<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>> avg;
    vector<vector<double>> dp;
    double solve(vector<int>& nums, int i, int k){
        if(i>=nums.size()){
            return 0;
        }
        if(k<=0){
            return -1e9;
        }
        if(dp[i][k]!=0){
            return dp[i][k];
        }
        double ret = 0;
        for(int s=0; i+s<nums.size(); s++){
            ret = max(ret, avg[i][i+s] + solve(nums, i+s+1, k-1));
        }
        return dp[i][k]=ret;
    }
    double largestSumOfAverages(vector<int>& nums, int k) {
        int n = nums.size();
        avg = vector(n, vector(n, 0.0));
        dp = vector(n, vector(k+1, 0.0));
        for(int i=0; i<n; i++){
            int s = 0;
            for(int j=i; j<n; j++){
                s += nums[j];
                avg[i][j] = (1.0) * s / (j-i+1);
            }
        }

        return solve(nums, 0, k);
    }
};
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