게임 of Sum

제목 링크: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 10문제: n개의 돌을 정하고 돌마다 점수를 매긴다. 현재 파트너 A와 파트너 B가 게임을 하고 있다. 파트너 A가 먼저 한다. 한 사람이 매번 처음부터 끝까지 k개의 돌을 선택할 수 있다. 만약에 두 사람이 매번 자신의 가장 좋은 상황(점수를 최대한 많이)에 따라 찾으면 마지막에 두 사람의 점수 차이는 얼마인지를 요구한다.
바둑 dp: 내가 매번 다 뽑은 후에 그가 뽑은 점수를 최대한 작게 해야 한다

#include
#include
#include
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++){//   
		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
",2*dfs(1,n)-sum[n]); 
	}
	return 0;
}