PAT甲题题解-1016. Phone Bills (25)-模拟、排序

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define ONLINE 0
#define OFFLINE 1
using namespace std;
const int maxn=1000+5;
const int DAYMINUTE=24*60;
const int HOURMINUTE=60;
int toll[24];
int n;
struct Record{
    char name[25];
    int time;
    int month;
    int day;
    int hour;
    int minute;
    int mark;
    bool operator<(const Record tmp)const{
        if(strcmp(name,tmp.name)==0)
            return time<tmp.time;
        else if(strcmp(name,tmp.name)<0)
            return true;
        else
            return false;
    }
}record[maxn];

struct Phone{
    char name[25];
    int month;
    int d1,h1,m1;
    int d2,h2,m2;
    int time1,time2;
}phone[maxn];
int cnt=0;
/**
求出第i个配对的通话时间费用
*/
int cal(int i){
    int sum=0;
    int start=phone[i].time1; //起始时间
    int ends=phone[i].time2;  //结束时间
    int tmp=0; //统计每个小时的时间段内，通话的分钟
    int idx; //对应时间段toll的索引
    for(int t=start;t<=ends;t++){
        if(t%60==0){
            if(t%DAYMINUTE==0)
                idx=(DAYMINUTE-1)/60;  //如果模为0，就应该是DAYMINUTE
            else
                idx=(t%DAYMINUTE-1)/60; //因为可能会有连续通话了好几天，所以得取模一下
            sum+=tmp*toll[idx];  //通话的分钟*该时间段的费用
            tmp=1; //为了方便起见，像01:06:00这种整时的，算作下一小时的
        }
        else
            tmp++;
    }
    if(tmp){
        //比如说1:08:03，由于1:08:00的时候算作8-9之间的，实际上统计的tmp=4，所以要-1
        sum+=(tmp-1)*toll[(ends%DAYMINUTE)/60];
    }
    return sum;
}
int main()
{
    char word[10];
    for(int i=0;i<24;i++)
        scanf("%d",&toll[i]);
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%s %d:%d:%d:%d %s",record[i].name,&record[i].month,&record[i].day,&record[i].hour,&record[i].minute,word);
        record[i].time=(record[i].day-1)*DAYMINUTE+record[i].hour*HOURMINUTE+record[i].minute;
        if(word[1]==‘n‘)
            record[i].mark=ONLINE;
        else
            record[i].mark=OFFLINE;
    }
    sort(record,record+n);
    int last=-1; //之前最近的一个on-line的索引
    //找出配对的on-line和off-line，存入phone数组，方便后序处理
    for(int i=0;i<n;i++){
        if(record[i].mark==ONLINE){
            last=i;
        }
        if(record[i].mark==OFFLINE && last!=-1){
            if(strcmp(record[i].name,record[last].name)==0){
                strcpy(phone[cnt].name,record[i].name);
                phone[cnt].month=record[i].month;
                phone[cnt].d1=record[last].day;
                phone[cnt].h1=record[last].hour;
                phone[cnt].m1=record[last].minute;
                phone[cnt].time1=record[last].time;
                phone[cnt].d2=record[i].day;
                phone[cnt].h2=record[i].hour;
                phone[cnt].m2=record[i].minute;
                phone[cnt].time2=record[i].time;
                cnt++;
                last=-1;
            }
        }
    }
    int tot=0,sum=0;
    for(int i=0;i<cnt;i++){
        if(i==0){
            printf("%s %02d\n",phone[i].name,phone[i].month);
            sum=cal(i);
            tot+=sum;
            int len=phone[i].time2-phone[i].time1;
            double cost=sum*1.0/100;
            printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2lf\n",phone[i].d1,phone[i].h1,phone[i].m1,phone[i].d2,phone[i].h2,phone[i].m2,len,sum*1.0/100);
        }
        else{
            if(strcmp(phone[i].name,phone[i-1].name)!=0){
                printf("Total amount: $%.2lf\n",tot*1.0/100);
                printf("%s %02d\n",phone[i].name,phone[i].month);
                tot=0;
            }
            sum=cal(i);
            tot+=sum;
            int len=phone[i].time2-phone[i].time1;
            int cost=sum*1.0/100;
            printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2lf\n",phone[i].d1,phone[i].h1,phone[i].m1,phone[i].d2,phone[i].h2,phone[i].m2,len,sum*1.0/100);
        }
    }
    if(tot){
        printf("Total amount: $%.2lf\n",tot*1.0/100);
    }
    return 0;
}