<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 f(int start, int end, vector<int>& nums, int k, vector<vector<vector<double>>>& dp) {
        if (dp[start][end][k] != -1) {
            return dp[start][end][k];
        }
        
        if (k == 1) {
            double x = 0;
            for (int i = start; i <= end; i++) {
                x += nums[i];
            }
            x /= (end - start + 1);
            return dp[start][end][k] = x;
        }

        if (start > end)
            return 0;

        if (start == end) 
            return dp[start][end][k] = nums[start];

        double sum = 0;
        double ans = 0;
        for (int i = start; i <= end; i++) {
            sum += nums[i];
            double temp = sum / (i - start + 1);
            if (i + 1 <= end) {
                ans = max(ans, temp + f(i + 1, end, nums, k - 1, dp));
            }
        }

        return dp[start][end][k] = ans;
    }

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

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