PAT 甲级 1016 Phone Bills (25 分) （结构体排序，模拟题，巧妙算时间，坑点太多，debug了好久）

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 (≤), 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-linerecord. 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

0  、1测试点：没有消费的用户不能输出

2测试点：on和off日期不在同一天却在同一个小时

3测试点：on和off日期同一天同一个小时

题意：

一个长途电话公司收费规则如下：

长途电话每分钟会有固定的收费，这取决于打电话时的时间。当一个顾客的长途电话接通时，这一刻的时间会记录下来，同样顾客挂断电话时也是这样。每个月，每分钟通话的电话账单就会寄给顾客。你的工作就是准备每月的账单，给出一系列电话通话记录。

每一个测试样例都有两个部分：费率结构和通话记录

费率结构包含了一行24个非负整数记录了一天24个小时每小时的长途电话费(美分/每分钟)。

下一行包含了一个正数N（≤1000），之后有N条记录，每一条电话通话记录包含了顾客的姓名(一个20个字符的字符串，没有空格) 时间（mm:dd:hh:mm）和文字on-line或者是off-line

对于每一个测试样例，所有的数据都是一个月内的。每一条on-line记录会有按时间先后排列的下一条记录（相同顾客名）状态为off-line和它构成一对。若有些on-line记录并没有可以配对的off-line记录，则忽视这条记录，同样若只有off-line记录，没有与之对应的on-line记录也忽视。可以保证至少有一条电话记录能够配对。你可以假设在同一时间同一顾客不会有两条记录。使用24小时时钟来记录时间

对于每一个测试样例，你需要为每一个顾客打印出电话账单

电话账单打印顺序按照顾客姓名的字母表顺序。对每一个顾客，首先打印出顾客名字和账单的月份。然后对每一个通话期间，用一行打印出开始时间和结束时间和日期（dd:hh:mm），持续时间(以分钟为单位) ，收费。所有通话按照时间先后顺序列出。最后打印出这个月的总费用

思路：

1.先把所有数据排个序，先按名字排，再时间

2.遍历，用个is_out标记是不是已经匹配且输出过(用于判断该不该打印Total money)，用last记录上一次的状态，如果上一次状态是on-line且这次是off-line，那么{

　　再判断有没有答应过名字(is_out)

　　然后开始算时间和钱：时间和钱不用1分钟1分钟遍历的算，先让on的天数和小时数和off的相等，然后再减去多算的小时数，最后考虑分钟

　　//分钟
　　money-=last.m*p[last.h];//多的减去，注意是 p[last.h]
　　money+=a[i].m*p[a[i].h];//少的加上，注意是 p[a[i].h]

}

3.字符串那里月份的处理卡了。。。哭泣

把整数加到字符串上

#include<bits/stdc++.h>
using namespace std;
int main(){

    string s="";
    s+="";
    int a=;
    s+=a+'';
    cout<<s<<endl;
    return ;
 } 

4.is_out是换一个名字就要初始化的，忘了。。。难过。。。

最终满分代码：

#include<bits/stdc++.h>
using namespace std;
struct node{
    string name;
    int d;
    int h;
    int m;
    string state;
}a[];
bool cmp(node &x,node &y){
    if(x.name==y.name){
        if(x.d==y.d){
            if(x.h==y.h){
                return x.m<y.m;
            }else{
                return x.h<y.h;
            }
        }else{
            return x.d<y.d;
        }
    }else{
        return x.name<y.name;
    }
}

int main()
{
    double p[];
    double sum_p=;
    for(int i=;i<=;i++){
        cin>>p[i];
        p[i]*=0.01;
        sum_p+=p[i]*;
    }
    int n;
    int month;
    cin>>n;
    for(int i=;i<=n;i++){
        cin>>a[i].name;
        scanf("%2d:%2d:%2d:%2d",&month,&a[i].d,&a[i].h,&a[i].m);
        cin>>a[i].state;
    }
    sort(a+,a++n,cmp);
    /*for(int i=1;i<=n;i++){
        cout<<a[i].name<<" "<<a[i].d<<" "<<a[i].h<<" "<<a[i].m<<" "<<a[i].state<<endl;
    }*/
    node last;
    last.name="";
    int is_out=;//坑点！是否有匹配上且输出的
    double totel=;
    for(int i=;i<=n;i++){
        if(i==){
            last.name=a[i].name;
            last.d=a[i].d;
            last.h=a[i].h;
            last.m=a[i].m;
            last.state=a[i].state;
            is_out=;
            continue;
        }
        if(a[i].name!=last.name){
            if(is_out)
            {
                //结算前一个人
                printf("Total amount: $%.2lf\n",totel);
                totel=;
            }
            is_out=; //这句导致第1，2个测试点过不了，名字一旦不同，不管有没有输出过，is_out要初始化
        }else{
            //看看状态对不对的上，对不上为"";
            if(last.state=="on-line"&&a[i].state=="off-line"){
                //计算时间
                int t=a[i].d-last.d;
                int t1=t**;
                int t2=(a[i].h-last.h)*;
                int t3=(a[i].m-last.m);
                //计算bill，不需要一分钟一分钟的跑，找到规律即可
                double money=;
                money+=t*sum_p;
                //小时
                if(a[i].h>last.h)//比大小
                {
                    for(int j=last.h;j<a[i].h;j++){
                        money+=p[j]*;
                    }
                }
                else{
                    for(int j=a[i].h;j<last.h;j++){
                        money-=p[j]*;
                    }
                }
                //分钟
                money-=last.m*p[last.h];//多的减去，注意是 p[last.h]
                money+=a[i].m*p[a[i].h];//少的加上，注意是 p[a[i].h]
                //如果这个人没有被输出过
                if(is_out==)
                {
                    cout<<a[i].name<<" ";
                    printf("%02d\n",month);
                    is_out=;//匹配上且输出了
                }
                printf("%02d:%02d:%02d %02d:%02d:%02d ",last.d,last.h,last.m,a[i].d,a[i].h,a[i].m);
                printf("%d $%.2lf\n",t1+t2+t3,money);
                totel+=money;
            }
        }
        last.name=a[i].name;
        last.d=a[i].d;
        last.h=a[i].h;
        last.m=a[i].m;
        last.state=a[i].state;
    }
    if(is_out){//匹配上的才输出
        printf("Total amount: $%.2lf\n",totel);
    }
    return ;
 } 

我自己编的测试数据

aaa ::: off-line
aaa ::: off-line
aaa ::: on-line
aaa ::: on-line

aaa ::: on-line
aaa ::: off-line
aaa ::: off-line
aaa ::: off-line

aaa ::: on-line
aaa ::: off-line

CYLL ::: on-line
CYLL ::: off-line
CYJJ ::: off-line
CYLL ::: off-line
CYJJ ::: on-line
aaa ::: on-line
aaa ::: on-line
CYLL ::: on-line
aaa ::: on-line
aaa ::: off-line

CYJJ ::: off-line
CYJJ ::: on-line
CYJJ ::: on-line
CYJJ ::: off-line

CYLL ::: on-line
CYLL ::: off-line
CYLL ::: on-line
CYLL ::: off-line
CYLL ::: on-line
CYLL ::: off-line
CYLL ::: on-line
CYLL ::: on-line
CYLL ::: off-line
CYLL ::: off-line
CYLL ::: off-line
CYLL ::: off-line
MQ ::: on-line
MQ ::: on-line
MQ ::: on-line
MQ ::: off-line
YF ::: on-line
YF ::: on-line

看着真舒服！！！