1016 Phone Bills (25分)

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<string>
 4 #include<vector>
 5 #include<map>
 6 #include<algorithm>
 7 using namespace std;
 8 double rate[25]={0};
 9 struct CALL{
10     string name;
11     int mon,dd,hh,mm;
12     int status;
13     int time;
14 };
15 int toMin(CALL call){//将时间转化为距离月初的时间 
16     return call.dd*24*60+call.hh*60+call.mm;
17 }
18 bool cmp(CALL a,CALL b){
19     if(a.name!=b.name)
20       return a.name<b.name;
21     else
22       return a.time<b.time;
23 }
24 double getCost(CALL x){
25     double ans=0;
26     ans=ans+rate[x.hh]*x.mm;//分钟 
27     for(int i=0;i<x.hh;i++){//小时 
28         ans=ans+rate[i]*60;
29     }
30     ans=ans+rate[24]*60*x.dd;
31     return ans; 
32 }
33 double Cost(CALL x,CALL y){
34     return (getCost(x)-getCost(y));
35 }
36 int main(){ 
37     
38     for(int i=0;i<24;i++){
39         cin>>rate[i];
40         rate[24]+=rate[i];//小技巧 
41     }
42     int N;
43     int mon,dd,hh,mm;
44     string name,onoff;
45     cin>>N;
46     
47     vector<CALL> A(N);
48     for(int i=0;i<N;i++){
49         cin>>A[i].name;
50         scanf("%d:%d:%d:%d",&A[i].mon,&A[i].dd,&A[i].hh,&A[i].mm);
51         cin>>onoff;
52         if(onoff=="on-line")
53            A[i].status=1;
54         else
55            A[i].status=0;
56         A[i].time=toMin(A[i]);
57     }
58     sort(A.begin(),A.end(),cmp);
59     double Total=0;
60     map<string,vector<CALL> > B;
61     for(int i=1;i<N;i++){
62         if(A[i].name==A[i-1].name&&A[i].status==0&&A[i-1].status==1){
63             B[A[i].name].push_back(A[i-1]);
64             B[A[i].name].push_back(A[i]);
65         }
66     }
67     for(map<string,vector<CALL> >::iterator it=B.begin();it!=B.end();it++){
68         cout<<it->first<<" ";
69         printf("%02d\n",A[0].mon);
70         vector<CALL> temp=it->second;
71         double total=0;
72         for(int i=0;i<temp.size();i++){
73             printf("%02d:%02d:%02d ",temp[i].dd,temp[i].hh,temp[i].mm);
74             if(i%2!=0)
75             { printf("%d ",temp[i].time-temp[i-1].time);
76               total+=Cost(temp[i],temp[i-1]);
77               printf("$%.2f\n",Cost(temp[i],temp[i-1])/100)    ;
78             }
79         }
80         printf("%s%.2f\n","Total amount: $",total/100);
81         
82     }
83    
84     return 0;
85 }