题目链接：https://cn.vjudge.net/problem/UVA-10891 

This is a two player game. Initially there are n integer numbers in an array and players A and B get chance to take them alternatively. Each player can take one or more numbers from the left or right end of the array but cannot take from both ends at a time. He can take as many consecutive numbers as he wants during his time. The game ends when all numbers are taken from the array by the players. The point of each player is calculated by the summation of the numbers, which he has taken. Each player tries to achieve more points from other. If both players play optimally and player A starts the game then how much more point can player A get than player B?

Input

The input consists of a number of cases. Each case starts with a line specifying the integer n (0 < n ≤100), the number of elements in the array. After that, n numbers are given for the game. Input is terminated by a line where n=0.

 

Output

For each test case, print a number, which represents the maximum difference that the first player obtained after playing this game optimally.


Sample Input                               

4
4 -10 -20 7
4
1 2 3 4
0

 Output for Sample Input
7
10
题意:给定n个石头，每个石头有一个分数，现在小伙伴A和小伙伴B进行一个游戏，小伙伴A先手，每个人每次可以选择从头或从尾取k个石头，要求出如果两个人每次都按自己最好的情况（分数尽量多）去取，最后两个人分数差是多少。

博弈dp：我每次取完之后，要让他取的分数尽量小

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[105],dp[105][105],sum[105];
bool vis[105][105];
int dfs(int l,int r){
	if(vis[l][r]) return dp[l][r];
	if(l>r) return 0;
	vis[l][r]=true;
	int ans=-999999999;
	for(int i=1;i<=r-l+1;i++){//取几张
		ans=max(ans,sum[r]-sum[l-1]-min(dfs(l+i,r),dfs(l,r-i)));
	}
	return dp[l][r]=ans;
}
int main(){
	int n;
	while(~scanf("%d",&n)&&n){
		memset(vis,0,sizeof(vis));
		for(int i=1;i<=n;i++){
			scanf("%d",&a[i]);
			sum[i]=sum[i-1]+a[i];
		}
		printf("%d\n",2*dfs(1,n)-sum[n]); 
	}
	return 0;
}