HDOJ/HDU 1982 Kaitou Kid - The Phantom Thief (1)(字符串处理)
2016-06-02 13:29
579 查看
Problem Description
Do you know Kaitou Kid? In the legend, Kaitou Kid is a master of disguise, and can take on the voice and form of anyone. He is not an evil person, but he is on the wrong side of the law. He’s the very elusive phantom thief who never miss his prey although he always uses word puzzles to announce his targets before action.
![](http://img.blog.csdn.net/20160602132407751)
You are the leader of a museum. Recently, you get several priceless jewels and plan to hold an exhibition. But at the moment, you receive Kid’s word puzzle… Fortunately, It seems Kid doesn’t want to trouble you, and his puzzle is very easy. Just a few minutes, You have found the way to solve the puzzle:
(1) change 1 to ‘A’, 2 TO ‘B’,..,26 TO ‘Z’
(2) change ‘#’ to a blank
(3) ignore the ‘-’ symbol, it just used to separate the numbers in the puzzle
Input
The first line of the input contains an integer C which means the number of test cases. Then C lines follow. Each line is a sentence of Kid’s word puzzle which is consisted of ‘0’ ~ ‘9’ , ‘-’ and ‘#’. The length of each sentence is no longer than 10000.
Output
For each case, output the translated text.
Sample Input
4
9#23-9-12-12#19-20-5-1-12#1-20#12-5-1-19-20#15-14-5#10-5-23-5-12
1-14-4#12-5-1-22-5#20-8-5#13-21-19-5-21-13#9-14#20#13-9-14-21-20-5-19
1-6-20-5-18#20-8-5#15-16-5-14-9-14-7#15-6#20-8-5#5-24-8-9-2-9-20-9-15-14
7-15-15-4#12-21-3-11
Sample Output
I WILL STEAL AT LEAST ONE JEWEL
AND LEAVE THE MUSEUM IN T MINUTES
AFTER THE OPENING OF THE EXHIBITION
GOOD LUCK
题意:
就是输入数字#和-,数字1-26分别对应A-Z.
#对应空格 -没有含义,就是把数字隔开
注意这一种输入:
1
###—##
两种方法:
一,常规方法:
二:利用Java中的replaceAll()方法:
Do you know Kaitou Kid? In the legend, Kaitou Kid is a master of disguise, and can take on the voice and form of anyone. He is not an evil person, but he is on the wrong side of the law. He’s the very elusive phantom thief who never miss his prey although he always uses word puzzles to announce his targets before action.
You are the leader of a museum. Recently, you get several priceless jewels and plan to hold an exhibition. But at the moment, you receive Kid’s word puzzle… Fortunately, It seems Kid doesn’t want to trouble you, and his puzzle is very easy. Just a few minutes, You have found the way to solve the puzzle:
(1) change 1 to ‘A’, 2 TO ‘B’,..,26 TO ‘Z’
(2) change ‘#’ to a blank
(3) ignore the ‘-’ symbol, it just used to separate the numbers in the puzzle
Input
The first line of the input contains an integer C which means the number of test cases. Then C lines follow. Each line is a sentence of Kid’s word puzzle which is consisted of ‘0’ ~ ‘9’ , ‘-’ and ‘#’. The length of each sentence is no longer than 10000.
Output
For each case, output the translated text.
Sample Input
4
9#23-9-12-12#19-20-5-1-12#1-20#12-5-1-19-20#15-14-5#10-5-23-5-12
1-14-4#12-5-1-22-5#20-8-5#13-21-19-5-21-13#9-14#20#13-9-14-21-20-5-19
1-6-20-5-18#20-8-5#15-16-5-14-9-14-7#15-6#20-8-5#5-24-8-9-2-9-20-9-15-14
7-15-15-4#12-21-3-11
Sample Output
I WILL STEAL AT LEAST ONE JEWEL
AND LEAVE THE MUSEUM IN T MINUTES
AFTER THE OPENING OF THE EXHIBITION
GOOD LUCK
题意:
就是输入数字#和-,数字1-26分别对应A-Z.
#对应空格 -没有含义,就是把数字隔开
注意这一种输入:
1
###—##
两种方法:
一,常规方法:
import java.util.Scanner; public class Main{ static char[] STR={'a','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'}; public static void main(String[] args) { Scanner sc = new Scanner(System.in); int t =sc.nextInt(); while(t-->0){ String str=sc.next(); String s[] = str.split("-"); for(int i=0;i<s.length;i++){ String strN = ""; for(int j=0;j<s[i].length();j++){ if(s[i].charAt(j)!='#'){ strN+=s[i].charAt(j); }else{ if(!strN.equals("")) System.out.print(STR[Integer.parseInt(strN)]); System.out.print(" "); strN=""; } } if(!strN.equals("")) System.out.print(STR[Integer.parseInt(strN)]); } System.out.println(); } } }
二:利用Java中的replaceAll()方法:
import java.util.Scanner; public class Main{ public static void main(String[] args) { Scanner sc = new Scanner(System.in); int t =sc.nextInt(); while(t-->0){ String str=sc.next(); str=str.replaceAll("#", " "); for(int i=26;i>=1;i--){ String a=""+(char)('A'+i-1); String b=""+i; str=str.replaceAll(b, a); } str=str.replaceAll("-", ""); System.out.println(str); } } }
相关文章推荐
- HDOJ/HDU 1982 Kaitou Kid - The Phantom Thief (1)(字符串处理)
- 排序(三)之直接插入排序Straight Insertion Sort
- 14.1.2 Checking InnoDB Availability 检查InnoDB 可用性:
- 14.1.2 Checking InnoDB Availability 检查InnoDB 可用性:
- 14.1.2 Checking InnoDB Availability 检查InnoDB 可用性:
- windows 7 64位 JMail注册
- USACO 2009 Feb Gold 1.Fair Shuttle 贪心
- Inpaint 2.4简体中文版 任何人都会…
- [leetcode] 70. Climbing Stairs
- CLOSE_WAIT状态的生成原因
- CS231n 2016 通关 第五章 Training NN Part1
- 553 relay check local fail. (本域用户必须通过验证) ----一个老项目发邮件问题
- leecode_220 Contains Duplicate III
- 并发注意事项(2)并发工具优先于wait和notify
- 03-树3 Tree Traversals Again
- Failed to connect to repository database. OMS will be automatically restarted once it identifies tha
- 扩展SplitContainer控件
- ExecutorService——shutdown方法和awaitTermination方法
- ExecutorService——shutdown方法和awaitTermination方法
- Unable to determine if the owner (Domain\UserName) of job JOB_NAME has server access