Kaitou Kid

Kaitou Kid - The Phantom Thief (1)
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3216 Accepted Submission(s): 1434

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

//首先说下map大法万岁 哈哈哈哈

#include<iostream>#include<string>#include<map>#include<stdio.h>using namespace std;int main(){    map<string,char> mymap;    mymap["1"] = 'A';    mymap["2"] = 'B';    mymap["3"] = 'C';    mymap["4"] = 'D';    mymap["5"] = 'E';    mymap["6"] = 'F';    mymap["7"] = 'G';    mymap["8"] = 'H';    mymap["9"] = 'I';    mymap["10"] = 'J';    mymap["11"] = 'K';    mymap["12"] = 'L';    mymap["13"] = 'M';    mymap["14"] = 'N';    mymap["15"] = 'O';    mymap["16"] = 'P';    mymap["17"] = 'Q';    mymap["18"] = 'R';    mymap["19"] = 'S';    mymap["20"] = 'T';    mymap["21"] = 'U';    mymap["22"] = 'V';    mymap["23"] = 'W';    mymap["24"] = 'X';    mymap["25"] = 'Y';    mymap["26"] = 'Z';    mymap["#"] = ' ';    int ncase;    scanf("%d",&ncase);    getchar();    while(ncase--)    {        char ch ;        int flag = 0;        string str="";        while((ch=getchar())!='\n')        {            if(ch=='-')            {                if(str!="")                    cout<<mymap[str];                str="";                continue;            }            if(ch=='#')            {                if(str!="")                    cout<<mymap[str];                cout<<' ';                str="";                flag = 1;            }            else            {                str+=ch;                flag = 1;            }        }        if(str!="")            cout<<mymap[str];        if(flag)//这里判断一下是否需要打印回车 如果字符串全是------那么就不需要打印了...一开始这里没加 这里WA            cout<<endl;    }    return 0;}