UVA

题意：AB两人分别拿一列n个数字，只能从左端或右端拿，不能同时从两端拿，可拿一个或多个，问在两人尽可能多拿的情况下，A最多比B多拿多少。

分析：

1、枚举先手拿的分界线，要么从左端拿，要么从右端拿，比较得最优解。

2、dp(i, j)---在区间(i, j)中A最多比B多拿多少。

3、tmp -= dfs(i + 1, r);//A拿了区间(l, i)，B在剩下区间里尽可能拿最优

tmp是A拿的，dfs(i + 1, r)是B比A多拿的，假设dfs(i + 1, r)=y-x，y是B拿的，x是A拿的

则tmp-dfs(i + 1, r) = tmp - y + x,也就是最终A比B多拿的。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 100 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int sum[MAXN];
int dp[MAXN][MAXN];
int dfs(int l, int r){
    if(dp[l][r] != INT_INF) return dp[l][r];
    int diff = sum[r] - sum[l - 1];
    for(int i = l; i < r; ++i){//先手从左端拿
        int tmp = sum[i] - sum[l - 1];
        tmp -= dfs(i + 1, r);//后手从右端拿
        if(tmp > diff) diff = tmp;
    }
    for(int i = l; i < r; ++i){
        int tmp = sum[r] - sum[i];
        tmp -= dfs(l, i);
        if(tmp > diff) diff = tmp;
    }
    return dp[l][r] = diff;
}
int main(){
    int n;
    while(scanf("%d", &n) == 1){
        if(!n) return 0;
        memset(dp, INT_INF, sizeof dp);
        sum[0] = 0;
        for(int i = 1; i <= n; ++i){
            scanf("%d", &sum[i]);
            sum[i] += sum[i - 1];
        }
        printf("%d
", dfs(1, n));
    }
    return 0;
}