立志用最少的代碼做最高效的表達(dá)

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PAT甲級(jí)最優(yōu)題解——>傳送門

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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.

Input Specification:
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 (≤1000), 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.

Output Specification:
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.

Sample Input:
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

Sample Output:
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

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注意

如果某一個(gè)人雖然有賬單，但是沒有找到有效的通話記錄，這個(gè)人不予輸出

輸入的一天內(nèi)各個(gè)小時(shí)段的話費(fèi)單位是美分/分鐘，所以在計(jì)算的時(shí)候一個(gè)小時(shí)的話費(fèi)應(yīng)該是60*price[i]，另外輸出費(fèi)用要按美元計(jì)算，所以要除以100，不妨在輸入一天內(nèi)各個(gè)小時(shí)段的話費(fèi)時(shí)就直接進(jìn)行除以100的操作

輸出月、日、時(shí)、分都必須有兩位數(shù)字不夠的要在高位補(bǔ)0

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#include<bits/stdc++.h> using namespace std; struct Time{//定義時(shí)間類int month,day,hour,minute,time,online=0;//月日時(shí)分信息、距0日0時(shí)0分的分鐘數(shù)、指明這一時(shí)間是online還是offlinebool operator <(const Time&t)const{//重載<運(yùn)算符return this->time<t.time;} }; double price[25];//存儲(chǔ)一天內(nèi)各個(gè)小時(shí)段的話費(fèi)單位是美元/分鐘 double compute(Time t,int day){//計(jì)算當(dāng)前時(shí)間到day日0時(shí)0分所用話費(fèi)double bill=0.0;for(int i=0;i<t.hour;++i)bill+=60*price[i];return bill+t.minute*price[t.hour]+price[24]*60*(t.day-day); } int main(){for(int i=0;i<24;++i){//讀取一天內(nèi)各個(gè)小時(shí)段的話費(fèi)scanf("%lf",&price[i]);price[i]/=100.0;//化為美元price[24]+=price[i];}int N;scanf("%d",&N);map<string,set<Time>>bill;//存儲(chǔ)每個(gè)人的名字和對(duì)應(yīng)的話費(fèi)時(shí)間for(int i=0;i<N;++i){//讀取賬單string s1,s2;Time t;cin>>s1;scanf("%d:%d:%d:%d",&t.month,&t.day,&t.hour,&t.minute);t.time=(t.day*24+t.hour)*60+t.minute;cin>>s2;if(s2=="on-line")t.online=1;bill[s1].insert(t);}for(auto i=bill.cbegin();i!=bill.cend();++i){//查找有效的話費(fèi)記錄并就算用時(shí)與話費(fèi)并輸出bool output=false;//該人的話費(fèi)記錄是否需要輸出/是否有有效的話費(fèi)記錄double sumBill=0.0;//該人的話費(fèi)總額for(auto j=(i->second).cbegin();j!=(i->second).cend();++j){//定義set的迭代器jauto jnext=j;//定義指向j指向的下一個(gè)對(duì)象的迭代器++jnext;for(;jnext!=(i->second).cend()&&!(j->online&&!jnext->online);++j,++jnext);//查找有效的話費(fèi)記錄if(jnext!=(i->second).cend()){//如果查找到了if(!output){//還沒有輸出過姓名的月份進(jìn)行輸出，并通過修改output變量標(biāo)明該人的話費(fèi)記錄需要輸出output=true;printf("%s %02d\n",(i->first).c_str(),j->month);}printf("%02d:%02d:%02d %02d:%02d:%02d ",j->day,j->hour,j->minute,jnext->day,jnext->hour,jnext->minute);//輸出話費(fèi)賬單起止時(shí)間int t=jnext->time-j->time;//計(jì)算話費(fèi)double bill=compute(*jnext,j->day)-compute(*j,j->day);//計(jì)算話費(fèi)賬單用時(shí)sumBill+=bill;//加到話費(fèi)總額上printf("%d $%.2f\n",t,bill);//輸出}}if(output)//如果該人的話費(fèi)記錄需要輸出，輸出話費(fèi)總額printf("Total amount: $%.2f\n",sumBill);}return 0; }

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耗時(shí)

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痛苦難道是白忍受的嗎？它應(yīng)該使我偉大！???????——托馬斯 ? 曼

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