f(nums, n - 1, 0, 0, k - 1);
    }
};


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