• PAT 1014. 福尔摩斯的约会


    1014. 福尔摩斯的约会 (20)

    大侦探福尔摩斯接到一张奇怪的字条:“我们约会吧! 3485djDkxh4hhGE 2984akDfkkkkggEdsb s&hgsfdk d&Hyscvnm”。大侦探很快就明白了,字条上奇怪的乱码实际上就是约会的时间“星期四 14:04”,因为前面两字符串中第1对相同的大写英文字母(大小写有区分)是第4个字母'D',代表星期四;第2对相同的字符是'E',那是第5个英文字母,代表一天里的第14个钟头(于是一天的0点到23点由数字0到9、以及大写字母A到N表示);后面两字符串第1对相同的英文字母's'出现在第4个位置(从0开始计数)上,代表第4分钟。现给定两对字符串,请帮助福尔摩斯解码得到约会的时间。

    输入格式:

    输入在4行中分别给出4个非空、不包含空格、且长度不超过60的字符串。

    输出格式:

    在一行中输出约会的时间,格式为“DAY HH:MM”,其中“DAY”是某星期的3字符缩写,即MON表示星期一,TUE表示星期二,WED表示星期三,THU表示星期四,FRI表示星期五,SAT表示星期六,SUN表示星期日。题目输入保证每个测试存在唯一解。

    输入样例:
    3485djDkxh4hhGE 
    2984akDfkkkkggEdsb 
    s&hgsfdk 
    d&Hyscvnm
    
    输出样例:
    THU 14:04
     1 #include<iostream>
     2 #include<map>
     3 #include<utility> 
     4 #include<algorithm>
     5 #include<iomanip>
     6 using namespace std;
     7 int Isletter(char ch){
     8     if(int(ch)>=65&&int(ch)<=90)
     9     return 1;
    10     else if((int(ch)>=97&&int(ch)<=122))
    11     return 1;
    12     else return 0;
    13 }
    14 int main(){
    15     string s1,s2,s3,s4;
    16     string Day,Hour;
    17     int flag=0,min;
    18     cin>>s1>>s2>>s3>>s4;
    19     map<char,string> day={{'A',"MON"},{'B',"TUE"},{'C',"WED"},
    20                         {'D',"THU"},{'E',"FRI"},{'F',"SAT"},{'G',"SUN"}};
    21     map<char,string> hour={{'0',"00:"},{'1',"01:"},{'2',"02:"},{'3',"03:"},
    22                         {'4',"04:"},{'5',"05:"},{'6',"06:"},{'7',"07:"},
    23                         {'8',"08:"},{'9',"09:"},{'A',"10:"},{'B',"11:"},{'C',"12:"},
    24                         {'D',"13:"},{'E',"14:"},{'F',"15:"},{'G',"16:"},{'H',"17:"},
    25                         {'I',"18:"},{'J',"19:"},{'K',"20:"},{'L',"21:"},{'M',"22:"},
    26                         {'N',"23:"}};                    
    27     int l=s1.length()<s2.length()?s1.length():s2.length();
    28     for(int i=0;i<l;i++){
    29     if(flag==0&&s1[i]==s2[i]&&day.find(s1[i])!=day.end())
    30     {Day=day[s1[i]]; flag++; i++;}
    31     if(flag==1&&s1[i]==s2[i]&&hour.find(s1[i])!=hour.end())
    32     {Hour=hour[s1[i]]; break;}
    33     }
    34     l=s3.length()<s4.length()?s3.length():s4.length();
    35     for(int i=0;i<l;i++){
    36      if(s3[i]==s4[i]&&Isletter(s3[i])==1)
    37      {min=i; break;}
    38     }
    39     cout<<Day<<" "<<Hour<<setw(2)<<setfill('0')<<min;            
    40     return 0; 
    41 }
    View Code
  • 相关阅读:
    Codeforces 845E Fire in the City 线段树
    Codeforces 542D Superhero's Job dp (看题解)
    Codeforces 797F Mice and Holes dp
    Codeforces 408D Parcels dp (看题解)
    Codeforces 464D World of Darkraft
    Codeforces 215E Periodical Numbers 容斥原理
    Codeforces 285E Positions in Permutations dp + 容斥原理
    Codeforces 875E Delivery Club dp
    Codeforces 888F Connecting Vertices 区间dp (看题解)
    Codeforces 946F Fibonacci String Subsequences dp (看题解)
  • 原文地址:https://www.cnblogs.com/A-Little-Nut/p/8056016.html
Copyright © 2020-2023  润新知