问题描述
| 试题编号： | 201712-3 |
| 试题名称： | Crontab |
| 时间限制： | 10.0s |
| 内存限制： | 256.0MB |
| 问题描述： |
样例输入
3 201711170032 201711222352
0 7 * * 1,3-5 get_up 30 23 * * Sat,Sun go_to_bed 15 12,18 * * * have_dinner 样例输出
201711170700 get_up
201711171215 have_dinner 201711171815 have_dinner 201711181215 have_dinner 201711181815 have_dinner 201711182330 go_to_bed 201711191215 have_dinner 201711191815 have_dinner 201711192330 go_to_bed 201711200700 get_up 201711201215 have_dinner 201711201815 have_dinner 201711211215 have_dinner 201711211815 have_dinner 201711220700 get_up 201711221215 have_dinner 201711221815 have_dinner |
1 #include<bits/stdc++.h> 2 #define pb emplace_back 3 using namespace std; 4 typedef vector<string> vecstr; 5 const char Month[14][4]={"","jan","feb","mar","apr","may","jun","jul","aug","sep","oct","nov","dec"}; 6 const char Week[8][4]={"sun","mon","tue","wed","thu","fri","sat"}; 7 int n,cnt,Daysofmon[14]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 8 map<string,int> weekmap,monthmap; map<string,vecstr > all; 9 string stime,etime; 10 inline void stdit(string &s){transform(s.begin(),s.end(),s.begin(),::tolower);} 11 //字符串中的字母全部变为小写字母 12 inline string i2s(int a){//将给定的整数变为字符串 13 ostringstream o;o<<a; 14 return o.str(); 15 } 16 inline int s2i(const string &s){//将给定的字符串变为整数 17 int num; 18 istringstream i(s);i>>num; 19 return num; 20 } 21 inline bool leap(int year){ 22 return (year%400==0)||(year%100!=0&&year%4==0); 23 } 24 string getweek(const string &cyyyy,const string &cmm,const string &cdd){ 25 //根据1970-01-01是星期四开始推算yyyymmdd为星期几 26 int year=s2i(cyyyy),mm=s2i(cmm),dd=s2i(cdd); 27 int sum=0; 28 for(int i=1970;i<year;i++) sum+=leap(i)?366:365; 29 Daysofmon[2]=leap(year)?29:28; 30 for(int i=1;i<mm;i++) sum+=Daysofmon[i]; 31 sum+=dd-1; 32 return i2s((4+sum)%7); 33 } 34 vecstr setvec(string &str,int type=0){ 35 //把时间(分钟、小时、年、月、日、周)分割成数组化的标准形式 36 // type一共三种，当key为1是，表示建立week，当key为2时，表示建立month 37 // 其他的为0 38 vecstr rev; 39 str+=','; 40 string curs,arry[2];int brd[2]; 41 for(int ix,fx;~(ix=str.find(','));){ 42 curs=str.substr(0,ix); 43 str=str.substr(ix+1); 44 fx=curs.find('-'); 45 if(~fx){//连续时间段 46 arry[0]=curs.substr(0,fx); 47 arry[1]=curs.substr(fx+1); 48 for(int b=0;b<2;b++){ 49 string &ss=arry[b]; 50 if(isalpha(ss[0])){ 51 brd[b]=(type==1)?weekmap[ss]:monthmap[ss]; 52 } 53 else{ 54 brd[b]=s2i(ss); 55 } 56 } 57 for(int i=brd[0];i<=brd[1];i++) rev.pb(i2s(i)); 58 } 59 else{//单个时间 60 if(isalpha(curs[0])){ 61 if(type==1) rev.pb(i2s(weekmap[curs])); 62 else rev.pb(i2s(monthmap[curs])); 63 } 64 else{ 65 rev.pb(curs); 66 } 67 } 68 } 69 return rev; 70 } 71 int main(){ 72 cin>>n; 73 cin>>stime>>etime; 74 int syear=s2i(stime.substr(0,4)),eyear=s2i(etime.substr(0,4)); 75 for(int i=0;i<7;i++) weekmap[Week[i]]=i; 76 for(int i=1;i<=12;i++) monthmap[Month[i]]=i; 77 for(string minutes,hours,d_of_m,month,d_of_w,command;n--;){ 78 cin>>minutes>>hours>>d_of_m>>month>>d_of_w>>command; 79 stdit(month); 80 stdit(d_of_w); 81 if(minutes=="*") minutes="0-59"; 82 vecstr vminutes=setvec(minutes); 83 if(hours=="*") hours="0-23"; 84 vecstr vhours=setvec(hours); 85 if(d_of_m=="*") d_of_m="1-31"; 86 vecstr vd_of_m=setvec(d_of_m); 87 if(month=="*") month="1-12"; 88 vecstr vmonth=setvec(month,2); 89 if(d_of_w=="*") d_of_w="0-6"; 90 vecstr vd_of_w=setvec(d_of_w,1); 91 set<string> one(vd_of_w.begin(),vd_of_w.end()); 92 //对前面的四项进行遍历，然后根据星期几和时间的合法性进行取舍 93 for(int nowyear=syear;nowyear<=eyear;nowyear++){ 94 Daysofmon[2]=leap(nowyear)?29:28; 95 for(auto &i:vmonth){ 96 for(auto &j:vd_of_m){ 97 string cyear=i2s(nowyear); 98 if(Daysofmon[s2i(i)]<s2i(j)||one.count(getweek(cyear,i,j))==0) continue; 99 // 判断日期是否合法：1、当前年月日是否存在 2、当前年月日星期几 100 for(auto &k:vhours){ 101 for(auto &m:vminutes){ 102 string ss=cyear 103 +(i.size()>1?i:"0"+i) 104 +(j.size()>1?j:"0"+j) 105 +(k.size()>1?k:"0"+k) 106 +(m.size()>1?m:"0"+m); 107 if(ss>=stime&&ss<etime) all[ss].pb(command); 108 //注意 [ ) 109 } 110 } 111 } 112 } 113 } 114 } 115 for(auto &i:all){ 116 set<string> haxi;//不判重 95分，不明白~~ 117 for(auto &k:i.second){ 118 if(haxi.count(k)) continue; 119 cout<<i.first<<" "<<k<<"\n"; 120 haxi.insert(k); 121 } 122 } 123 return 0; 124 }