hdoj-1982-Kaitou Kid - The Phantom Thief (1)

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 
     
     
     
     

 

字符串题，给你一些数字和一些符号然后让你转换成英文，-是分隔符，#是空格

一个个去判断就好了 ，水题

#include<stdio.h>
#include<string.h>
int main()
{
    char s[10005];
    int k,x,i;
    scanf("%d",&k);
    while(k--)
    {
        scanf("%s",&s);
        x=strlen(s);
        for(i=0;i<x;i++)
        {
            if(s[i]>='0'&&s[i]<='9')
            {
                if(s[i+1]=='-')
                {
                    putchar(s[i]+'A'-'1');
                    i++;
                }
                else if(s[i+1] >='0'&&s[i+1]<='9')
                {

                    putchar((s[i]-'0')*10 + s[i+1]-'0'+'A'- 1);
                    i++;
                }
                else putchar(s[i]+ 'A'-'1');
            }
            else if(s[i] =='#')
                putchar(' ');
        }
        printf("\n");
        }
        return 0;
    }