HDOJ 4690 EBCDIC 模拟

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EBCDIC

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 811    Accepted Submission(s): 368

Problem Description

A mad scientist found an ancient message from an obsolete IBN System/360 mainframe. He believes that this message contains some very important secret about the Stein's Windows Project. The IBN System/360 mainframe uses Extended Binary Coded Decimal Interchange Code (EBCDIC). But his Artificial Intelligence Personal Computer (AIPC) only supports American Standard Code for Information Interchange (ASCII). To read the message, the mad scientist ask you, his assistant, to convert it from EBCDIC to ASCII.
Here is the EBCDIC table.
      
Here is the ASCII table.

 

Input

The input of this problem is a line of uppercase hexadecimal string of even length. Every two hexadecimal digits stands for a character in EBCDIC, for example, "88" stands for 'h'.

 

Output

Convert the input from EBCDIC to ASCII, and output it in the same format as the input.

 

Sample Input

   
   
   
   

    
    
    
    C59340D7A2A840C3969587999696
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    456C2050737920436F6E67726F6F


    
    
    
    
 
     
 
      Hint 
     
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Author

Zejun Wu (watashi)

 

Source

2013 Multi-University Training Contest 9

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>

using namespace std;

map<string,int> mp;

string EBCDIC[10000]=
{"NUL","SOH","STX","ETX","  ","HT","  ","DEL","  ","  ","  ","VT","FF","CR","SO","SI",
"DLE","DC1","DC2","DC3","  ","  ","BS","  ","CAN","EM","  ","  ","IFS","IGS","IRS","IUS ITB",
"  ","  ","  ","  ","  ","LF","ETB","ESC","  ","  ","  ","  ","  ","ENQ","ACK","BEL",
"  ","  ","SYN","  ","  ","  ","  ","EOT","  ","  ","  ","  ","DC4","NAK","  ","SUB",
"SP","  ","  ","  ","  ","  ","  ","  ","  ","  ","  ",".","<","(","+","|",
"&","  ","  ","  ","  ","  ","  ","  ","  ","  ","!","$","*",")",";","  ",
"-","/","  ","  ","  ","  ","  ","  ","  ","  ","  ",",","%","_",">","?",
"  ","  ","  ","  ","  ","  ","  ","  ","  ","`",":","#","@","'","=","\"",
"  ","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","  ","  ","  ","  ","  ","  ",
"^","  ","  ","  ","  ","  ","  ","  ","  ","  ","[","]","  ","  ","  ","  ",
"{","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","  ","  ","  ","  ","  ","  ",
"0","1","2","3","4","5","6","7","8","9","  ","  ","  ","  ","  ","  "};

string ASC[10000]=
{
"NUL","SOH","STX","ETX","EOT","ENQ","ACK","BEL","BS","HT","LF","VT","FF","CR","SO","SI",
"DLE","DC1","DC2","DC3","DC4","NAK","SYN","ETB","CAN","EM","SUB","ESC","IFS","IGS","IRS","IUS ITB",
"SP","!","\"","#","$","%","&","'","(",")","*","+",",","-",".","/",
"0","1","2","3","4","5","6","7","8","9",":",";","<","=",">","?",
"@","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","[","\\","]","^","_",
"`","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","{","|","}","~","DEL"
};

void init()
{
    for(int i=0;i<8;i++)
    {
        for(int j=0;j<16;j++)
        {
            mp[ASC[i*16+j]]=i*16+j;
        }
    }
}

int num(char c)
{
    if(c>='0'&&c<='9')
        return c-'0';
    return c-'A'+10;
}

char CHAR(int x)
{
    if(x<=9) return x+'0';
    return x-10+'A';
}

int main()
{
    string READ;
    init();
    while(cin>>READ)
    {
        for(int i=0,sz=READ.size();i<sz;i+=2)
        {
            int c1=num(READ[i]),c2=num(READ[i+1]);
            string ck=EBCDIC[c1*16+c2];
            int id=mp[ck];
            printf("%c%c",CHAR(id/16),CHAR(id%16));
        }
        putchar(10);
    }
    return 0;
}