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;elsereturn 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;elseC[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;}