标签：lse nal while sum cin precision spec tar one
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.
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.
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.
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
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）
标签：lse nal while sum cin precision spec tar one
原文地址：https://www.cnblogs.com/ruruozhenhao/p/13188753.html
