HDU 1982 字符串模拟水题

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作为博物馆馆长，你收到了怪盗基德的挑战信，但他似乎并不想为难你，留下的谜题异常简单。只需将数字转换为字母，忽略特定符号，即可揭示他的计划。

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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

ACcode：

#include <bits/stdc++.h>
using namespace std;
string str[10010];
string str1[10010];
int main ()
{
    int n;
    while(cin>>n)
    {
        for(int i=0;i<10010;i++)
        {
            str[i]="";
            str1[i]="";
        }
        for(int i=0;i<n;i++)
        {
            cin>>str[i];
            str[i]+='-';
        }
        int sum=0;
        for(int i= 0;i<n;i++)
        {

            for(int j=0;j<str[i].length();j++)
            {
                if((str[i][j]>='0' && str[i][j]<='9') || str[i][j]=='#')
                {
                    if(str[i][j]>='0' && str[i][j]<='9')
                    {
                      sum=sum*10+str[i][j]-'0';
                      if(!(str[i][j+1]>='0' && str[i][j+1]<='9'))
                      {
                          str1[i]+=('A'+sum-1);
                          sum=0;
                      }
                    }
                    else
                        str1[i]+=' ';
                }
            }
        }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<str1[i].length();j++)
            {
                    cout<<str1[i][j];
            }
            puts("");
        }
    }
    return 0;
}