E - Tickets HDU - 1260 （动态规划）

E - Tickets HDU - 1260 

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. 

InputThere 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. 
OutputFor 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

这题真心是想吐血不知道要怎么说，很快就写完了蛮简单了  就是题目题目题目我读错了或者说是看不懂吧  心痛痛到无法呼吸 后来百度下知道题目了心中有千万只XXXX在狂奔

是一道简单的动态规划的问题

代码如下 ：

#include
#include
#include
#include
using namespace std;
int num[2005];
int total[2005];
int dp[2005];
int main()
{
    int n,k,i,j;
    scanf("%d",&n);
    int sum;
    while(n--)
    {
    	scanf("%d",&k);
    	for(i=1;i<=k;i++)
    	scanf("%d",&num[i]);
    	for(i=2;i<=k;i++)
    	scanf("%d",&total[i]);
    	memset(dp,0,sizeof(dp));
        if(k==1)
        sum=num[1];
        else if(k==2)
        sum=min(total[2],num[1]+num[2]);
    	else 
    	{
    	  	dp[0]=0;
			dp[1]=num[1];
			for(i=2;i<=k;i++)	
    		{
    			dp[i]=min(dp[i-2]+total[i],dp[i-1]+num[i]);
    		}
    		sum=dp[k];
    	}
        int hh,mm,ss;
        ss=sum%60;
        mm=sum/60%60;
        hh=sum/3600+8;
        if(hh<=12)
           printf("%02d:%02d:%02d am\n",hh,mm,ss);
        else 
           printf("%02d:%02d:%02d pm\n",hh-12,mm,ss);
    }    
	return 0;
}