1016 Phone Bills (25分)

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。

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;

    // }





    

 

要对每月的话费金额进行析并绘制图形展示,我们可以使用Python的数据析库pandas和数据可视化库matplotlib。首先,假设你有一个包含每个月话费数据的CSV文件或列表,数据结构可能是这样的: ```python data = { 'Month': ['Jan', 'Feb', 'Mar', ..., 'Nov', 'Dec'], # 月份 'Phone_Fees': [50, 75, 100, ..., 80, 95] # 每月话费 } ``` 以下是一步步操作的示例: 1. **导入必要的库**: ```python import pandas as pd import matplotlib.pyplot as plt ``` 2. **加载数据** (这里假设已经读取了CSV文件): ```python df = pd.read_csv('monthly_phone_bill.csv') # 替换为你的文件路径 ``` 3. **检查数据**: ```python print(df.head()) # 查看前几行数据 ``` 4. **描述性统计析**: ```python summary_stats = df['Phone_Fees'].describe() # 计算费用布的统计数据(如平均值、标准差等) print(summary_stats) ``` 5. **绘制直方图**: ```python plt.figure(figsize=(10, 6)) # 设置图表大小 plt.hist(df['Phone_Fees'], bins=range(0, max(df['Phone_Fees']) + 50, 50), # 布区间范围和间距 alpha=0.5, # 阴影效果 edgecolor='black', # 边框颜色 label='Monthly Phone Fees') plt.title('Distribution of Monthly Phone Bills') plt.xlabel('Amount') plt.ylabel('Frequency') plt.legend() plt.show() ``` 6. **绘制箱线图**: ```python plt.boxplot(df['Phone_Fees'], vert=False, labels=df['Month']) plt.title('Box plot of Monthly Phone Bills') plt.yticks(range(1, len(df['Month']) + 1), df['Month']) plt.grid(axis='x') plt.show() ``` 这将生成一个显示话费布的图形,你可以根据实际需求选择合适的图表类型。如果你的数据不在CSV文件中,也可以直接使用pandas DataFrame的方式处理。
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