<p>Alice and Bob continue their games with piles of stones. There are a number of piles <strong>arranged in a row</strong>, and each pile has a positive integer number of stones <code>piles[i]</code>. The objective of the game is to end with the most stones.</p>

<p>Alice and Bob take turns, with Alice starting first.</p>

<p>On each player&#39;s turn, that player can take <strong>all the stones</strong> in the <strong>first</strong> <code>X</code> remaining piles, where <code>1 &lt;= X &lt;= 2M</code>. Then, we set <code>M = max(M, X)</code>. Initially, M = 1.</p>

<p>The game continues until all the stones have been taken.</p>

<p>Assuming Alice and Bob play optimally, return the maximum number of stones Alice can get.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">piles = [2,7,9,4,4]</span></p>

<p><strong>Output:</strong> <span class="example-io">10</span></p>

<p><strong>Explanation:</strong></p>

<ul>
	<li>If Alice takes one pile at the beginning, Bob takes two piles, then Alice takes 2 piles again. Alice can get <code>2 + 4 + 4 = 10</code> stones in total.</li>
	<li>If Alice takes two piles at the beginning, then Bob can take all three piles left. In this case, Alice get <code>2 + 7 = 9</code> stones in total.</li>
</ul>

<p>So we return 10 since it&#39;s larger.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">piles = [1,2,3,4,5,100]</span></p>

<p><strong>Output:</strong> <span class="example-io">104</span></p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= piles.length &lt;= 100</code></li>
	<li><code>1 &lt;= piles[i]&nbsp;&lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
private:
    int pile_size = 0;
    vector<vector<int>> dp;
    vector<int> cum_sum;
    int recfun(int m, int curr_idx){
        if(curr_idx >= pile_size) return 0;
        if(dp[m][curr_idx] != -1) return dp[m][curr_idx];
        int ret = 0;
        for(int x = 1; x <= 2*m; ++x){
            if(curr_idx+x > pile_size) continue;
            int prev = 0;
            if(curr_idx != 0) prev = cum_sum[curr_idx-1];
            int curr = cum_sum[curr_idx+x-1] - prev;
            // cout << curr_idx << " " << x << " " << curr << endl;
            int future = cum_sum[pile_size-1] - cum_sum[curr_idx+x-1] - recfun(max(m, x), curr_idx+x);
            ret = max(ret, curr+future);
        }
        return dp[m][curr_idx] = ret;
    }
public:
    int stoneGameII(vector<int>& piles) {
        pile_size = piles.size();
        dp.resize(64, vector<int>(pile_size, -1));
        cum_sum.resize(pile_size, 0);
        cum_sum[0] = piles[0];
        for(int i = 1; i < piles.size(); ++i){
            cum_sum[i] = cum_sum[i-1]+piles[i];
        }
        int ret = recfun(1, 0);
        return ret;
    }
};
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