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 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没有特别复杂的数据结构，主要是思想上要理清楚，代码中为了方便比较和统计时间，全部转换为以分钟为单位进行表示
代码如下：
#include<iostream>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;
struct record
{
int dd,hh,mm;
int t;//换算为分钟后的总时间
string tag;//on,off
};
bool cmp(record a,record b)
{
return a.t<b.t;
}
int n;//记录个数
int price[24];
map<string,vector<record>>M;//每一个用户的记录
int main()
{
for(int i=0;i<24;i++)scanf("%d",&price[i]);
int month;//根据题目要求，所有记录都是在一个月内的
scanf("%d",&n);
for(int i=0;i<n;i++)
{
string name,tag;
int dd,hh,mm;
char c;//用来吃掉输入时的“：”
cin>>name>>month>>c>>dd>>c>>hh>>c>>mm>>tag;
record temp;
temp.dd=dd;
temp.hh=hh;
temp.mm=mm;
temp.tag=tag;
temp.t=dd*1440+hh*60+mm;//换算成用分钟表示
M[name].emplace_back(temp);//存入该用户的记录容器中
}
//开始匹配
for(auto it=M.begin();it!=M.end();++it)
{
auto V=it->second;//first获取的是name，second获取记录的容器
sort(V.begin(),V.end(),cmp);//将所有记录按时间顺序进行排序
double total=0;//月度费用
for(int i=0;i<V.size();)
{
if(i+1<V.size()&&V[i].tag>V[i+1].tag)//匹配到了on-off
{
if(!total)//第一次匹配到的时候total为0，此时输出用户名
{
cout<<it->first;
printf(" %02d\n",month);
}
int time1=V[i].t;
int time2=V[i+1].t;
double part=0;
for(int time=time1;time<time2;time++)
{
part+=price[time%1440/60];
//价格是按每分钟来算的，我们就一分钟一分钟地加，time%1440先算出一天当中占多少分钟，再/60求出是哪个小时区间
}
printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2f\n",V[i].dd,V[i].hh,V[i].mm,V[i+1].dd,V[i+1].hh,V[i+1].mm,V[i+1].t-V[i].t,part/100);
total+=part;
i+=2;
}
else{
i+=1;
}
}
if(total)printf("Total amount: $%.2f\n",total/100);
}
return 0;
}运行结果如下：
?