<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) {
        int n = nums.size();
        vector<vector<double>> dp(n + 1, vector<double>(K + 1, 0)); //dp[i][k] is largest sum of nums [0:i) divide to k subarray
        vector<vector<double>> avg(n + 1, vector<double>(n + 1, 0));
        nums.insert(nums.begin(), 0);
        dp[0][0] = 0;
        for (int i = 1; i <= n; i++)
            dp[i][0] = INT_MIN;
 
        for (int i = 1; i <= n; i++)
            nums[i] += nums[i-1];
            
        for (int i = 1; i <= n; i++) {
            for (int j = i; j <= n; j++) {
                avg[i][j] = 1.0 * (nums[j] - nums[i-1]) / (j - i + 1);
            }
        }

        for (int i = 1; i <= n; i++) {
            for (int k = 1; k <= min(i, K); k++) {
                //double sum = 0;
                for (int j = k; j <= i; j++) {
                    //sum += nums[j];
                    dp[i][k] = max(dp[i][k], dp[j-1][k-1] + avg[j][i]);
                }
            }
        }

        return dp[n][K];
    }
};


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