PKU_campus_2018_A Wife

思路：

题目链接http://poj.openjudge.cn/practice/C18A/

先说一个结论，每一天要么7要么0，由此提供一种状态压缩dp的解法。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 const int MAXN = , INF = 0x3f3f3f3f;
 int a[MAXN], dp[][ << ];
 int main()
 {
     int t, n;
     cin >> t;
     while (t--)
     {
         cin >> n;
         int msk = ( << ) - ;
         memset(dp, 0x3f, sizeof dp);
         for (int i = ; i <= n; i++) cin >> a[i];
         for (int i = ; i <  << ; i++) dp[][i] = ;
         for (int i = ; i < n; i++)
         {
             memset(dp[i +  & ], 0x3f, sizeof dp[i +  & ]);
             for (int j = ; j <  << ; j++)
             {
                 int tmp = j <<  & msk;
                 dp[i +  & ][tmp | ] = min(dp[i +  & ][tmp | ],
                                              dp[i & ][j] +  * a[i + ]);
                 if (i >=  && !tmp) continue;
                 dp[i +  & ][tmp] = min(dp[i +  & ][tmp], dp[i & ][j]);
             }
         }
         int minn = INF;
         for (int i = ; i <  << ; i++) minn = min(minn, dp[n & ][i]);
         cout << minn << endl;
     }
     return ;
 }