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, FRIfor Friday, SAT for Saturday, and SUN for Sunday. It is guaranteed that the result is unique for each case.

Sample Input:

1
2
3
4

3485djDkxh4hhGE 
2984akDfkkkkggEdsb 
s&hgsfdk 
d&Hyscvnm

Sample Output:

1

THU 14:04

题解

思路

• 有一点我一定要提一下
• 这道题说的重复字符
• 它们的下标也要是相同的
• 举个例子，题目里的意思是，第二个字符串出现的第一个在第一个字符串中的“A-G”字符是“D”。然而这样的意思是错的。
• 但实际上，题目里的真实含义是，第二个和第一个字符串下标相同且字符相同并且字符在“A-G”的第一个字符是“D”。
• 这样就简化了很多，原来拿哈希集合怼半天不对，还不知道问题错在哪。
• 这道题属实弟弟哦。

数据结构

• day_of_week 是一个字典，将字母映射到礼拜几
• a,b,c,d分别是原始的四个字符串
• day,hour,minute 是礼拜几，小时数和分钟数
• 在遍历中，i是下标，j是值
• start 记录一下前两个字符串第一次出现重复字符且字符在“A-G”中的下标

算法

• 照着题目意思来就行了

代码

• 由于使用Python3 能AC，因此只放了Python的代码。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31

day_of_week = {
    'A': "MON",
    'B': "TUE",
    'C': "WED",
    "D": "THU",
    "E": "FRI",
    "F": "SAT",
    "G": "SUN"
}

a = input()
b = input()
for i, j in enumerate(b):
    if a[i] == b[i] and j in "ABCDEFG":
        day = day_of_week[j]
        start = i
        break
for i, j in enumerate(b[start+1:]):
    if a[i+start+1] == j and (j.isdigit() or ord("A") <= ord(j) <= ord("N")):
        hour = int(j) if j.isdigit() else ord(j) - ord("A") + 10
        break

c = input()
d = input()
for i, _ in enumerate(d):
    if d[i] == c[i] and d[i].isalpha():
        minute = i
        break
print(day, end=" ")
print("%02d:%02d" % (hour, minute))