PAT 1061 Dating（20分）未完成

原题链接：PAT 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

题目大意：

给了你四行字符串，要求从里面解出日期信息：

前两个字符串中第一个出现的相同大写字母（A-G）代表星期
前两个字符串中第二个出现的相同数字（0-9）或者大写字母（A-N）代表小时，注意要补0
后两个字符串中第一个出现相同字母（不管大小写）的位置代表分钟，注意要补0

最后按格式输出即可

方法一：字符串

C++代码：

// pat 1061 测试点2和5还是没过 
#include <iostream>
#include <cstring>
using namespace std;string s[5];
string week[7] = {"MON", "TUR", "WED", "THU", "FRI", "SAT", "SUN"};int main(){for(int i = 1; i <= 4; i++)getline(cin, s[i]);// 前两个字符判断星期、小时数 bool flag = 0;	// flag为0时表示判断星期 for(int i = 0; i < min(s[1].size(), s[2].size()); i++ ){// 判断星期 if(s[1][i] == s[2][i] && s[1][i] >= 'A' && s[1][i] <= 'G' && flag == 0){cout << week[s[1][i] - 'A'] << " "; flag = 1; continue;} // 判断小时if(s[1][i] == s[2][i] && s[1][i] >= '0' && s[1][i] <= '9' && flag == 1){printf("%02d:", s[1][i] - '0');break;}if(s[1][i] == s[2][i] && s[1][i] >= 'A' && s[1][i] <= 'N' && flag == 1){ cout << s[1][i] - 'A' + 10 << ":";break;}}// 后两个字符判断分钟数 for(int i = 0; i < min(s[3].size(), s[4].size()); i++ ){// 这里不管大小写只要相同即可 if(s[3][i] == s[4][i] && ((s[3][i] > 'a' && s[3][i] < 'z') || (s[3][i] > 'A' && s[3][i] < 'Z'))){printf("%02d\n", i);break; } }return 0;
}