Tickets

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12331    Accepted Submission(s): 6096

Problem Description

Jesus, what a great movie! Thousands of people are rushing to the cinema. However, this is really a tuff time for Joe who sells the film tickets. He is wandering when could he go back home as early as possible.

A good approach, reducing the total time of tickets selling, is let adjacent people buy tickets together. As the restriction of the Ticket Seller Machine, Joe can sell a single ticket or two adjacent tickets at a time.

Since you are the great JESUS, you know exactly how much time needed for every person to buy a single ticket or two tickets for him/her. Could you so kind to tell poor Joe at what time could he go back home as early as possible? If so, I guess Joe would full of appreciation for your help.

 

Input

There are N(1<=N<=10) different scenarios, each scenario consists of 3 lines:

1) An integer K(1<=K<=2000) representing the total number of people;

2) K integer numbers(0s<=Si<=25s) representing the time consumed to buy a ticket for each person;

3) (K-1) integer numbers(0s<=Di<=50s) representing the time needed for two adjacent people to buy two tickets together.

 

Output

For every scenario, please tell Joe at what time could he go back home as early as possible. Every day Joe started his work at 08:00:00 am. The format of time is HH:MM:SS am|pm.

Sample Input

2220 254018

Sample Output

08:00:40 am08:00:08 am

题目大意与分析

有k个人，给出他们买自己的票各自需要多少时间，以及相邻两个人一块买需要多少时间，求从八点开始买，最快什么时候能卖完

状态转移方程：

　　　　dp[i]=min(dp[i-1]+a1[i],dp[i-2]+a2[i-1]);

需要注意的是，前补零的写法：%02d

#include
     
      using namespace std;int dp[2005],a1[2005],a2[2005],i,n,k,times,h,m,s;int main(){    cin>>n;    while(n--)    {        cin>>k;        for(i=1;i<=k;i++)        {            cin>>a1[i];        }        for(i=1;i<=k-1;i++)        {            cin>>a2[i];        }        dp[1]=a1[1];        dp[0]=0;        for(i=2;i<=k;i++)        {            dp[i]=min(dp[i-1]+a1[i],dp[i-2]+a2[i-1]);        }        times=dp[k];        s=times%60;        m=(times%3600)/60;        h=8+times/3600;        if(h<12)        printf("%02d:%02d:%02d am\n",h,m,s);        else        printf("%02d:%02d:%02d pm\n",h-12,m,s);    }}