<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<double>> dp;
  vector<int> nums;
  int k;

  const double inf = 1e19;

  double solve(int idx, int partition) {
    if (partition >= k && idx != nums.size())
      return -inf;

    if (idx >= nums.size()) {
      return 0;
    }

    auto& ret = dp[idx][partition];
    if (ret != -1)
      return ret;

    double sum = 0; int c = 0;
    for (int i = idx; i < nums.size(); ++i) {
      sum += nums[i]; c++;
      ret = max(ret, sum / c + solve(i + 1, partition + 1));
    }
    return ret;
  }


  double largestSumOfAverages(vector<int>& nums_, int k_) {
    nums = nums_, k = k_;
    dp.resize(nums.size(), vector<double>(k, -1));
    return solve(0, 0);
  }
};
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