标签：ges lock osi ecif ted case red minutes color
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.
Each input file contains one test case. Each case has two parts: the rate structure, and the phone call records.
The rate structure consists of a line with 24 non-negative integers denoting the toll (cents/minute) from 00:00 - 01:00, the toll from 01:00 - 02:00, and so on for each hour in the day.
The next line contains a positive number N (≤), followed by N lines of records. Each phone call record consists of the name of the customer (string of up to 20 characters without space), the time and date (mm:dd:hh:mm), and the word on-line or off-line.
For each test case, all dates will be within a single month. Each on-line record is paired with the chronologically next record for the same customer provided it is an off-line record. Any on-line records that are not paired with an off-line record are ignored, as are off-line records not paired with an on-line record. It is guaranteed that at least one call is well paired in the input. You may assume that no two records for the same customer have the same time. Times are recorded using a 24-hour clock.
For each test case, you must print a phone bill for each customer.
Bills must be printed in alphabetical order of customers‘ names. For each customer, first print in a line the name of the customer and the month of the bill in the format shown by the sample. Then for each time period of a call, print in one line the beginning and ending time and date (dd:hh:mm), the lasting time (in minute) and the charge of the call. The calls must be listed in chronological order. Finally, print the total charge for the month in the format shown by the sample.
10 10 10 10 10 10 20 20 20 15 15 15 15 15 15 15 20 30 20 15 15 10 10 10
10
CYLL 01:01:06:01 on-line
CYLL 01:28:16:05 off-line
CYJJ 01:01:07:00 off-line
CYLL 01:01:08:03 off-line
CYJJ 01:01:05:59 on-line
aaa 01:01:01:03 on-line
aaa 01:02:00:01 on-line
CYLL 01:28:15:41 on-line
aaa 01:05:02:24 on-line
aaa 01:04:23:59 off-line
CYJJ 01
01:05:59 01:07:00 61 $12.10
Total amount: $12.10
CYLL 01
01:06:01 01:08:03 122 $24.40
28:15:41 28:16:05 24 $3.85
Total amount: $28.25
aaa 01
02:00:01 04:23:59 4318 $638.80
Total amount: $638.80
模拟题，有点麻烦
1 #include<iostream> 2 #include<cstring> 3 #include<vector> 4 #include<algorithm> 5 using namespace std; 6 int rate[24]; 7 int n; 8 vector<string> namelist; 9 struct Record{ 10 string name; 11 int month; 12 int day; 13 int hour; 14 int minute; 15 bool line; 16 }record[1001]; 17 18 struct Bill{ 19 int sday; 20 int shour; 21 int sminute; 22 int eday; 23 int ehour; 24 int eminute; 25 }bill[501]; 26 27 int minutes; 28 float money; 29 30 void insertName(string name){ //将名字按字母序排列，并清除相同名字 31 if(namelist.size()==0){ 32 namelist.push_back(name); 33 }else if(namelist.size()==1){ 34 if(namelist[0]==name){ 35 return; 36 }else if(namelist[0]<name){ 37 namelist.push_back(name); 38 }else{ 39 namelist.insert(namelist.begin(),name); 40 } 41 }else{ 42 if(namelist.front()>name){ 43 namelist.insert(namelist.begin(),name); 44 }else if(namelist.back()<name){ 45 namelist.push_back(name); 46 }else{ 47 for(vector<string>::iterator it=namelist.begin();it!