1061 Dating (20分)

最新推荐文章于 2022-08-16 22:44:10 发布

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Sherlock Holmes收到的神秘字符串暗示了一个约会时间。通过解析字符串中的共同字符，可以揭示出隐藏的星期几、小时和分钟。本文提供了一个算法解决方案，用于解码这种类型的谜题。

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

题意：题目真的很简单，但是题意来说，出大问题！

星期几：前两个字符串中，第一个位置相同&&字符相同&&大写字母A～G分别代表周一～周日

小时：前两个字符串中，在上述位置以后的位置中满足第一个位置相同&&字符相同&&（数字0～9 || 大写字母A～N）分别代表0～23时

分钟：最后两个字符串中，第一个位置相同&&字符相同（区分大小写）的位置代表0～59分

直接莽就好了～

#include<iostream>
#include<vector>
using namespace std;

int main(){
    string s1, s2, s3, s4;
    string s[7] = {"MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"};
    cin >> s1 >> s2 >> s3 >> s4;
    int first = 0;
    int sen = 0;
    int flag = 0;
    if(s1.length() > s2.length()){
        for(int i = 0; i < s2.length(); i++){
            if(s1[i] == s2[i]){
                if(((s1[i] >= '0' && s1[i] <= '9') || (s1[i] >= 'A' && s1[i] <= 'N')) && first){
                    sen++;
                    if(sen == 1){
                        flag = i;
                    }
                }
                if((s1[i] >= 'A' && s1[i] <= 'G') && !first){
                    first++;
                    cout << s[s1[i] - 'A'];

                }
                if(first == 1 && sen == 1){
                    if((s1[flag] >= '0' && s1[flag] <= '9')){
                        printf(" %02d", s1[flag] - '0');
                        break;
                    }else
                    {
                        printf(" %02d", s1[flag] - 'A' + 10);
                        break;
                    }
                    
                    
                }
                
            }
        }
    }else
    {
        for(int i = 0; i < s1.length(); i++){
            if(s1[i] == s2[i]){
                if(((s1[i] >= '0' && s1[i] <= '9') || (s1[i] >= 'A' && s1[i] <= 'N')) && first){
                    sen++;
                    if(sen == 1){
                        flag = i;
                    }
                }
                if((s1[i] >= 'A' && s1[i] <= 'G') && !first){
                    first++;
                    cout << s[s1[i] - 'A'];

                }
                if(first == 1 && sen == 1){
                    if((s1[flag] >= '0' && s1[flag] <= '9')){
                        printf(" %02d", s1[flag] - '0');
                        break;
                    }else
                    {
                        printf(" %02d", s1[flag] - 'A' + 10);
                        break;
                    }
                    
                    
                }
                
            }
        }
    }


    if(s3.length() > s4.length()){
        for(int i = 0; i < s4.length(); i++){
            if(s3[i] == s4[i] && isalpha(s3[i])){
                printf(":%02d\n", i);
                break;
            }
        }
    }else
    {
        for(int i = 0; i < s3.length(); i++){
            if(s3[i] == s4[i] && isalpha(s3[i])){
                printf(":%02d\n", i);
                break;
            }
        }
    }
    
    return 0;
}