<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 dp[105][105];
int f(int i,int m,vector<int>&piles,vector<int>&pre){
    // cout<<i<<" "<<m<<" "<<turn<<endl;
    if(i>=piles.size())return 0;
    if(dp[i][m]!=-1)return dp[i][m];
   int ans=INT_MIN;
        // int sum=piles[i];
        int sum=0;
        for(int k=i+1;k<=min(2*m+i,(int)piles.size());k++){
            sum+=piles[k-1];
            ans=max(ans,sum+pre.back()-pre[k-1]-f(k,max(m,k-i),piles,pre));
       
        }
        // cout<<i<<" "<<m<<" "<<turn <<" "<<ans<<endl;
        return dp[i][m]=ans;
}
    int stoneGameII(vector<int>& piles) {
        // memset(dp,-1,sizeof(dp));
        int n=piles.size();
        vector<int>pre(piles.size());
        pre[0]=piles[0];
        for(int i=1;i<pre.size();i++){
            pre[i]+=pre[i-1]+piles[i];
        }
        // return f(0,1,piles,pre);

vector<vector<int>>dp(n+1,vector<int>(105));
for(int i=n-1;i>=0;i--){
    for(int m=1;m<105;m++){
        int ans=INT_MIN;
        int sum=0;
        for(int k=i+1;k<=min(i+2*m,n);k++){
            sum+=piles[k-1];
            ans=max(ans,sum-dp[k][max(m,k-i)]+pre.back()-pre[k-1]);
        }
        dp[i][m]=ans;
    }
}
return dp[0][1];


        
    }
};
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