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-linerecords 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
#include<iostream>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
int charge_standard[25] = {};
struct record{
string name;
int mon;
int day;
int hour;
int min;
int last_time;
int total_min;
bool flag;
};
bool cmp_name_time(record a, record b){
return (a.name != b.name) ? a.name < b.name : a.total_min < b.total_min;
}
bool cmp_total_time(vector<record> a, vector<record> b){
return a.front().total_min < b.front().total_min;
}
double total_charge_from_zero(record temp){
double total_charge_from_zero = 0;
total_charge_from_zero = temp.day * charge_standard[24] * 60;
for(int i = 0; i < temp.hour; i++){
total_charge_from_zero += 60 * charge_standard[i];
}
total_charge_from_zero += charge_standard[temp.hour] * temp.min;
return total_charge_from_zero;
}
int main(){
vector<record> input_records;
map<string, vector<record> > output_records;
for(int i = 0; i < 24; i++){
int temp;
cin >> temp;
charge_standard[i] = temp;
charge_standard[24] += temp;
}
int n;
cin >> n;
for(int i = 0; i < n; i++){
record temp_record;
string name;
int mon, day, hour, min;
string on_of;
cin >> name;
scanf("%d:%d:%d:%d", &mon, &day, &hour, &min);
cin >> on_of;
temp_record.name = name;
temp_record.mon = mon;
temp_record.day = day;
temp_record.hour = hour;
temp_record.min = min;
temp_record.total_min = day * 24 * 60 + hour * 60 + min;
temp_record.flag = (on_of == "on-line") ? true : false;
input_records.push_back(temp_record);
}
sort(input_records.begin(), input_records.end(), cmp_name_time);
for(int i = 1; i < input_records.size(); i++){
if(input_records[i - 1].name == input_records[i].name && input_records[i - 1].flag && !input_records[i].flag){
output_records[input_records[i - 1].name].push_back(input_records[i - 1]);
output_records[input_records[i].name].push_back(input_records[i]);
}
}
// sort(output_records.begin(), output_records.end(), cmp_total_time);
map<string, vector<record> >::iterator it;
for(it = output_records.begin(); it != output_records.end(); it++ ){
double total_charges = 0;
vector<record> temp_output = it -> second;
cout << it -> first;
printf(" %02d", temp_output.front().mon);
cout << endl;
for(int i = 1; i < temp_output.size(); i += 2){
printf("%02d:%02d:%02d ", temp_output[i - 1].day, temp_output[i - 1].hour, temp_output[i - 1].min);
printf("%02d:%02d:%02d ", temp_output[i].day, temp_output[i].hour, temp_output[i].min);
int total_min = temp_output[i].total_min - temp_output[i - 1].total_min;
cout << total_min;
printf(" $%.2f", (total_charge_from_zero(temp_output[i]) - total_charge_from_zero(temp_output[i - 1])) / 100.0);
cout << endl;
total_charges += ((total_charge_from_zero(temp_output[i]) - total_charge_from_zero(temp_output[i - 1])) / 100.0);
}
printf("Total amount: $%.2f", total_charges);
cout << endl;
}
return 0;
}
// for(int i = 0; i < input_records.size(); i++){
// cout << input_records[i].name << " " << input_records[i].total_min << " " << input_records[i].flag << endl;
// }
本文详细介绍了如何设计一个电话计费系统,该系统能够根据不同的时间段收取不同的费用,并且能够处理大量的电话记录,准确地计算每个月的费用。文章通过一个具体的示例,展示了如何读取电话记录,解析时间信息,匹配呼叫的开始和结束,以及如何根据费率结构计算费用。
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