PAT(甲级)1016. Phone Bills (25)

Phone Bills (25)

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 12.10Totalamount:  12.10
CYLL 01
01:06:01 01:08:03 122 24.4028:15:4128:16:0524  3.85
Total amount: 28.25aaa0102:00:0104:23:594318  638.80
Total amount: $638.80

题目大意：给出n个电话记录和24个小时每个小时单位时间花费 保证每个人的电话记录在一个月内 找出相互匹配的记录 计算通话时间和花费 保证至少存在一个匹配的通话记录
分析：模拟 写了两遍 第一遍15分 但是没找到错。。。。然后从头换思路写 看到25分的一瞬间很惊喜 不敢相信。。。。

#include 
#include 
#include 
#include 
#include 
#define maxn 1010
using namespace std;
int n,sal[24];
struct node{
    char name[30];
    int mm,dd,hh,mn;//记录月份 日期 小时 分钟
    int state;//记录开始 结束 电话
    int flag;//记录是否输出 以及开始 结束电话
    int totaltime;//一通电话的时间
    double spend;//一通电话的花费
};
node rec[maxn];
int cmp(node a,node b){//排序
    if(strcmp(a.name, b.name) == 0){
        if(a.mm == b.mm){
            if(a.dd == b.dd){
                if(a.hh == b.hh)
                    return a.mn < b.mn;
                return a.hh < b.hh;
            }
            return a.dd < b.dd;
        }
        return a.mm < b.mm;
    }
    return strcmp(a.name, b.name) < 0;
}
void caculat(int s,int e){//计算一通电话的时间 花费
    int sumtime = (rec[e].dd - rec[s].dd) * 24 * 60 + (rec[e].hh - rec[s].hh) * 60 + rec[e].mn - rec[s].mn;
    rec[e].totaltime = sumtime;
    int j=0,sd=rec[s].dd,sh=rec[s].hh,sm=rec[s].mn;
    double sum=0;
    for(int i = 0; i < sumtime; ++i){
        j++, sm++;
        if(sm == 60){
            sum += (j * sal[sh] / 100.0);
            sh++;
            sm = 0;
            j = 0;
        }
        if(sh == 24){
            sd++;
            sh = 0;
        }
    }
    sum += (j * sal[sh] / 100.0);
    rec[e].spend = sum;
    return ;
}
int judge(double x){//判0
    if(x < 1e-5 && x > -1e-5)
        return 0;
    return 1;
}
int main()
{
    char temstate[20];
    for(int i = 0; i < 24; ++i){
        scanf("%d", &sal[i]);
    }
    scanf("%d", &n);
    for(int i = 0; i < n; ++i){
        scanf("%s", rec[i].name);
        scanf("%d:%d:%d:%d", &rec[i].mm, &rec[i].dd, &rec[i].hh, &rec[i].mn);
        scanf("%s", temstate);
        if(temstate[1] == 'n')
            rec[i].state = 1;
        else
            rec[i].state = -1;
        rec[i].flag = 0;
        rec[i].spend = 0;
    }
    sort(rec, rec + n, cmp);
    for(int i = 0; i < n; ++i){
        while(i < n && strcmp(rec[i].name, rec[i + 1].name) == 0){
            if(rec[i].state == 1 && rec[i + 1].state == -1){
                caculat(i, i + 1);
                if(judge(rec[i + 1].spend)){
                    rec[i].flag = 1;//开始电话
                    rec[i + 1].flag = -1;//结束电话
                }
                i += 2;
            }
            else
                i++;
        }
    }
    for(int i = 0; i < n; ++i){
        int flag = 0, j = i;
        double sum=0;
        while(j < n && strcmp(rec[i].name, rec[++j].name) == 0);
        while( i < j){
            if(!flag && rec[i].flag == 1){
                flag=1;
                printf("%s %02d\n", rec[i].name, rec[i].mm);
            }
            if(rec[i].flag == 1)
                printf("%02d:%02d:%02d ", rec[i].dd, rec[i].hh, rec[i].mn);
            if(rec[i].flag == -1){
                sum += rec[i].spend;
                printf("%02d:%02d:%02d %d $%.2lf\n", rec[i].dd, rec[i].hh, rec[i].mn, rec[i].totaltime, rec[i].spend);
            }
            i++;
        }
        if(flag){
            printf("Total amount: $%.2lf\n", sum);
        }
        i--;
    }
    return 0;
}