PAT1014
Suppose a bank has N windows open for service.
一个银行有N个服务的窗口
There is a yellow line in front of the windows which devides the waiting area into two parts.
在每个窗口前面都一根黄线用来划分等待区的两个部分
The rules for the customers to wait in line are:
对于顾客来说等待线的规则是这样的
- The space inside the yellow line in front of each window is enough to contain a line with M customers.
在每个窗口前面的黄线以内能够包含M个顾客
Hence when all the N lines are full, all the customers after (and including) the (NM+1)st one will have to wait in a line behind the yellow line.
因此,当所有N条线,所有的顾客中第N*M+1位顾客必须在黄线后等待
- Each customer will choose the shortest line to wait in when crossing the yellow line.
当越过黄线之后,每个顾客将会选择最短的队伍去等待。
If there are two or more lines with the same length, the customer will always choose the window with the smallest number.
如果有两个或者更多的同样长度的队伍,顾客会总是会选择更小编号的队伍。
- Customer[i] will take T[i] minutes to have his/her transaction processed.
顾客i将会花费Ti分钟处理自己的交易流程
- The first N customers are assumed to be served at 8:00am.
前面N个顾客假定从早上8点开始被服务
Now given the processing time of each customer, you are supposed to tell the exact time at which a customer has his/her business done.
现在给出每个顾客的交易时间,要求你求出每个顾客完成业务的准确时间
For example, suppose that a bank has 2 windows and each window may have 2 custmers waiting inside the yellow line.
例如,一个银行有两个窗口,每个一个窗口能有两个顾客等在黄线内
There are 5 customers waiting with transactions taking 1, 2, 6, 4 and 3 minutes, respectively.
现在我5位顾客,花费的时间分别是12643
At 08:00 in the morning, customer1 is served at window1 while customer2 is served at window2.
在早上8点,顾客1在窗口1,顾客2在窗口2
Customer3 will wait in front of window1 and customer4 will wait in front of window2.
顾客3等在窗口1,顾客4等在窗口2
Customer5 will wait behind the yellow line.
顾客5等在黄线后
At 08:01, customer1 is done and customer5 enters the line in front of window1 since that line seems shorter now.
8点01分,顾客1完成了,顾客5进入线内,到了队伍最短的窗口1
Customer2 will leave at 08:02, customer4 at 08:06, customer3 at 08:07, and finally customer5 at 08:10.
顾客2,在8点02离开,顾客4.。。。。。。。
Input
Each input file contains one test case. Each case starts with a line containing 4 positive integers: N (<=20, number of windows), M (<=10, the maximum capacity of each line inside the yellow line), K (<=1000, number of customers), and Q (<=1000, number of customer queries).
每个输入文件包含一个测试用例,每个测试用例由4个整数开始,N, M, K, Q
The next line contains K positive integers, which are the processing time of the K customers.
下面K个正整数表示K顾客花费的时间
The last line contains Q positive integers, which represent the customers who are asking about the time they can have their transactions done. The customers are numbered from 1 to K.
最后一行Q个整数,代表那些想要询问他们想要询问他们到底花费了多少时间的人。顾客的编号从1到K
Output
For each of the Q customers, print in one line the time at which his/her transaction is finished, in the format HH:MM where HH is in [08, 17] and MM is in [00, 59]. Note that since the bank is closed everyday after 17:00, for those customers who cannot be served before 17:00, you must output "Sorry" instead.
对于q个顾客,打印一行时间,他完成的时间,格式是HH:MM,HH是08-17,MM是00-59。记住银行每天17:00关门,对于那些不能再17:00完成的业务的,你需要输出soory
Sample Input
2 2 7 5
1 2 6 4 3 534 2
3 4 5 6 7
Sample Output
08:07
08:06
08:10
17:00
Sorry
题目的意思已经很明显了,银行排队,已经很明显了,数据结构用队列就行了,每个窗口一个队列,然后到时间了完成了出队,然入队,主要考验模拟时的逻辑能力。用queue会简单一些。
PAT1014的更多相关文章
- PAT1014——福尔摩斯的约会
大侦探福尔摩斯接到一张奇怪的字条:“我们约会吧! 3485djDkxh4hhGE 2984akDfkkkkggEdsb s&hgsfdk d&Hyscvnm”.大侦探很快就明白了,字条 ...
- pat1014. Waiting in Line (30)
1014. Waiting in Line (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Suppo ...
- PAT-1014 Waiting in Line (30 分) 优先队列
Suppose a bank has N windows open for service. There is a yellow line in front of the windows which ...
- PAT甲级1014. Waiting in Line
PAT甲级1014. Waiting in Line 题意: 假设银行有N个窗口可以开放服务.窗前有一条黄线,将等候区分为两部分.客户要排队的规则是: 每个窗口前面的黄线内的空间足以包含与M个客户的一 ...
- PAT 1011-1020 题解
早期部分代码用 Java 实现.由于 PAT 虽然支持各种语言,但只有 C/C++标程来限定时间,许多题目用 Java 读入数据就已经超时,后来转投 C/C++.浏览全部代码:请戳 本文谨代表个人思路 ...
随机推荐
- xml 和json 数据格式及解析
来源:http://blog.jobbole.com/79252/ 引言 NOKIA 有句著名的广告语:“科技以人为本”.任何技术都是为了满足人的生产生活需要而产生的.具体到小小的一个手机,里面蕴含的 ...
- Shell脚本中让进程休眠的方法(sleep用法)
有时候写Shell的脚本,用于顺序执行一系列的程序. 有些程序在停止之后并没能立即退出,就例如有一个 tomcat 挂了,就算是用 kill -9 命令也还没瞬间就结束掉. 这么如果 shell 还没 ...
- 12C CLONE PDB and config service_listener
Clone PDB PtestDEV to Ptestuat in testuat 1) Clone PtestDEV to Ptestuat C:\Windows\system32> ...
- Struts2--课程笔记1
第一个Struts程序: 在开发Struts程序之前,首先要导入额外的jar包,基本需求的是14个jar包,关于14个ja包是什么,有什么作用,此处不讲述. 还要配置web.xml文件,注册Strut ...
- U盘启动安装Ubuntu
1.UltraISO制作USB启动盘 2.打开U盘目录下的\syslinux\syslinux.cfg, 将default vesamenu.c32注释为 # default vesamenu.c32
- android CTS测试
CTS认证是获得Google推出的Android系统中Android Market服务的前提 CTS兼容性测试的主要目的和意义在于使得用户在Android系统的应用过程中,有更好的用户体验,并展现出A ...
- Word试卷文档模型化解析存储到数据库
最近在搞一套在线的考试系统,有许多人反映试题的新增比较麻烦(需要逐个输入),于是呼就整个了试卷批量导入了 poi实现word转html 模型化解析html html转Map数组 Map数组(数组的操作 ...
- DOM操作-引用同级的元素
代码: ———————————————————————————————— <script type="text/javascript"> //获取 ...
- 第十四章:使用CSS3进行增强
1.为不支持某些属性的浏览器使用polyfill:如果想弥合较弱的浏览器和较强的浏览器之间的功能差异,可以使用polyfill(通常又称作垫片),通常用js实现.但是有些较弱的浏览器运行JS的速度要慢 ...
- js基础和工具库
/* * 作者: 胡乐 * 2015/4/18 * js 基础 和 工具库 * * * */ //根据获取对象 function hGetId(id){ return document.getElem ...