<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;
double solve(vector<int>&sum,int s,int k){
    int n= sum.size();
    double ans=0.0;
    if(s==n-1){
        return 0;
    }
    if(k==0){
        double a= (double)(sum[n-1]-sum[s])/(double)(n-1-s);
        return a;
    }
    if(dp[s][k]!=-1.0){
        return dp[s][k];
    }
    for(int i=s+1;i<n;i++){
        double a= (double)(sum[i]-sum[s])/(double)(i-s)+ solve(sum,i,k-1);
        // cout<<a<<" ";
        ans= max(a,ans);
    }
    cout<<s<<" "<<k<<" "<<ans<<" ";
    return dp[s][k]= ans;
}
    double largestSumOfAverages(vector<int>& nums, int k) {
        int n= nums.size();
        vector<int>sum(n+1,0);
        int s=0;
        sum[0]=0;
        for(int i=0;i<n;i++){
            s+=nums[i];
            sum[i+1]= s;
        }
        dp.resize(102, vector<double>(102, -1));
        return solve(sum,0,k-1);
    }
};
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