博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
A1016 Phone Bills (25 分)
阅读量:4921 次
发布时间:2019-06-11

本文共 3624 字,大约阅读时间需要 12 分钟。

A long-distance telephone company charges its customers by the following rules:

Making a long-distance call costs a certain amount per minute, depending on the time of day when the call is made. When a customer starts connecting a long-distance call, the time will be recorded, and so will be the time when the customer hangs up the phone. Every calendar month, a bill is sent to the customer for each minute called (at a rate determined by the time of day). Your job is to prepare the bills for each month, given a set of phone call records.

Input Specification:

Each input file contains one test case. Each case has two parts: the rate structure, and the phone call records.

The rate structure consists of a line with 24 non-negative integers denoting the toll (cents/minute) from 00:00 - 01:00, the toll from 01:00 - 02:00, and so on for each hour in the day.

The next line contains a positive number N (1000), followed by N lines of records. Each phone call record consists of the name of the customer (string of up to 20 characters without space), the time and date (mm:dd:hh:mm), and the word on-line or off-line.

For each test case, all dates will be within a single month. Each on-line record is paired with the chronologically next record for the same customer provided it is an off-line record. Any on-line records that are not paired with an off-line record are ignored, as are off-line records not paired with an on-line record. It is guaranteed that at least one call is well paired in the input. You may assume that no two records for the same customer have the same time. Times are recorded using a 24-hour clock.

Output Specification:

For each test case, you must print a phone bill for each customer.

Bills must be printed in alphabetical order of customers' names. For each customer, first print in a line the name of the customer and the month of the bill in the format shown by the sample. Then for each time period of a call, print in one line the beginning and ending time and date (dd:hh:mm), the lasting time (in minute) and the charge of the call. The calls must be listed in chronological order. Finally, print the total charge for the month in the format shown by the sample.

Sample Input:

10 10 10 10 10 10 20 20 20 15 15 15 15 15 15 15 20 30 20 15 15 10 10 1010CYLL 01:01:06:01 on-lineCYLL 01:28:16:05 off-lineCYJJ 01:01:07:00 off-lineCYLL 01:01:08:03 off-lineCYJJ 01:01:05:59 on-lineaaa 01:01:01:03 on-lineaaa 01:02:00:01 on-lineCYLL 01:28:15:41 on-lineaaa 01:05:02:24 on-lineaaa 01:04:23:59 off-line

Sample Output:

CYJJ 0101:05:59 01:07:00 61 $12.10Total amount: $12.10CYLL 0101:06:01 01:08:03 122 $24.4028:15:41 28:16:05 24 $3.85Total amount: $28.25aaa 0102:00:01 04:23:59 4318 $638.80Total amount: $638.80 注意点: 1.没有配对的用户不能输出(测试点3)。needPrint 2.名字必须相同(测试点2)。rec[j].name==rec[j+1].name
#include
using namespace std;const int maxn=100010;struct Record{ string name; int month; int day; int hh; int mm; int flag; string status;};Record rec[maxn];int rate[24];int n;bool cmp(Record a,Record b){ if(a.name!=b.name) return a.name
>rate[i]; cin>>n; char c; Record temp; for(int i=0;i
>temp.name>>temp.month>>c>>temp.day>>c>>temp.hh>>c>>temp.mm >>temp.status; if(temp.status=="on-line") temp.flag=1; else temp.flag=2; rec[i]=temp; } sort(rec,rec+n,cmp); // print(); // int count1=0;// int money1=Calculate(rec[7].day,rec[7].hh,rec[7].mm,rec[8].day,rec[8].hh,rec[8].mm,count1); // cout<
<<" "<
<

  

 

转载于:https://www.cnblogs.com/moranzju/p/11095088.html

你可能感兴趣的文章
GitHub创建项目,保存代码。
查看>>
sed 小结
查看>>
LeetCode #3 Longest Substring Without Repeating Characters (M)
查看>>
Python 乱码问题解决办法
查看>>
单例模式易错分析
查看>>
【BZOJ4827】【HNOI2017】礼物
查看>>
Struts2
查看>>
编写高质量代码改善C#程序的157个建议——建议63:避免“吃掉”异常
查看>>
MySQL性能调优与架构设计——第9章 MySQL数据库Schema设计的性能优化
查看>>
python 操作 excel
查看>>
XML解析之DOM ,SAX解析区别
查看>>
如何把一篇Word文档里的所有换行符去掉?
查看>>
改造vim
查看>>
C++编写Node.js插件(Addon)
查看>>
Excel-漏斗图分析(差异分析)
查看>>
ansible使用5-Variables
查看>>
SpringBoot2.0源码分析(二):整合ActiveMQ分析
查看>>
pickle 模块
查看>>
eclipse中的项目Java build path (Java创建路径)详解
查看>>
Codeforces Round #439 (Div. 2) 题解
查看>>