1016 Phone Bills
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 (≤), 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 10
10
CYLL 01:01:06:01 on-line
CYLL 01:28:16:05 off-line
CYJJ 01:01:07:00 off-line
CYLL 01:01:08:03 off-line
CYJJ 01:01:05:59 on-line
aaa 01:01:01:03 on-line
aaa 01:02:00:01 on-line
CYLL 01:28:15:41 on-line
aaa 01:05:02:24 on-line
aaa 01:04:23:59 off-line
Sample Output:
CYJJ 01
01:05:59 01:07:00 61 $12.10
Total amount: $12.10
CYLL 01
01:06:01 01:08:03 122 $24.40
28:15:41 28:16:05 24 $3.85
Total amount: $28.25
aaa 01
02:00:01 04:23:59 4318 $638.80
Total amount: $638.80
题意:
计算每一个用户每一个月用在打电话上的开销。
思路:
模拟
Code:
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 vector<int> costs(24);
6 int costOfDay;
7
8 struct People {
9 string name;
10 vector<string> on_line;
11 vector<string> off_line;
12 };
13
14 bool cmp1(People a, People b) { return a.name < b.name; }
15 bool cmp2(string s1, string s2) { return s1 < s2; }
16
17 pair<int, double> parse(string str1, string str2) {
18 int day1 = stoi(str1.substr(3, 2));
19 int day2 = stoi(str2.substr(3, 2));
20 pair<int, int> time1 = {stoi(str1.substr(6, 2)), stoi(str1.substr(9, 2))};
21 pair<int, int> time2 = {stoi(str2.substr(6, 2)), stoi(str2.substr(9, 2))};
22 int numOfDays = 0;
23 if (time1.first * 60 + time1.second < time2.first * 60 + time2.second) {
24 numOfDays = day2 - day1;
25 } else {
26 numOfDays = day2 - day1 - 1;
27 }
28 double cost = 0.0;
29 int totalMinute = 0;
30 if (time1.first == time2.first) {
31 cost += (time2.second - time1.second) * costs[time1.first];
32 totalMinute += time2.second - time1.second;
33 } else {
34 cost += (60 - time1.second) * costs[time1.first] +
35 time2.second * costs[time2.first];
36 totalMinute += 60 - time1.second + time2.second;
37 for (int i = time1.first + 1; i < time2.first; ++i) {
38 cost += 60 * costs[i];
39 totalMinute += 60;
40 }
41 }
42 totalMinute += numOfDays * 24 * 60;
43 cost += numOfDays * costOfDay;
44 double dollar = cost / 100.0;
45 return {totalMinute, dollar};
46 }
47
48 int main() {
49 for (int i = 0; i < 24; ++i) {
50 cin >> costs[i];
51 costOfDay += costs[i] * 60;
52 }
53 int N;
54 cin >> N;
55 getchar();
56 string record;
57 map<string, People> m;
58 for (int i = 0; i < N; ++i) {
59 getline(cin, record);
60 int space1, space2;
61 space1 = record.find(' ');
62 space2 = record.find(' ', space1 + 1);
63 string name = record.substr(0, space1);
64 string date = record.substr(space1 + 1, space2 - space1 - 1);
65 string status = record.substr(space2 + 1);
66 if (status == "on-line") {
67 m[name].on_line.push_back(date);
68 } else {
69 m[name].off_line.push_back(date);
70 }
71 }
72 vector<People> peoples;
73 for (auto it : m) {
74 it.second.name = it.first;
75 peoples.push_back(it.second);
76 }
77 sort(peoples.begin(), peoples.end(), cmp1);
78 for (People p : peoples) {
79 sort(p.on_line.begin(), p.on_line.end(), cmp2);
80 sort(p.off_line.begin(), p.off_line.end(), cmp2);
81 vector<string> start, end;
82 int index = 0, len1 = p.on_line.size(), len2 = p.off_line.size();
83 for (int i = 0; i < len2 && index < len1; ++i) {
84 while (index < len1 && p.on_line[index] < p.off_line[i]) index++;
85 if (index > 0) {
86 start.push_back(p.on_line[index - 1]);
87 end.push_back(p.off_line[i]);
88 }
89 }
90 cout << p.name << " " << start[0].substr(0, 2) << endl;
91 double totalCost = 0.0;
92 for (int i = 0; i < start.size(); ++i) {
93 cout << start[i].substr(3) << " " << end[i].substr(3) << " ";
94 pair<int, double> ans = parse(start[i], end[i]);
95 totalCost += ans.second;
96 cout << ans.first << " $" << fixed << setprecision(2) << ans.second
97 << endl;
98 }
99 cout << "Total amount: $" << totalCost << endl;
100 }
101
102 return 0;
103 }
写了好久,最后通过了一组测试点,溜了……(待我刷完PAT最后一题,再来补你)
1016 Phone Bills的更多相关文章
- PAT 1016 Phone Bills[转载]
