HDU——1982Kaitou Kid - The Phantom Thief (1)（坑爹string题）

Kaitou Kid - The Phantom Thief (1)

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2694    Accepted Submission(s): 1182

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

 

49#23-9-12-12#19-20-5-1-12#1-20#12-5-1-19-20#15-14-5#10-5-23-5-121-14-4#12-5-1-22-5#20-8-5#13-21-19-5-21-13#9-14#20#13-9-14-21-20-5-191-6-20-5-18#20-8-5#15-16-5-14-9-14-7#15-6#20-8-5#5-24-8-9-2-9-20-9-15-147-15-15-4#12-21-3-11

 

题目挺坑，多个#不能无视，多个-要视为1个-，WA+PE数次之后终于对了

#include<iostream>#include<string>#include<algorithm>#include<sstream>using namespace std;int main(void){    int n,num,i,j;    string s,ans;    while (cin>>n)    {        getchar();        while (n--)        {                        getline(cin,s);            ans="";for(i=0; i<s.size(); i++){if(s[i]=='-')//第一种为-continue;else if(isdigit(s[i]))//第二种为数字{if(isdigit(s[i+1])){ans=ans+char((s[i]-'0')*10+s[i+1]-'0'+64);i++;//既然已经判断过了I+1那循环就要从下下次开始，下同}else if(s[i+1]=='-'){ans=ans+char(s[i]-'0'+64);i++;}else if(s[i+1]=='#'){ans=ans+char(s[i]-'0'+64);ans=ans+" ";i++;}else ans=ans+char(s[i]-'0'+64);}else if(s[i]=='#')//第三种为#{ans+=" ";}}            cout<<ans<<endl;        }            }    return 0;}