Tickets

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

 

Sample Output

 

Source

浙江工业大学第四届大学生程序设计竞赛

一开始状态方程考虑的非常复杂 一直在想着如何利用前一个状态推下一个状态，所以开始我的一个状态里有三种情况，然后这种复杂的情况有点驾驭不了，没想到直接可以用前两种状态对当前状态。悲剧

#include<iostream>using namespace std;int a[2001],b[2001],dp[2001];#define min(x,y) (x)<(y)? (x):(y)int n;void pre(){    int i,j;    cin>>n;    for(i=1;i<=n;i++) cin>>a[i];    for(i=2;i<=n;i++) cin>>b[i];}void solve(){    int i,j;    dp[0]=0;dp[1]=a[1];    for(i=2;i<=n;i++)        dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i]);    int num=dp[n];    int h,m,s;    h=num/60/60;num-=h*60*60;    m=num/60;num-=m*60;    s=num;    printf("%02d:%02d:%02d",h+8>12? h+8-12:h+8,m,s);    printf(" %s\n",h+8>12? "pm":"am");}int main(void){    int t,i,j;    while(cin>>t){        while(t--){            pre();                    solve();        }    }    return 0;}