PAT A 1061. Dating (20)

题目

Sherlock Holmes received a note with some strange strings: "Let's date!  3485djDkxh4hhGE  2984akDfkkkkggEdsb  s&hgsfdk  d&Hyscvnm".  It took him only a minute to figure out that those strange strings are actually referring to the coded time "Thursday 14:04" -- since the first common capital English letter (case sensitive) shared by the first two strings is the 4th capital letter 'D', representing the 4th day in a week; the second common character is the 5th capital letter 'E', representing the 14th hour (hence the hours from 0 to 23 in a day are represented by the numbers from 0 to 9 and the capital letters from A to N, respectively); and the English letter shared by the last two strings is 's' at the 4th position, representing the 4th minute.  Now given two pairs of strings, you are supposed to help Sherlock decode the dating time.

Input Specification:

Each input file contains one test case.  Each case gives 4 non-empty strings of no more than 60 characters without white space in 4 lines.

Output Specification:

For each test case, print the decoded time in one line, in the format "DAY HH:MM", where "DAY" is a 3-character abbreviation for the days in a week -- that is, "MON" for Monday, "TUE" for Tuesday, "WED" for Wednesday, "THU" for Thursday, "FRI" for Friday, "SAT" for Saturday, and "SUN" for Sunday.  It is guaranteed that the result is unique for each case.

Sample Input:

3485djDkxh4hhGE 2984akDfkkkkggEdsb s&hgsfdk d&Hyscvnm

Sample Output:

THU 14:04

 

题目表达不清晰。

不符合星期的字符跳过后在小时计算中是否要计算没有表达清楚。

实际也是跳过，即从取得表达星期的字符后查询第一个符合小时表达的字符。

所谓的相同时指相同序号位置上的字符相同，如示例。

 

代码：

#include <iostream>#include <cstdio>#include <string>using namespace std;int main(){string s1,s2,s3,s4;//输入字符串cin>>s1>>s2>>s3>>s4;string week[7]={"MON","TUE","WED","THU","FRI","SAT","SUN"};//星期int d,h,m;//天，小时，分int i;for(i=0;i<s1.size();i++)//获取第一个，有范围限制{if(s1[i]==s2[i]&&s1[i]>='A'&&s1[i]<='G'){d=s1[i]-'A';break;}}for(i++;i<s1.size();i++)//小时，获取第一个后面的“第二个”，大坑货……{if(s1[i]==s2[i]&&((s1[i]>='A'&&s1[i]<='N')||(s1[i]>='0'&&s1[i]<='9'))){if(s1[i]>='0'&&s1[i]<='9')h=s1[i]-'0';elseh=s1[i]-'A'+10;break;}}for(i=0;i<s3.size();i++)//获取分钟{if(s3[i]==s4[i]&&((s3[i]>='a'&&s3[i]<='z')||(s3[i]>='A'&&s3[i]<='Z'))){m=i;break;}}cout<<week[d]<<" ";printf("%02d:%02d",h,m);return 0;}