pat 1061

解密约会代码

最新推荐文章于 2025-08-31 12:34:04 发布

原创最新推荐文章于 2025-08-31 12:34:04 发布 · 549 阅读

0 ·

CC 4.0 BY-SA版权

通过寻找字符串中的共同字符来解码隐藏的约会时间。利用特定字符映射到星期几、小时和分钟，本代码实现了从四组字符串中提取并解析出预定格式的日期时间。

1061. Dating (20)

时间限制

50 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 <stdio.h>
#include <string.h>

char str1[80], str2[80], str3[80], str4[80];
char map[8][10] = {
"MON",
"TUE",
"WED",
"THU",
"FRI",
"SAT",
"SUN"
};

int main()
{
	int i;

	while(scanf("%s%s%s%s", str1, str2, str3, str4) != EOF)
	{
		int len1 = strlen(str1);
		int len2 = strlen(str2);

		for(i = 0; i < len1 && i < len2; i++)
		{
			if(str1[i] == str2[i] && str1[i] >= 'A' && str1[i] <= 'G')
				break;
		}

		printf("%s ", map[str1[i] - 'A']);

		i++;

		for(; i < len1 && i < len2; i++)
		{
			if(str1[i] == str2[i] && str1[i] >= 'A' && str1[i] <= 'N')
				break;
			else if(str1[i] == str2[i] && str1[i] >= '0' && str1[i] <= '9')
				break;
		}

		if(str1[i] >= 'A' && str1[i] <= 'N')
			printf("%d:", str1[i] - 'A' + 10);
		else if(str1[i] >= '0' && str1[i] <= '9')
			printf("%02d:", str1[i] - '0');

		int len3 = strlen(str3), len4 = strlen(str4);

		for(i = 0; i < len3 && i < len4; i++)
		{
			if(str3[i] == str4[i] && (str3[i] >= 'a' && str3[i] <= 'z' || str3[i] >= 'A' && str3[i] <= 'Z'))
				break;
		}

		printf("%02d\n", i);
	}
	return 0;
}