PAT甲级 1061.Dating (20 分)

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

分析：

本题（与PAT乙级 1014.福尔摩斯的约会 (20 分)是同一道题）有以下四个需要注意的地方：

（1）前面两个字符串中第 1 对相同的大写英文字母表示星期。一个星期7天，故其取值范围为A-G，而不是A-Z。

（2）前面两个字符串中第 2 对相同的字符（第 1 对相同的大写英文字母出现之后）表示钟点（小时）。一天的 0 点到 23 点由数字 0 到 9、以及大写字母 A 到 N 表示。

（3）后面两个字符串中第 1 对相同的英文字母出现的位置（从 0 开始计数）表示分钟。

（4）小时和分钟不足两位数时，需要在前面添0。

#include <cstdio>
#include <cstring>

int main() {
	char week[7][4] = { "MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN" };
	char str[4][61];
	for (int i = 0; i < 4; i++) {
		scanf("%s", str[i]);
	}
	int len0 = strlen(str[0]), len1 = strlen(str[1]);
	int len2 = strlen(str[2]), len3 = strlen(str[3]);
	int i = 0, j = 0;
	while (i < len0 && i < len1) {
		// 第一对相同的大写英文字母，取值范围为A-G
		if (str[0][i] == str[1][i] && 'A' <= str[0][i] && str[0][i] <= 'G') {
			printf("%s ", week[str[0][i] - 'A']);
			break;
		}
		i++;
	}
	i += 1;
	while (i < len0 && i < len1) {
		// 第二对相同的字符，取值范围为0-9及A-N
		if (str[0][i] == str[1][i]
				&& (('A' <= str[0][i] && str[0][i] <= 'N')
						|| ('0' <= str[0][i] && str[0][i] <= '9'))) {
			if ('A' <= str[0][i] && str[0][i] <= 'N') {
				printf("%d:", str[0][i] - 'A' + 10);
			} else {
				printf("%02d:", str[0][i] - '0');
			}
			break;
		}
		i++;
	}
	while (j < len2 && j < len3) {
		// 第一对相同的英文字母
		if (str[2][j] == str[3][j]
				&& (('A' <= str[2][j] && str[2][j] <= 'Z')
						|| ('a' <= str[2][j] && str[2][j] <= 'z'))) {
			printf("%02d\n", j);
			break;
		}
		j++;
	}
	return 0;
}