HDOJ/HDU 1982 Kaitou Kid - The Phantom Thief (1)(字符串处理)

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.

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);
        }
    }

}