Tickets

Tickets

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

题意是这样的，有一队人买票，可以一张一张的买，也可以一次买两张，求最短花费的时间。

简单dp，设一次买一张票的钱为数组a[n], 一次卖两张的为数组b[n]以及结果数组dp[],将a[0],a[1],b[0]初始化之后

循环判断dp[i] = min((a[i] + dp[i - 1]),(b[i] + dp[i - 2]));即可。

ac代码

#include <iostream>
#include <string>
#include <cmath>
#include <cstdio>
#include <algorithm>
using namespace std;

int main()
{
    int o;
    cin >> o;
    while(o--)
    {
        int k, a[2005] = {0}, b[2005] = {0}, dp[2005] = {0}, i;
        cin >> k;
        for(i = 0; i < k; i++)
            cin >> a[i];
        for(i = 1; i < k; i++)
            cin >> b[i];
        dp[0] = a[0];
        dp[1] = min(a[0] + a[1], b[1]);
        for(i = 2; i < k; i++)
        {
            dp[i] = min((a[i] + dp[i - 1]),(b[i] + dp[i - 2]));
        }
        int h=dp[k - 1]/3600+8;
        int m=dp[k - 1]/60%60;
        int s=dp[k - 1]%60;
        if(h % 24 > 12)
            printf("%02d:%02d:%02d %s\n",h%24,m,s,"pm");
        else
            printf("%02d:%02d:%02d %s\n",h%24,m,s,"am");
    }
    return 0;
}