PAT甲级 1061. Dating (20)

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, "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

思路：

这里题目没有说清：

1）在日期上，选择前两个字符串中在A~G范围内有相等的字符（代表1~7）；

2）小时上，则是在日期之后，前两个字符串中0~9和A~N范围内相等的字符（代表0~23）；

3）分钟上，后两个字符串中英语字符（a~z和A~Z）相等的字符位置。

此外，输出要考虑有效字符，不足则0补足。

代码：

#include<iostream>#include<string>#include<algorithm>using namespace std;string day[7] = {"MON","TUE","WED","THU","FRI","SAT","SUN"};int main(){ifstream cin;cin.open("case1.txt");string s1, s2, s3, s4;cin >> s1 >> s2 >> s3 >> s4;int i,d,h,m;for (i = 0; i < s1.size() && i < s2.size(); ++i){if (s1[i] == s2[i] && s1[i] >= 'A' && s1[i] <= 'G'){d = (s1[i]-'A');break;}}++i;for (; i < s1.size() && s2.size(); ++i){if (s1[i] == s2[i]){if (s1[i] >= '0' && s1[i] <= '9'){h = s1[i] - '0';break;}else{if (s1[i] >= 'A' && s1[i] <= 'N'){h = s1[i] - 'A' + 10;break;}}}}for (i = 0; i < s3.size() && i < s4.size(); ++i){if (s3[i] == s4[i]){if ((s3[i] >= 'a' && s3[i] <= 'z') || (s3[i] >= 'A' && s3[i] <= 'Z')){m = i;break;}}}printf("%s %02d:%02d\n",day[d].c_str(),h,m);system("pause");return 0;}