(PAT 1061) Dating (字符串处理)

刺骨的言语ヽ痛彻心扉 2022-03-21 04:14 166阅读 0赞

1061 Dating （20 point(s)）

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

解题思路:

算法比较简单,字符串处理筛选字符就行,难在范围的控制上

日期范围:A-G (大写字母)

小时范围:0-9 A-N (大写字母)

分钟范围:大小写字母

#include <iostream>
    #include <algorithm>
    #include <string>
    using namespace std;
    
    string Date[7] = { "MON","TUE","WED","THU","FRI","SAT","SUN" };
    int main() {
    	string codes[4];
    	for (int i = 0; i < 4; ++i) {
    		cin >> codes[i];
    	}
    	int pointer1 = 0;
    	int pointer2 = 0;
    	int pointer3 = 0;
    	int pointer4 = 0;
    	int num1 = 0, num2 = 0, num3 = 0;
    	while (pointer1 < codes[0].length() && pointer2 < codes[1].length()) {
    		if (isupper(codes[0][pointer1]) && isupper(codes[1][pointer2]) && codes[0][pointer1] <= 71 && codes[0][pointer1] == codes[1][pointer2]) {
    			num1 = codes[0][pointer1] - 65;
    			pointer1++;
    			pointer2++;
    			break;
    		}
    		pointer1++;
    		pointer2++;
    	}
    
    	while (pointer1 < codes[0].length() && pointer2 < codes[1].length()) {
    		if ((isupper(codes[0][pointer1]) && isupper(codes[1][pointer2]) && codes[0][pointer1] <= 78) || (isdigit(codes[0][pointer1]) && isdigit(codes[1][pointer2])) && codes[0][pointer1] == codes[1][pointer2]) {
    			if (isdigit(codes[0][pointer1])) {
    				num2 = codes[0][pointer1] - 48;
    			}
    			else {
    				num2 = codes[0][pointer1] - 65 + 10;
    			}
    			break;
    		}
    		pointer1++;
    		pointer2++;
    	}
    
    	while (pointer3 < codes[2].length() && pointer4 < codes[3].length()) {
    		if (isalpha(codes[2][pointer3]) && isalpha(codes[3][pointer4]) && codes[2][pointer3] == codes[3][pointer4]) {
    			num3 = pointer3;
    			break;
    		}
    		pointer3++;
    		pointer4++;
    	}
    
    	printf("%s %02d:%02d", Date[num1].c_str(), num2, num3);
    
    	return 0;
    }