[PAT-甲级]1016.Phone Bills

1016. Phone Bills (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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-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

解题思路：自定义排序题，详见cmp函数，排序后进行模拟计算，注意代码中needPrint和next的含义，needPrint标志当前用户中是否存在有效的记录，next表示下一用户首条通话记录在数组中的位置。其次计算费用的函数getAns，从通话开始，循环到通话结束，话费一分钟一分钟的加，注意minute和hour的进位。题目比较繁琐。

代码如下：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int rate[26];struct Record{char name[22];int month;int day;int hour;int minute;bool flag;}record[1010], tmp;bool cmp(Record a, Record b){if(strcmp(a.name, b.name) != 0)return strcmp(a.name, b.name) < 0;else if(a.month != b.month)return a.month < b.month;else if(a.day != b.day)return a.day < b.day;else if(a.hour != b.hour)return a.hour < b.hour;elsereturn a.minute < b.minute;}void getAns(int on, int off, int &time, int &money){tmp = record[on];// 直到通话结束，一分钟一分钟计算money while(tmp.day < record[off].day || tmp.hour < record[off].hour ||tmp.minute < record[off].minute){time ++;money += rate[tmp.hour];tmp.minute ++;//当前通话增加一分钟 // minute满60进位到hour if(tmp.minute >= 60){tmp.minute = 0;tmp.hour ++;}// hour满24进位到day if(tmp.hour >= 24){tmp.hour = 0;tmp.day ++;}}}int main(){freopen("D://input.txt", "r", stdin);for(int i = 0; i < 24; i ++)scanf("%d", &rate[i]);int n;scanf("%d", &n);char flag[10];for(int i = 0; i < n; i ++){scanf("%s", record[i].name);scanf("%d:%d:%d:%d", &record[i].month, &record[i].day, &record[i].hour, &record[i].minute);scanf("%s", flag);if(strcmp(flag, "on-line") == 0)record[i].flag = true;elserecord[i].flag = false;}// 排序 sort(record, record+n, cmp);// on, off是配对的通话记录，next为下一用户的首条通话在数组中的位置 int on = 0, off, next;while(on < n){int needPrint = 0;next = on;while(next < n && strcmp(record[next].name, record[on].name) == 0){if(needPrint == 0 && record[next].flag == true)needPrint = 1;// 找到on,needPrint = 1 else if(needPrint == 1 && record[next].flag == false)needPrint = 2;// 找到off,needPrint = 2 next ++;}// 没有找到配对 if(needPrint < 2){on = next;continue;}int allMoney = 0;printf("%s %02d\n", record[on].name, record[on].month); while(on < next){while(on < next - 1&& !(record[on].flag == true && record[on+1].flag==false)){on ++;}off = on + 1;if(off == next){on = next;break;}printf("%02d:%02d:%02d ", record[on].day, record[on].hour, record[on].minute);printf("%02d:%02d:%02d ", record[off].day, record[off].hour, record[off].minute);int time = 0, money = 0;getAns(on, off, time, money);allMoney += money;printf("%d $%.2f\n", time, money/100.0);on = off + 1;}printf("Total amount: $%.2f\n", allMoney/100.0);}return 0;}