PAT甲级——A1016 Phone Bills-白红宇

PAT甲级——A1016 Phone Bills

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发布时间：2019-06-08

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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

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-linerecord. 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 最麻烦的就是计算费用 其实也不麻烦，你就算每天从00：00开始计费，然后将下线总费用与上线总费用相减就行 时间同样是这么计算

1 #include 
     
       2 #include 
      
        3 #include 
        4 #include 
        
          5 using namespace std; 6 int N; 7 vector
         
          cost(25); 8  9 double calMoney(string str, int& time)10 {11     int d, h, m;12     double money = 0;13     d = (str[3] - '0') * 10 + str[4] - '0';14     h = (str[6] - '0') * 10 + str[7] - '0';15     m = (str[9] - '0') * 10 + str[10] - '0';16     money += cost[24] * 60 * d + cost[h] * m;17     for (int i = 0; i < h; ++i)18         money += cost[i] * 60;19     time = 24 * 60 * d + h * 60 + m;20     return money;21 }22 int main()23 {    24     for (int i = 0; i < 24; ++i)25     {26         cin >> cost[i];27         cost[24] += cost[i];28     }29     cin >> N;30     map
          
           ,less
           
            >data;31 //外部用名字排序,名字是用降序排序，内部用时间排序,时间时用默认升序排序32 for (int i = 0; i < N; ++i)33 {34 string name, time, type;35 cin >> name >> time >> type;36 data[name][time] = type;37 }38 for (auto ptr = data.begin(); ptr != data.end(); ++ptr)39 {40 int f = 1, st = 0, et = 0;41 double s = 0, sum = 0;42 string sl,el;43 for (auto it = ptr->second.begin(); it != ptr->second.end(); ++it)44 {45 if (it == ptr->second.begin()) continue;46 auto pt = it;47 if (it->second == "off-line" && (--pt)->second == "on-line")48 {49 if (f==1)50 {51 cout << ptr->first << " " << it->first[0] << it->first[1] << endl;52 f = 0;53 }54 s = calMoney(it->first, et) - calMoney(pt->first, st);55 sl.assign(pt->first.begin() + 3, pt->first.end());56 el.assign(it->first.begin() + 3, it->first.end());57 cout << sl << " " << el << " " << (et - st) << " ";58 printf("$%0.2f\n", s/100);59 sum += s;60 }61 }62 if(f==0)63 printf("Total amount: $%0.2f\n", sum/100);64 }65 return 0;66 }

转载于:https://www.cnblogs.com/zzw1024/p/11198406.html

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