1061 Dating

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：

　　在前两个字符串中找出天数，和小时数。在后两个字符串中找出分钟数。天数所对应的是（‘A’ - ‘G’），小时数所对应的是（0-9 ‘A’ - ‘N’）所以不能单纯的用isupper()来判断，分钟数所对应的时第三组和第四组字符串中第一个相等英文字符的下标。

思路：

　　模拟。

Code：

 1 #include <bits/stdc++.h>
 2
 3 using namespace std;
 4
 5 int main() {
 6     string str1, str2, str3, str4;
 7     cin >> str1 >> str2 >> str3 >> str4;
 8     string week[7] = {"MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"};
 9     int i = 0, j = 0, hour;
10     string day;
11     while (i < str1.length() && i < str2.length()) {
12         if (str1[i] == str2[i] && (str1[i] >= 'A' && str1[i] <= 'G')) {
13             day = week[str1[i] - 'A'];
14             break;
15         }
16         i++;
17     }
18     i = i + 1;
19     while (i < str1.length() && i < str2.length()) {
20         if (str1[i] == str2[i]) {
21             if (str1[i] >= 'A' && str1[i] <= 'N') {
22                 hour = str1[i] - 'A' + 10;
23                 break;
24             } else if (isdigit(str1[i])) {
25                 hour = str1[i] - '0';
26                 break;
27             }
28         }
29         i++;
30     }
31     while (j < str3.length() && j < str4.length()) {
32         if (str3[j] == str4[j] && isalpha(str3[j])) break;
33         j++;
34     }
35     cout << day;
36     printf(" %02d:%02d\n", hour, j);
37     return 0;
38 }