PAT 甲级 1016. Phone Bills (25)

题目：点击打开链接

思路：1.根据输入，将通话信息根据用户名、时间进行排序

            2.排序完后，进行信息筛选：在计算单次通话话费时，先判断两条信息是否属于同一用户，再判断其是否为一on-line ,一off-line的情况；

            3.在计算话费时，注意时间跨度

代码：

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<cstdio>
using namespace std;

//int toll[24]={10,10,10,10,10,10,20,20,20,15,15,15,15,15,15,15,20,30,20,15,15,10,10,10};
int toll[24];
struct customer
{
string name;
string month;
string t;
int time;     //时间记录
bool record; //拨打or挂断
bool operator<(const customer &a)const
{
if(name!=a.name)
return name<a.name;
else
return time<a.time;
}

};

//计算时间
int cal_time(string a)
{
int time=0;
time=(a[9]-'0')*10+a[10]-'0';
time+=((a[6]-'0')*10+a[7]-'0')*60;
time+=((a[3]-'0')*10+a[4]-'0')*60*24;
return time;
}
//计算话费,这里要注意时间隔天的情况
int cal_cost(int begin_t,int end_t)
{
int c=0;
int h1=(begin_t/60)%24;   //起始小时
int t=end_t-begin_t;
begin_t-=(begin_t/60/24*60*24);
end_t=begin_t+t;
int h2;
while(h1<(end_t/60))
{
h2=h1%24;
c+=toll[h2]*((h1+1)*60-begin_t);
++h1;
begin_t=h1*60;
}
h2=h1%24;
c+=toll[h2]*(end_t-begin_t);
return c;

}

int main()
{
//输入
int i=0;
for(;i<24;++i)
scanf("%d",toll+i);
int N;
cin>>N;
vector<customer> C(N);
//信息输入
string T,R;
for(i=0;i<N;++i)
{
cin>>C[i].name>>T>>R;
C[i].time=cal_time(T);
C[i].month=T.substr(0,2);    //这两部分保存，便于输出
C[i].t=T.substr(3,8);
if(R=="on-line")
C[i].record=0;
else
C[i].record=1;
}

sort(C.begin(),C.end()); //排序

string name;
int total_time=0;
int total_cost=0;
int cost;
bool flag=0;
for(i=1;i<N;++i)
{
if(C[i].record==1 && C[i-1].record==0 && C[i].name==C[i-1].name)//表明信息有效
{
if(C[i].name!=name)     //出现换人的情况
{
if(flag)            //判断是否为首次输出，若为首次输出，则这一部分不必输出
{
cout<<"Total amount: $";
printf("%d.%02d\n",total_cost/100,total_cost%100);
total_cost=0;
}
name=C[i].name;
cout<<name<<" "<<C[i].month<<endl;
flag=1;
}
total_time=C[i].time-C[i-1].time;
cost=cal_cost(C[i-1].time,C[i].time);
total_cost+=cost;
cout<<C[i-1].t<<" "<<C[i].t<<" ";
printf("%d $%d.%02d\n",total_time,cost/100,cost%100);

}

}
cout<<"Total amount: $";
printf("%d.%02d\n",total_cost/100,total_cost%100);
system("pause");
return 0;
}