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HDOJ 4690 EBCDIC 模拟


把图片用ORC转化成文字,用vim录个宏很容易就可以字母们格式化.....

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
E.html download 方便图中文字复制 http://pan.baidu.com/share/link?shareid=453447595&uk=352484775
 

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;
}



HDOJ 4690 EBCDIC 模拟