NOI 9268 酒鬼

http://noi.openjudge.cn/ch0206/9268/

题意：

输入n个酒瓶，每个酒瓶中有不同体积的酒，问最多能喝到多少酒，前提是不能连续喝三瓶酒。

思路：

d[i]的意思是分析到第i个酒瓶时所能喝的最多酒。

状态转移方程为：

d[i] = max(d[i - 1], max(d[i - 2] + a[i], d[i - 3] + a[i] + a[i - 1]));

#include<iostream> 
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;

int n;
int a[705];
int d[705];

int main()
{
    //freopen("D:\txt.txt", "r", stdin);
    while (cin >> n && n)
    {
        for (int i = 1; i <= n; i++)
            cin >> a[i];
        d[0] = 0;
        for (int i = 1; i <= n; i++)
        {
            if (i <= 2)    d[i] = d[i - 1] + a[i];
            else
            {
                d[i] = max(d[i - 1], max(d[i - 2] + a[i], d[i - 3] + a[i] + a[i - 1]));
            }
        }
        cout << d[n] << endl;
    }
    return 0;
}