1016 Phone Bills （25 分）
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思路
计算电话费，每个小时段收费不一样。计算时，只有同一个用户，前一个on-line、后一个off-line才算配对成功，这段时间才可以算。
可以先按照名字排序，再按照时间排序，前一个on-line、后一个off-line就是通话时长，就可以计算
注意这里收费是（cents/minutes），cent换算成dollar，要除以100。
计算时长！推荐用之前的时间循环加1min直到达到之后的时间。
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <vector>
using namespace std;
struct call
{
char name[21], status[10];
int month, D, H, M, time, flag;
};
bool cmp(call c1, call c2)
{ /* 按名字升序，如果相等再按时间升序 */
return strcmp(c1.name, c2.name) == 0 ? c1.time < c2.time : strcmp(c1.name, c2.name) < 0;
}
int main(int argc, char const *argv[])
{
int prize[25], N, countnameflag = 0;
prize[24] = 0;
vector <call> v; /* 存放配对的call */
for (int i = 0; i < 24; i++)
{
scanf("%d", &prize[i]);
prize[24] += prize[i]; /* 用于存放一天总和 */
}
scanf("%d", &N);
call B[N]; /* 电话账单数组，保存每一条记录 */
for (int i = 0; i < N; i++)
{
scanf("%s %d:%d:%d:%d %s", B[i].name, &B[i].month, &B[i].D, &B[i].H, &B[i].M, B[i].status);
B[i].time = B[i].D * 24 * 60 + B[i].H * 60 + B[i].M;
if (strcmp(B[i].status, "on-line") == 0)
B[i].flag = 1; /* on-line */
else
B[i].flag = 0; /* off-line */
}
sort(B, B+N, cmp); /* 按题意排序 */
for (int i = 1; i < N; i++)
if (strcmp(B[i].name, B[i-1].name) == 0) /* 如果是同一个用户 */
if (B[i-1].flag && !B[i].flag) /*如果前后正好配对 */
{ /* 将结果存入向量中 */
v.push_back(B[i-1]);
v.push_back(B[i]);
i++; /* 下一次比较 跳过 i+1与i */
}
char name[21];
strcpy(name, v[0].name);
printf("%s %02d\n", v[0].name, v[0].month);
double TotalSum = 0;
for (int i = 0; i < v.size(); i = i+2)
{
if (strcmp(v[i].name, name) == 0)
countnameflag = 1; /* 将flag置为1 防止重复打印相同的名字 */
else /* 如果当前客户的名字不同于前面客户的名字 */
{
countnameflag = 0;
strcpy(name, v[i].name);
}
if (countnameflag == 0) /* 如果是新用户，打印名字信息并且打印上个用户的total */
{
printf("Total amount: $%.2lf\n", TotalSum);
TotalSum = 0;
printf("%s %02d\n", v[i].name, v[i].month);
}
int p = 0;
double sum = 0;
if (v[i].H < v[i+1].H || v[i].D < v[i+1].D) /* 如果通话小时跨度超过2或跨越一天 */
{
/* 先把零头计算一下 */
sum += prize[v[i].H]*(60-v[i].M) + prize[v[i+1].H]*v[i+1].M;
if (v[i].D < v[i+1].D)
{ /* 如果电话的D值之差超过1，也即不是在一天内打完（比如昨晚23点到今天1点） */
for (int j = v[i].H+1; j < 24; j++)
p += prize[j];
for (int j = 0; j < v[i+1].H; j++)
p += prize[j];
p += (v[i+1].D - v[i].D - 1) * prize[24];
}
else
for (int j = v[i].H + 1; j < v[i+1].H; j++)
p += prize[j];
sum += p * 60;
}
else
sum += prize[v[i].H] * (v[i+1].M - v[i].M);
sum /= 100; /* 将cent->dollar */
printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2lf\n", v[i].D, v[i].H, v[i].M, v[i+1].D, v[i+1].H, v[i+1].M, v[i+1].time-v[i].time, sum);
TotalSum += sum;
}
printf("Total amount: $%.2lf\n", TotalSum);
return 0;
}
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
using namespace std;
struct customer
{
string name;//名字
int month,dd,hh,mm;//月 天 时 分
bool status;
};
bool cmp(customer a, customer b){
if(a.name != b.name)
return a.name<b.name;
else if(a.month!=b.month)
return a.month<b.month;
else if (a.dd!=b.dd)
return a.dd<b.dd;
else if(a.hh!=b.hh)
return a.hh<b.hh;
else
return a.mm<b.mm;
}
int prize[24];
double get_ans(customer a, customer b,int& time){
int money=0;
while (a.dd>b.dd||a.hh>b.hh||a.mm>b.mm)
{
time++;
money+=prize[b.hh];
b.mm++;
if (b.mm>=60)
{
b.mm=0;
b.hh++;
}
if (b.hh>=24)
{
b.hh=0;
b.dd++;
}
}
return money/100.0;
}
int main()
{
for (int i=0;i<24;i++)
cin>>prize[i];
int n;
string name,status;
int month,dd,hh,mm;
cin>>n;
customer data[n];
for(int i=0;i<n;i++)
{
cin>>name;
scanf("%d:%d:%d:%d",&month,&dd,&hh,&mm);
cin>>status;
data[i].name=name;
data[i].month=month;
data[i].dd=dd;
data[i].hh=hh;
data[i].mm=mm;
if(status.compare("on-line")==0)
data[i].status=true;
else
data[i].status=false;
}
sort(data,data+n,cmp);
//先读取第一个人
customer temp;
temp=data[0];
int on=-1,off;
bool flag=false;
if (temp.status)
on=0;
double total=0.0;
for (int i=1;i<n;i++)
{
if (data[i].name==temp.name)//若此数据名字和上一个数据相同
{
if (data[i].status)
on=i;//记录on下标
else
off=i;//记录off下标
if (off==on+1)//若此数据为off 上一个数据为on
{
if (!flag)
{
cout<<data[i].name<<" ";
printf("%02d\n",data[i].month);
flag=true;
}
int time=0;
double bills=0.0;
bills=get_ans(data[i],temp,time);
total += bills;
printf("%02d:%02d:%02d %02d:%02d:%02d",temp.dd,temp.hh,temp.mm,data[i].dd,data[i].hh,data[i].mm);
cout<<" "<<time;
printf(" $%.2f\n",bills);
}
}
else
{
if (flag)
{
printf("Total amount: $%.2f\n",total);
flag=false;
total=0.0;
}
}
temp=data[i];
if (data[i].status)
on=i;//记录on下标
else
off=i;//记录off下标
if (i==n-1)
{
if (flag)
printf("Total amount: $%.2f\n",total);
}
}
return 0;
}