<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 n;
    int dp[101][101][101];
    vector<int> suffix;
    int answer(int pos, vector<int> &piles, int m,bool chance){
        if(pos>=n) return 0;
        if(dp[pos][m][chance]!=-1) return dp[pos][m][chance];
        int ans=0;
        if(chance){
            for(int i=pos,count=1;i<n && count<=2*m;++i,++count){
                int t=min(max(count,m),n);
                ans=max(ans,suffix[i+1]-suffix[pos]+answer(i+1,piles,t,!chance));
            }

        }else{
            ans=INT_MAX;
            for(int i=pos,count=1;i<n && count<=2*m;++i,++count){
                int t=min(max(count,m),n);
                ans=min(ans,answer(i+1,piles,t,!chance));
            }
        }
        return dp[pos][m][chance]=ans;

    }
    int stoneGameII(vector<int>& piles) {
        n=piles.size();
        memset(dp,-1,sizeof(dp));
        suffix=vector<int> (n+1);
        for(int i=1;i<=n;i++) suffix[i]=piles[i-1]+suffix[i-1];
        return answer(0,piles,1,true);
    }
};
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