=namelist.end()-1;it++){ 48 vector<string>::iterator nex=it+1; 49 if(name==*it||name==*nex){ 50 return; 51 }else if(name>*it&&name<*nex){ 52 namelist.insert(nex,name); 53 return; 54 } 55 } 56 } 57 } 58 } 59 60 bool cmp(Record a,Record b){ //根据时间先后快排 61 int x=a.month*1000000+a.day*10000+a.hour*100+a.minute; 62 int y=b.month*1000000+b.day*10000+b.hour*100+b.minute; 63 if(x<y){ 64 return true; 65 }else{ 66 return false; 67 } 68 } 69 70 float calculateMoney(int sd,int sh,int sm,int ed,int eh,int em){ //计算时间与费用 71 if(ed-sd==0){ //当天开始当天结束 72 if(eh-sh==0){ 73 money=rate[sh]*(em-sm); 74 minutes=em-sm; 75 }else{ 76 minutes=60-sm+em+(eh-sh-1)*60; 77 money=(60-sm)*rate[sh]+em*rate[eh]; 78 for(int i=sh+1;i<eh;i++){ 79 money+=rate[i]*60; 80 } 81 } 82 }else{ //当天开始，非当天结束 83 if(sh==23){ 84 money=rate[23]*(60-sm); 85 minutes=60-sm; 86 }else{ 87 minutes=60-sm+(23-sh)*60; 88 money=(60-sm)*rate[sh]; 89 for(int i=sh+1;i<24;i++){ 90 money+=rate[i]*60; 91 } 92 } 93 if(eh==0){ 94 money+=rate[0]*em; 95 minutes+=em; 96 }else{ 97 minutes+=eh*60+em; 98 money+=rate[eh]*em; 99 for(int i=0;i<eh;i++){ 100 money+=rate[i]*60; 101 } 102 } 103 for(int i=sd+1;i<ed;i++){ 104 minutes+=60*24; 105 for(int j=0;j<24;j++){ 106 money+=rate[j]*60; 107 } 108 } 109 } 110 } 111 112 int main(){ 113 for(int i=0;i<24;i++){ 114 cin>>rate[i]; 115 } 116 cin>>n; 117 for(int i=0;i<n;i++){ 118 string l; 119 cin>>record[i].name; 120 insertName(record[i].name); 121 scanf("%d:%d:%d:%d",&record[i].month,&record[i].day,&record[i].hour,&record[i].minute); 122 cin>>l; 123 if(l=="on-line"){ 124 record[i].line=true; 125 }else{ 126 record[i].line=false; 127 } 128 } 129 sort(record,record+n,cmp); //根据时间对记录快排 130 for(vector<string>::iterator it=namelist.begin();it!=namelist.end();it++){ //根据姓名和月份，记录开始、结束的时间 131 for(int i=1;i<=12;i++){ 132 memset(bill,0,sizeof(bill)); 133 int c=0; 134 bool on=true; 135 for(int j=0;j<n;j++){ 136 if(record[j].name==*it&&record[j].month==i&&on==true&&record[j].line==true){ 137 bill[c].sday=record[j].day; 138 bill[c].shour=record[j].hour; 139 bill[c].sminute=record[j].minute; 140 on=false; 141 }else if(record[j].name==*it&&record[j].month==i&&on==false&&record[j].line==true){ 142 bill[c].sday=record[j].day; 143 bill[c].shour=record[j].hour; 144 bill[c].sminute=record[j].minute; 145 }else if(record[j].name==*it&&record[j].month==i&&on==false&&record[j].line==false){ 146 bill[c].eday=record[j].day; 147 bill[c].ehour=record[j].hour; 148 bill[c].eminute=record[j].minute; 149 on=true; 150 c++; 151 } 152 } 153 if(c>0){ //开始输出结果 154 cout<<*it<<" "; 155 if(i<10){ 156 cout<<"0"<<i<<endl; 157 }else{ 158 cout<<i<<endl; 159 } 160 float sum=0.0; 161 for(int j=0;j<c;j++){ 162 minutes=0; 163 money=0.0; 164 calculateMoney(bill[j].sday,bill[j].shour,bill[j].sminute,bill[j].eday,bill[j].ehour,bill[j].eminute); 165 if(bill[j].sday<10){ 166 cout<<"0"<<bill[j].sday<<":"; 167 }else{ 168 cout<<bill[j].sday<<":"; 169 } 170 if(bill[j].shour<10){ 171 cout<<"0"<<bill[j].shour<<":"; 172 }else{ 173 cout<<bill[j].shour<<":"; 174 } 175 if(bill[j].sminute<10){ 176 cout<<"0"<<bill[j].sminute<<" "; 177 }else{ 178 cout<<bill[j].sminute<<" "; 179 } 180 if(bill[j].eday<10){ 181 cout<<"0"<<bill[j].eday<<":"; 182 }else{ 183 cout<<bill[j].eday<<":"; 184 } 185 if(bill[j].ehour<10){ 186 cout<<"0"<<bill[j].ehour<<":"; 187 }else{ 188 cout<<bill[j].ehour<<":"; 189 } 190 if(bill[j].eminute<10){ 191 cout<<"0"<<bill[j].eminute<<" "; 192 }else{ 193 cout<<bill[j].eminute<<" "; 194 } 195 printf("%d $%.2f\n",minutes,money/100); 196 sum+=money/100; 197 } 198 printf("Total amount: $%.2f\n",sum); 199 } 200 } 201 } 202 return 0; 203 }
标签：ges lock osi ecif ted case red minutes color
原文地址：https://www.cnblogs.com/orangecyh/p/10308245.html
