【题解搬运】PAT_A1016 Phone Bills

从我原来的博客上搬运。原先blog作废。

题目

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.

题解

繁琐，但是不难。合理运用封装能够减少代码量。

剩下的是两个注意事项：1.是连续的on-off算作一个记录。一个例子：

01 on 02 on 03 off 04 off 那么02-03算作一个记录。

2.必须要注意这个情况，就是如果没有合法的记录的话就不打印账单！！！不论是一开始的还是tot Amount！我第二个一开始忘了不打印了！！

以上。

代码

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

#define f1(x,y) for(int x=1;x<=y;++x)
#define f0(x,y) for(int x=0;x!=y;++x)
#define bf1(x,y,z) for(int x=y;x>=z;--x)
#define bf0(x,y,z) for(int x=y;x!=z;--x)
typedef long long ll;
typedef unsigned long long ull;
int fee[25];
int ohrfee[25]={0};//fee of 00:00~i:00
int dayfee=0;
struct Record
{
    int month;
    int recMin;
    bool isOn;
    Record(int _m=1,int _r=0,bool _o=true):
        month(_m),recMin(_r),isOn(_o)
        {}
    bool operator < (const Record& rhs) const
    {
        return recMin<rhs.recMin;
    }
};
typedef pair<Record,Record> Call;
struct User
{
    string name;
    int month;
    vector<Record> rec;
    vector<Call> cal;
    User(string _n):name(_n) {}
    void calc()
    {
        sort(rec.begin(),rec.end());
        int marked[1005];
        memset(marked,-1,sizeof(marked));
        month=rec[0].month;
        bool isOk=false;
        f0(i,rec.size()) if(rec[i].isOn && i+1<rec.size() && !(rec[i+1].isOn))
        {
            cal.push_back(make_pair(rec[i],rec[i+1]));
        }
        return;
    }
};
vector<User> u;
map<string,int> idx;
int main()
{
    int f;
    f0(i,24)
    {
        scanf("%d",&f);
        fee[i]=f;
        if(i==0) ohrfee[i]=0;
        else ohrfee[i]=ohrfee[i-1]+fee[i-1]*60;
        dayfee+=fee[i]*60;
    }
    int n; scanf("%d",&n);
    f1(i,n)
    {
        char name[25],status[10];
        int mon,d,h,minute;
        scanf("%s %d:%d:%d:%d %s",name,&mon,&d,&h,&minute,status);
        if(idx.find(string(name))==idx.end())
        {
            u.push_back(User(string(name)));
            idx[string(name)]=u.size()-1;
            u.back().rec.push_back(Record(mon,d*1440+h*60+minute,status[1]=='n'));
        }
        else u[idx[string(name)]].rec.push_back(Record(mon,d*1440+h*60+minute,status[1]=='n'));
    }
    for(auto iter=u.begin();iter!=u.end();++iter) iter->calc();
    for(auto iter=idx.begin();iter!=idx.end();++iter)
    {
        int num=iter->second;
        string nam=iter->first;
        User& usr=u[num];
        vector<Call>& c=usr.cal;
        if(c.size()!=0) printf("%s %02d\n",nam.c_str(),usr.month);
        int tot=0;//100cents->1$
        f0(i,c.size())
        {
            int fd,fh,fm,sd,sh,sm,cost;
            fd=c[i].first.recMin/1440;fh=(c[i].first.recMin%1440)/60;fm=(c[i].first.recMin%60);
            sd=c[i].second.recMin/1440;sh=(c[i].second.recMin%1440)/60;sm=(c[i].second.recMin%60);
            cost=(sd-fd)*dayfee+(ohrfee[sh]+sm*fee[sh])-(ohrfee[fh]+fm*fee[fh]);
            //printf("2:ohrfee[%d]=%d,sm=%d,fee[%d]=%d.\n",sh,ohrfee[sh],sm,sh,fee[sh]);
            //printf("1:ohrfee[%d]=%d,sm=%d,fee[%d]=%d.\n",fh,ohrfee[fh],fm,fh,fee[fh]);
            printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2f\n",fd,fh,fm,
                                                              sd,sh,sm,
                                                              c[i].second.recMin-c[i].first.recMin,cost/100.0);
            tot+=cost;
        }
        if(c.size()!=0) printf("Total amount: $%.2f\n",tot/100.0);
    }
    return 0;
}