PAT甲级1016题（电话账单）

字符串处理题

PAT

发布日期: 2021-02-23

文章字数: 331

阅读时长: 1 分

阅读次数:

1.题目介绍

1016

2.考察点，难度

字符串处理类，vector容器与结构体，字符串格式化转换函数，map容器，，难度难

3.解题代码

#include <bits/stdc++.h>
using namespace std;


struct Person{
    string name;
    string state;
    string format_time;
    int month;
    int time;
    bool operator<(const Person& t) const{
        return time<t.time;
    }
};

const int M=1440*31+10;
int fee[24];
double cost[M];
map<string,vector<Person>> m;

int main(){

    freopen("D:\\PAT\\Clion\\in.txt","r",stdin);
    for(int i=0;i<24;i++)   cin>>fee[i];
    for(int i=1;i<M;i++)    cost[i]=cost[i-1]+fee[(i-1)%1440/60]/100.0;

    int N;
    cin>>N;
    while(N--){
        char a[21],b[10],formaltime[20];
        int month,day,hour,minute,all_time;
        scanf("%s %d:%d:%d:%d %s",a,&month,&day,&hour,&minute,b);
        sprintf(formaltime,"%02d:%02d:%02d",day,hour,minute);
        all_time=(day-1)*1440+hour*60+minute;
        m[a].push_back({a,b,formaltime,month,all_time});
    }

    for(auto p:m){
        int total=0;
        double sum=0;
        sort(p.second.begin(),p.second.end());
        for(int i=0;i+1<p.second.size();i++){

            if(p.second[i].state=="on-line"&&p.second[i+1].state=="off-line"){
                if(total==0)   {
                    printf("%s %02d\n",p.first.c_str(),p.second[i].month);
                    total++;
                }
                cout<<p.second[i].format_time<<" "<<p.second[i+1].format_time;
                double out=cost[p.second[i+1].time]-cost[p.second[i].time];
                sum+=out;
                printf(" %d $%.2lf\n",p.second[i+1].time-p.second[i].time,out);
            }
        }
        if(total)   printf("Total amount: $%.2f\n",sum);

    }
    return 0;
}