1016 Phone Bills (25)(25 分)提问 A long-distance telephone company charges its customers by the followi ...
- 1016 Phone Bills (25 分)
1016 Phone Bills (25 分) A long-distance telephone company charges its customers by the following rul ...
- PAT甲级1016. Phone Bills
PAT甲级1016. Phone Bills 题意: 长途电话公司按以下规定向客户收取费用: 长途电话费用每分钟一定数量,具体取决于通话时间.当客户开始连接长途电话时,将记录时间,并且客户挂断电话时也 ...
- PAT 1016 Phone Bills(模拟)
1016. Phone Bills (25) 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 A long-di ...
- 1016. Phone Bills (25)——PAT (Advanced Level) Practise
题目信息: 1016. Phone Bills (25) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue A l ...
- PAT 甲级 1016 Phone Bills (25 分) (结构体排序,模拟题,巧妙算时间,坑点太多,debug了好久)
1016 Phone Bills (25 分) A long-distance telephone company charges its customers by the following r ...
- PAT A 1016. Phone Bills (25)【模拟】
题目:https://www.patest.cn/contests/pat-a-practise/1016 思路:用结构体存储,按照名字和日期排序,然后先判断是否有效,然后输出,时间加减直接暴力即可 ...
- PAT 1016. Phone Bills
A long-distance telephone company charges its customers by the following rules: Making a long-distan ...
- 1016. Phone Bills (25) -vector排序(sort函数)
题目如下: A long-distance telephone company charges its customers by the following rules: Making a long- ...
- 1016 Phone Bills (25)(25 point(s))
problem A long-distance telephone company charges its customers by the following rules: Making a lon ...
随机推荐
- 五大自动化测试的Python框架
1.Robot Framework 作为最重要的Python测试框架之一,Robot Framework主要被用在测试驱动(test-driven)类型的开发与验收中.虽然是由Python开发而来,但 ...
- Flannel和Calico网络插件工作流程对比
Flannel和Calico网络插件对比 Calico简介 Calico是一个纯三层的网络插件,calico的bgp模式类似于flannel的host-gw Calico方便集成 OpenStac ...
- 控制流程-if/while/for
目录 一.控制流程之if判断 1.单分支结构 2.双分支结构 3.多分支结构 二.控制流程之while循环 1.基本使用 2.break 3.continue 三.流程控制之for循环 1.break ...
- 你要是还学不会,请提刀来见 Typora+PicGo+Gitee + node.js 打造个人高效稳定优雅图床
你要是还学不会,请提刀来见 Typora+PicGo+Gitee + node.js 打造个人高效稳定优雅图床 经过前面两弹的介绍,相信大家对图床都不陌生了吧, 但是小魔童觉得这样做法还是不方便,使用 ...
- go 在crontab里面运行报错 解决方案
问题背景 你高高兴兴的写好了一个go脚本,放到你的服务器上,打算定期运行这个脚本,你打开crontab -e, 然后输入: */1 * * * * go run /root/test/main.go ...
- [Elementary Mechanics Using Python-02]Feather in tornado
Problem 9.17 Feather in tornado. In this project you will learn to use Newton's laws and the force m ...
- Codeforces Round #683 (Div. 2, by Meet IT)
A 初始情况\(1\) ~ \(n\)堆分别有 \(1\) ~ \(n\) 个糖果,第\(i\)次操作给除了所选堆的糖果数 \(+ i\), 找到一种方案可以使得所有堆糖果数相同,输出操作次数和每次选 ...
- BZOJ_2243 [SDOI2011]染色 【树链剖分+线段树】
一 题目 [SDOI2011]染色 二 分析 感觉树链剖分的这些题真的蛮考验码力的,自己的码力还是不够啊!o(╯□╰)o 还是比较常规的树链剖分,但是一定记得这里的线段树在查询的时候一定要考虑链于链相 ...
- Android Studio 之 BaseAdapter 学习笔记
•前行必备--ListView的显示与缓存机制 我们知道 ListView.GridView 等控件可以展示大量的数据信息. 假如下图中的 ListView 可以展示 100 条信息,但是屏幕的尺寸是 ...
- python自动统计zabbix系统监控覆盖率
脚本主要功能: 1)通过zabbix api接口采集所有监控主机ip地址: 2)通过cmdb系统(蓝鲸)接口采集所有生产主机IP地址.主机名.操作系统.电源状态: 3)以上2步返回数据对比,找出未监控 ...