大侦探福尔摩斯接到一张奇怪的字条:“我们约会吧! 3485djDkxh4hhGE 2984akDfkkkkggEdsb s&hgsfdk d&Hyscvnm”。大侦探很快就明白了,字条上奇怪的乱码实际上就是约会的时间“星期四 14:04”,因为前面两字符串中第1对相同的大写英文字母(大小写有区分)是第4个字母'D',代表星期四;第2对相同的字符是'E',那是第5个英文字母,代表一天里的第14个钟头(于是一天的0点到23点由数字0到9、以及大写字母A到N表示);后面两字符串第1对相同的英文字母's'出现在第4个位置(从0开始计数)上,代表第4分钟。现给定两对字符串,请帮助福尔摩斯解码得到约会的时间。
输入格式:
输入在4行中分别给出4个非空、不包含空格、且长度不超过60的字符串。
输出格式:
在一行中输出约会的时间,格式为“DAY HH:MM”,其中“DAY”是某星期的3字符缩写,即MON表示星期一,TUE表示星期二,WED表示星期三,THU表示星期四,FRI表示星期五,SAT表示星期六,SUN表示星期日。题目输入保证每个测试存在唯一解。
输入样例:
3485djDkxh4hhGE
2984akDfkkkkggEdsb
s&hgsfdk
d&Hyscvnm
输出样例:
THU 14:04
java解法:
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
String[] weekarr = {"MON ", "TUE ","WED ","THU ", "FRI ", "SAT ", "SUN "};
Scanner scan = new Scanner(System.in);
String str1 = scan.next();
String str2 = scan.next();
String str3 = scan.next();
String str4 = scan.next();
char res1 = 'A', res2 = '0';
int res3 = 0;
int cou = 1;
int len1 = Math.min(str1.length(), str2.length());
int len2 = Math.min(str3.length(), str4.length());
for(int i = 0; i < len1; i++)
{
char chr1 = str1.charAt(i);
char chr2 = str2.charAt(i);
if(cou == 1)
{
if(chr1 == chr2 && chr1 <= 'Z' && chr2 >= 'A')
{ res1 = chr1; cou++; continue; }
}
else if(cou == 2)
{
if(chr1 == chr2 && ((chr1 <= '9' && chr1 >= '0')
|| (chr1 <= 'N' && chr1 >= 'A')))
{ res2 = chr2; break; }
}
}
for(int i = 0; i < len2; i++)
{
char chr1 = str3.charAt(i);
char chr2 = str4.charAt(i);
if(chr1 == chr2 && ((chr1 < 'z' && chr1 > 'a') ||
(chr1 < 'Z' && chr2 > 'A')))
{ res3 = i; break; }
}
String s1 = weekarr[res1 - 'A'];
String s2,s3;
if(res2 <= '9') { s2 = "0" + Integer.toString(res2 - '0'); }
else { s2 = Integer.toString(res2 - 'A' + 10); }
if(res3 <= 9) { s3 = "0" + Integer.toString(res3); }
else { s3 = Integer.toString(res3); }
String finalresult = s1 + s2 + ":" + s3;
System.out.print(finalresult);
}
}
结果超时。
C++解法:
#include<iostream>
#include<string>
using namespace std;
int main()
{
string weekarr[] = { "MON ", "TUE ", "WED ", "THU ", "FRI ", "SAT ", "SUN " };
string str1, str2, str3, str4;
getline(cin, str1);
getline(cin, str2);
getline(cin, str3);
getline(cin, str4);
char res1 = 'A', res2 = '0';
int res3 = 0, cou = 1;
int len1 = str1.length() < str2.length() ? str1.length() : str2.length();
int len2 = str3.length() < str4.length() ? str3.length() : str4.length();
for (int i = 0; i < len1; i++)
{
char chr1 = str1.at(i);
char chr2 = str2.at(i);
if (cou == 1)
{
if (chr1 == chr2 && chr1 <= 'G' && chr1 >= 'A')
{
res1 = chr1; cou++; continue;
}
}
else if (cou == 2)
{
if (chr1 == chr2 && ((chr1 <= '9' && chr1 >= '0') || (chr1 <= 'N' && chr2 >= 'A')))
{
res2 = chr2; break;
}
}
}
for (int i = 0; i < len2; i++)
{
char chr1 = str3.at(i);
char chr2 = str4.at(i);
if (chr1 == chr2 && ((chr1 <= 'z' && chr1 >= 'a') || (chr1 <= 'Z' && chr1 >= 'A')))
{
res3 = i; break;
}
}
string s1, s2, s3;
s1 = weekarr[res1 - 'A'];
cout << s1;
if (res2 <= '9')
{
printf("0%d:", res2 - '0');
}
else printf("%d:", res2 - 'A' + 10);
if (res3 <= 9)
{
printf("0%d", res3);
}
else printf("%d", res3);
return 0;
}
注意判断边界条件
if (chr1 == chr2 && chr1 <= 'G' && chr1 >= 'A')
表示星期的数要在'A' 到 'G'之间
if (chr1 == chr2 && ((chr1 <= '9' && chr1 >= '0') || (chr1 <= 'N' && chr2 >= 'A')))
表示小时的数要在 '0'到‘9’或'A'到'N'之间
表示分钟的数由于每个字符串不超过60,因此不需要另外加判断语句