[PAT A1016]Phone Bills

本文详细介绍了如何设计和实现一个电话计费系统,包括读取费率结构、处理电话记录、计算通话费用及输出账单的过程。通过使用C++语言,作者分享了一种计算通话时间及其费用的有效算法。

[PAT A1016]Phone Bills

题目描述

1016 Phone Bills (25 分)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.

输入格式

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.

输出格式

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.

输入样例

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

输出样例

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

解析

  1. 题目大意:题目首先输入24个数,其中第i个数表示时间在[I-1,1]这个时间段每分钟的电话费,例如,第一个数是10,表示在[0,1]小时内,电话费是1毛钱每分钟。然后输入电话记录数,对于每个人的电话记录数按照时间排序,如果某一个电话记录是on-line然后下一个电话是off-line,这就说明他们构成一次完整的通话。如果某一个电话是on-line状态而他的下一个电话是on-line,这样就不构成一个完整的通话;
  2. 我们要做的就是按照姓名从低到高输出他们的账单,分别是每次通话的起始时间和终止时间,后面跟着的是此次通话的时长和此次通话的价格,然后我们需要统计他们这个月的电话账单,并输出他们这个月的总电话费
  3. 这个题目很复杂,我第一次做的时候总是有两组数据不能通过,后面我使用了map,才把逻辑关系理的更加清楚,这里要说的是我计算通话时间的算法,我觉得需要描述一下,帮助读者理解。我首先输入两个时间,然后用小的时间不断一分钟一分钟地加上去,然后每加一分钟就增加该时间段一分钟的价格,直到两个时间相等为止,那就是总的通话时间和通话费用。
  4. 具体代码和注释如下:
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
struct record
{
 string name, state; //姓名和状态(on-line或者off-line)
 int month, day, hour, minute;
};
int price[24]; //记录每个时间段的电话费单价
bool cmp(record r1, record r2)  //自己写的比较函数,按照名字聚合,然后内部再排
{
 if (r1.name != r2.name) return r1.name < r2.name;
 if (r1.month != r2.month) return r1.month < r2.month;
 if (r1.day != r2.day) return r1.day < r2.day;
 if (r1.hour != r2.hour) return r1.hour < r2.hour;
 else return r1.minute < r2.minute;
}
double pay(int d1, int h1, int m1, int d2, int h2, int m2)  //计算价格函数
{
 int time = 0;
 double money = 0;
 while (d1 != d2 || h1 != h2 || m1 != m2) {
  time++;
  m1++;
  money += price[h1];
  if (m1 == 60) {
   m1 = 0;
   h1++;
  }
  if (h1 == 24) {
   h1 = 0;
   d1++;
  }
 }
 printf("%d $%.2f\n", time, money / 100);
 return money / 100;
}
int main()
{
 int n;
 vector<record> res;   //记录所有账单
 for (int i = 0; i < 24; i++) scanf("%d", &price[i]);
 scanf("%d", &n);
 for (int i = 0; i < n; i++) {
  record r;
  cin >> r.name;
  scanf("%d:%d:%d:%d",&r.month, &r.day, &r.hour, &r.minute);
  cin >> r.state;
  res.push_back(r);
 }
 sort(res.begin(), res.end(), cmp);
 map<string, vector<record>> customer;  //对于每一个string name都对应一个vector数组,记录他的账单
 for (int i = 1; i < res.size(); i++) {
  if (res[i - 1].name == res[i].name && res[i - 1].state == "on-line" && res[i].state == "off-line") {
   customer[res[i - 1].name].push_back(res[i - 1]);
   customer[res[i - 1].name].push_back(res[i]);
  }
 }
 for (auto it : customer) {  //遍历图
  double total = 0;
  printf("%s %02d\n", it.first.c_str(), it.second[0].month);
  vector<record> r = it.second;
  for (int i = 1; i < r.size(); i+=2) {
   printf("%02d:%02d:%02d %02d:%02d:%02d ",r[i-1].day, r[i - 1].hour, r[i - 1].minute, r[i].day, r[i].hour, r[i].minute);
   total += pay(r[i - 1].day, r[i - 1].hour, r[i - 1].minute, r[i].day, r[i].hour, r[i].minute);
  }
  printf("Total amount: $%.2f\n", total);
 }
 return 0;
}
水平有限,如果代码有任何问题或者有不明白的地方,欢迎在留言区评论;也欢迎各位提出宝贵的意见!
### PAT 甲级 真题 1172 解析 对于PAT甲级真题1172,该题目名为“Phone Bill”,主要考察字符串处理以及简单的数据结构应用能力。此题目的背景设定为客户通话记录统计问题。 #### 题目描述 给定一组电话号码及其对应的拨打时间和持续时间,计算每位用户的月账单总额。每条通话记录包含三个字段:电话号码、起始时间和结束时间。要求按照输入顺序输出每个客户的总费用,并保留两位小数[^1]。 #### 输入格式说明 - 第一行给出正整数N (≤10^5),表示有N次呼叫; - 接下来N行,每行提供一次呼叫的信息:“手机号码 起始时刻 结束时刻”。其中,“起始时刻”和“结束时刻”的格式均为HH:MM:SS; #### 输出格式说明 - 对于每一个客户,先打印其手机号码,再跟上冒号和空格,最后是当月话费金额(精确到分),单位为元人民币RMB。 #### 示例代码实现 ```cpp #include <iostream> #include <map> #include <iomanip> using namespace std; int main() { int n; cin >> n; map<string, double> bills; while(n--) { string number; char start_time[9], end_time[9]; scanf("%s %s %s", &number[0], start_time, end_time); // Convert time strings to seconds since midnight. sscanf(start_time, "%*d:%*d:%d", &start_seconds); sscanf(end_time, "%*d:%*d:%d", &end_seconds); // Calculate duration and update bill accordingly. int duration = end_seconds - start_seconds; if(duration > 0){ bills[number] += ceil((double)duration / 60 * 0.01); } } for(auto& entry : bills){ cout << entry.first << ": " << fixed << setprecision(2) << entry.second << endl; } return 0; } ``` 上述C++程序实现了对输入数据的读取与处理逻辑,通过`<map>`容器来存储并累加各个用户的通话时长及相应费用。需要注意的是,在实际比赛中应当更加严谨地验证输入的有效性和边界条件。
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