<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 {
public:
    int stoneGameII(vector<int>& piles) {
        if(piles.size() == 1) return piles[0];
        vector<int> p_sum;
        int p_size = piles.size();
        int acc = 0;
        p_sum.push_back(0);
        for(auto &p : piles) {
            acc += p;
            p_sum.push_back(acc);
        }

        vector<vector<vector<int>>> dp(2, vector<vector<int>>(p_size*2, vector<int>(p_size*2, -1)));

        return recursive_take(0, 0, 1, p_sum, dp);
        
    }

    int recursive_take(int p, int took, int M, vector<int>& p_sum, vector<vector<vector<int>>>& dp) {
        if(took >= p_sum.size()) return 0;
        // cout << p << " " << took << " " << M << " s: " << p_sum.size() <<endl;
        if(dp[p%2][took][M] != -1) return dp[p%2][took][M];
        if(p % 2 == 0) {
            int max_p = 0;
            for(int i = 1 ; took+i < p_sum.size() && i <= M*2 ; i++) {
                // cout << took+i << endl;
                int acc_p = p_sum[took+i] - p_sum[took];
                max_p = max(max_p, acc_p+recursive_take(p+1, took+i, max(M, i), p_sum, dp));
            }
            dp[0][took][M] = max_p;
            return max_p;
        } else {
            int min_p = INT_MAX;
            for(int i = 1 ; took+i < p_sum.size() && i <= M*2 ; i++) {
                // cout << took+i << endl;
                min_p = min(min_p, recursive_take(p+1, took+i, max(M, i), p_sum, dp));
            }
            if(min_p == INT_MAX) return 0;
            dp[1][took][M] = min_p;
            return min_p;
        }
    }
};
28438