1061 Dating （20 分）

蔚落 2022-03-22 07:29 222阅读 0赞

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

#include <bits/stdc++.h>
    using namespace std;
    
    string days[7]={"MON","TUE","WED","THU","FRI","SAT","SUN"};
    int main()
    {
        string a,b,c,d;
        cin>>a>>b>>c>>d;
        int day=0,hour=0,minute=0;
        int i;
        for(i=0;i<a.size()&&i<b.size();i++){
            if(a[i]==b[i]&&a[i]>='A'&&a[i]<='G'){
                day=a[i]-'A';
                break;
            }
        }
        for(i=i+1;i<a.size()&&i<b.size();i++){
            if(a[i]==b[i]){
                if(isdigit(a[i]))hour=a[i]-'0';
                else if(a[i]>='A'&&a[i]<='N'){
                    hour=a[i]-'A'+10;
                }
                else continue;
                break;
            }
        }
        for(int j=0;j<c.size()&&j<d.size();j++){
            if(isalpha(c[j])&&c[j]==d[j]){
                minute=j;
                break;
            }
        }
        cout<<days[day];
        printf(" %02d:%02d",hour,minute);
        return 0;
    }