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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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

  1. 如何让图片在div里面剧中显示

    你可能有很多种方式,但是这种方式我觉得更加简单,供大家参考. 用一个 display:inline-block 的helper容器高度为height: 100% 并且vertical-align: m ...

  2. mysql创建计算列

    mysql> create table t(id int auto_increment not null,c1 int,c2 int,c3 int as (c1+c2),primary key( ...

  3. 如何为CentOS 7配置静态IP地址

    问题:在CentOS 7上,我想要将我其中一个网络接口从DHCP改为静态IP地址配置,如何才能永久为CentOS或RHEL 7上的网络接口分配静态IP地址? 如果你想要为CentOS 7中的某个网络接 ...

  4. Android自动测试之Monkey工具

    Monkey工具 前言: 最 近开始研究Android自动化测试方法,对其中的一些工具.方法和框架做了一些简单的整理,其中包括android测试框架.CTS.Monkey. Monkeyrunner. ...

  5. SmartImageView的使用

    对于Android智能图像查看 SmartImageView是一个简易替换为Android标准的ImageView这还可以让图像从URL或用户的联系人地址簿加载.图片被缓存到内存和磁盘的超快速装载. ...

  6. 理解 php 中& 引用

    php中引用&的真正理解-变量引用.函数引用.对象引用 php的引用(就是在变量或者函数.对象等前面加上&符号) //最重要就是 删除引用的变量 ,只是引用的变量访问不了,但是内容并没 ...

  7. LeetCode OJ 141. Linked List Cycle

    Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using ext ...

  8. ORACLE里锁有以下几种模式,v$locked_object,locked_mode【转】

    ORACLE里锁有以下几种模式:0:none1:null 空2:Row-S 行共享(RS):共享表锁,sub share 3:Row-X 行独占(RX):用于行的修改,sub exclusive 4: ...

  9. 浅谈测试rhel7新功能时的感受及遇到的问题【转载】

    半夜起来看世界杯,没啥激情,但是又怕错误意大利和英格兰的比赛,就看了rhel7 相关新功能的介绍. rhel7的下载地址: https://access.redhat.com/site/downloa ...

  10. android studio导入包后无法import

    android studio导入jar包的方法: 1.将jar包放到module的libs目录下 2.在所导入的jar包上右键,选择“Add as library”. 其中,第二点跟eclipse不同 ...