Tickets（DP）

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
2
2
20 25
40
1
8

Sample Output
08:00:40 am
08:00:08 am

题意：k个人买票，每一个人买票都花费一定的时间。每个人既可以自己单独买票，也可以和旁边的人一起买（最多两人一起 买）。告诉每个人单独买票的时间和邻近的两人一起买票的时间，求这些人最短需要多久才可以都买完票。

分析：用dp[i]表示到第i个人花费的时间，因此有状态转移方程dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i])。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int n,dp[2005],hh,mm,ss;
    int k,i,j,a[2005],b[2005];//a存第i个人自己买所花费时间。
    scanf("%d",&n);           //b存第i个与第i-1个人一起买花费的时间。
    while(n--)
    {
        memset(dp,0,sizeof(dp));

        scanf("%d",&k);
        for(i=1;i<=k;i++)
            scanf("%d",&a[i]);

        for(i=2;i<=k;i++)//从2开始存
            scanf("%d",&b[i]);

        dp[1]=a[1];//初始化。
        dp[2]=min(dp[1]+a[2],b[2]);

        for(i=3;i<=k;i++)
            dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i]);

        int temp=dp[k];

        hh=8+temp/3600;//处理时间
        mm=temp%3600/60;
        ss=temp%60;

        if(hh<12)
            printf("%02d:%02d:%02d am\n",hh,mm,ss);
        else
            printf("%02d:%02d:%02d am\n",hh-12,mm,ss);
    }
    return 0;
}