Tickets

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Appoint description: 
System Crawler  (2015-09-06)

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
 
 
 
#include <iostream>
#include <cstdio>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <climits>
using namespace std; const int SIZE = ;
int DP[SIZE],S[SIZE],H[SIZE]; int main(void)
{
int t,n; scanf("%d",&t);
while(t --)
{
memset(DP,,sizeof(DP));
scanf("%d",&n);
for(int i = ;i <= n;i ++)
scanf("%d",&S[i]);
for(int i = ;i <= n - ;i ++)
scanf("%d",&H[i]); if(n == )
printf("08:00:%02d am\n",S[]);
else
{
for(int i = ;i <= n;i ++)
DP[i] = min(DP[i - ] + S[i],DP[i - ] + H[i - ]);
int sec = DP[n];
int hour = sec / ( * );
sec -= hour * * ;
int min = sec / ;
sec -= min * ;
hour += ; bool flag_am = true;
if(hour >= )
{
flag_am = false;
hour -= ;
}
printf("%02d:%02d:%02d %s\n",hour,min,sec,flag_am ? "am" : "pm");
}
} return ;
}
 

怒刷DP之 HDU 1260的更多相关文章

  1. 怒刷DP之 HDU 1257

    最少拦截系统 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Statu ...

  2. 怒刷DP之 HDU 1160

    FatMouse's Speed Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Su ...

  3. 怒刷DP之 HDU 1176

    免费馅饼 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status  ...

  4. 怒刷DP之 HDU 1087

    Super Jumping! Jumping! Jumping! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64 ...

  5. 怒刷DP之 HDU 1114

    Piggy-Bank Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit S ...

  6. 怒刷DP之 HDU 1069

    Monkey and Banana Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u S ...

  7. 怒刷DP之 HDU 1024

    Max Sum Plus Plus Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u S ...

  8. 怒刷DP之 HDU 1029

    Ignatius and the Princess IV Time Limit:1000MS     Memory Limit:32767KB     64bit IO Format:%I64d &a ...

  9. 【DP】HDU 1260

    HDU 1260 Tickets 题意:有N个人要买票,你可以一个一个人卖票,时间分别为Xs,也可以相邻两个人一起卖票,时间为Ys,从早上八点开始卖票,问你何时最早将N个人的票卖完. 思路:解决情况是 ...

随机推荐

  1. SPSS二次开发

    在以前关于SPSS二次开发文章中留下过自己联系方式,差不多一年的时间,零零散散的和我取得联系的人也有几十位,看来对于SPSS二次开发的需求不少. Web SPSS系统是利用SPSS二次开发技术,使用户 ...

  2. jquery单选框 复选框表格高亮 选中

    单选框: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/T ...

  3. Angular JS 使用的注意事项

    标签中ng-app,什么情况需要指定名称,什么时候又不需要呢? 1.ng-app="" <div ng-app="" ng-controller='myC ...

  4. Linux下getopt()函数的简单使用

    最近在弄Linux C编程,本科的时候没好好学啊,希望学弟学妹们引以为鉴. 好了,虽然啰嗦了点,但确实是忠告.步入正题: 我们的主角----getopt()函数. 英雄不问出处,getopt()函数的 ...

  5. Lockless Ring Buffer Design

    https://www.kernel.org/doc/Documentation/trace/ring-buffer-design.txt Lockless Ring Buffer Design == ...

  6. vs2010代码注释自动生成api文档

    最近做了一些接口,提供其他人调用,要写个api文档,可是我想代码注释已经写了说明,能不能直接把代码注释生成api?于是找到以下方法 环境:vs2010 先下载安装Sandcastle 和Sandcas ...

  7. 内存不足(OutOfMemory)的调试分析

    32位操作系统的寻址空间是4G,其中有2G被操作系统占用,也就是说留给用户进程的内存只有2G(其中还要扣除程序加载时映像占用的部分空间,一般只有1.6G~1.8G左右可以使用). 如果进程运行中需要申 ...

  8. Java面试葵花宝典

    面向对象的特征有哪些方面  1. 抽象:抽象就是忽略一个主题中与当前目标2. 无关的那些方面,3. 以便更充分地注意与当前目标4. 有关的方面.抽象并不5. 打算了解全部问题,而6. 只是选择其中的一 ...

  9. Codeforces Good Bye 2015 A. New Year and Days 水题

    A. New Year and Days 题目连接: http://www.codeforces.com/contest/611/problem/A Description Today is Wedn ...

  10. poj 3635 Full Tank? ( 图上dp )

    题意: 已知每一个点的加油站的油价单位价格(即点权).每条路的长度(边权). 有q个询问.每一个询问包含起点s.终点e和油箱容量. 问从起点走到终点的最小花费.假设不可达输出impossible,否则 ...