<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 dp[101][101];
public:
  int stoneGameII(vector<int>& piles) {
    // dp[i][j]: maximum number of stones the first player get
    // for the first ith piles and M = j + 1
    // dp[i][j] = min{dp[i - X][j]} + sum

    const int n = piles.size();
    for (int i = 0; i <= n; ++i) {
      for (int j = 0; j <= n; ++j) {
        dp[i][j] = INT_MIN;
      }
    }

    function<int(int, int)> dfs = [&](int x, int m) {
      if (x == n) {
        return 0;
      }

      if (dp[x][m] != INT_MIN) {
        return dp[x][m];
      }

      int& best = dp[x][m];
      int take = 0;

      // Try to take 1 ~ 2M stones
      for (int i = x; i < min(n, x + 2 * m); ++i) {
        take += piles[i];
        best = max(best, take - dfs(i + 1, min(n, max(m, i - x + 1))));
      }

      return best;
    };

    int total = accumulate(piles.begin(), piles.end(), 0);
    return (total + dfs(0, 1)) / 2;
  }
};

11361