PAT 甲级 1061 Dating (20 分) 我的代码+别人的代码

缺乏、安全感 2024-02-19 20:59 42阅读 0赞

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

第一次写，没读懂题意，过了三个测试点。

题目大意： 给你两对字符串，按照如下规则从中解码出时间——  
1. 第一对儿字符串的第一个相同的大写字母，其所在字母表顺序表示星期数，如A是第一个字母，表示第一天，即星期一。我们也可以由此推出，这个所谓的第一个相同的大写字母有两个隐含条件，第一个是位置也要相同，第二个是有效取值范围在A~G之间（毕竟一周只有7天）；  
2.第一对儿字符串的第二个相同的大写字母，其对应24进制（0 ~ 9，A ~ N）的代表小时数，如字母A，表示10点；同1，这里隐含一个条件，即有效的取值范围；  
3. 第二对儿字符串第一个相同的字母，其所在位置代表分钟数。这里也隐含了一个条件，并不区分大小写。  
4. 输出格式：按照DAY HH:MM的形式输出，注意补零。

## 我的代码： ##

#include<iostream>
    #include<cstdio>
    #include<ctype.h>
    #include<map>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int main()
    {
    	string w[8]={"","MON","TUE","WED","THU","FRI","SAT","SUN"};
    	string a,b,c,d;
    	char t[3];
    	int i,j,cnt=0,k;
    	cin>>a>>b>>c>>d;
    	int len=a.length()>b.length()?b.length():a.length();
    	for(i=0;i<len;i++)
    	{
    		if(a[i]==b[i])
    		if(cnt<1)
    		{
    			if(a[i]>='A'&&a[i]<='G')
    			t[cnt++]=a[i];
    		}
    		else
    		{
    			if((a[i]>='A'&&a[i]<='N')||(a[i]>='0'&&a[i]<='9'))
    			t[cnt++]=a[i];
    		}
    		if(cnt>1)
    		break;
    	}
    	cout<<w[t[0]-'A'+1];
    	printf(" %02d:",t[1]>='A'?t[1]-'A'+10:t[1]-'0');
    	len=c.length()>d.length()?d.length():c.length();
    	for(i=0;i<len;i++)
    	{
    		if(c[i]==d[i]&&(c[i]>='a'&&c[i]<='a'+25||c[i]>='A'&&c[i]<='A'+25))
    		{
    			break;
    		}
    	}
    	printf("%02d\n",i);
    	return 0;
    }

别人的代码：

#include<bits/stdc++.h>
    using namespace std;
    map<char,string>mp;
    void init()
    {
    	mp['A'] = "MON";mp['B'] = "TUE";
    	mp['C'] = "WED";mp['D'] = "THU";
    	mp['E'] = "FRI";mp['F'] = "SAT";
    	mp['G'] = "SUN";
    }
    int main()
    {
    	string s1,s2,s3,s4;
    	cin>>s1>>s2>>s3>>s4;
    	init();
    	int minn = min(s1.length(),s2.length()),i;
    	for(i = 0; i < minn ;i++)
    	{
    		if(s1[i] == s2[i] && s1[i] >='A' && s1[i] <= 'G')
    		{
    			cout<<mp[s1[i]];
    			break;
    		}
    	}
    	for(++i; i < minn;i++)
    	{
    		if(s1[i] == s2[i])
    		{
    			if(s1[i] >= 'A' && s1[i] <= 'N')
    			{
    				printf(" %02d:",(10+s1[i]-'A'));
    				break;
    			}	
    			else if(isdigit(s1[i]))
    			{
    				printf(" %02d:",s1[i]-'0');
    				break;
    			}	
    		}
    	}
    	minn = min(s3.length(),s4.length());
    	for(int i = 0; i< minn ;i++)
    	{
    		if(s3[i] == s4[i] && isalpha(s3[i]))
    		{
    			printf("%02d\n",i);
    			return 0;
    		}	
    	}
    	return 0;
    }