HDU 4690 EBCDIC

EBCDIC

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)

Total Submission(s): 169    Accepted Submission(s): 87

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

 

 

多校第九场了，今天唯一过的一道题（别再唯一了好吗T_T）。。。

就是把上面表格中字符的标号对应到下面的标号上去，毫无难度的签道题，就是写起来费点事

另外注意某些字符在C/C++中的表示，比如'\\'，'\?'，'\"'，'\''等

还有就是那些不可打印的字符，我用了#define的方法把它们处理掉了

 

#include<stdio.h>

#define NUL 0x0
#define SOH 0x1
#define STX 0x2
#define ETX 0x3
#define EOT 0x4
#define ENQ 0x5
#define ACK 0x6
#define BEL 0x7
#define BS 0x8
#define HT 0x9
#define LF 0xA
#define VT 0xB
#define FF 0xC
#define CR 0xD
#define SO 0xE
#define SI 0xF
#define DLE 0x10
#define DC1 0x11
#define DC2 0x12
#define DC3 0x13
#define DC4 0x14
#define NAK 0x15
#define SYN 0x16
#define ETB 0x17
#define CAN 0x18
#define EM 0x19
#define SUB 0x1A
#define ESC 0x1B
#define IFS 0x1C
#define IGS 0x1D
#define IRS 0x1E
#define IUS_ITB 0x1F
#define SP 0x20
#define DEL 0x7F

const char c[16][16]={{NUL ,SOH ,STX ,ETX ,'\0', HT ,'\0',DEL ,'\0','\0','\0', VT , FF , CR , SO , SI },
                      {DLE ,DC1 ,DC2 ,DC3 ,'\0','\0', BS ,'\0',CAN , EM ,'\0','\0',IFS ,IGS ,IRS ,IUS_ITB},
                      {'\0','\0','\0','\0','\0', LF ,ETB ,ESC ,'\0','\0','\0','\0','\0',ENQ ,ACK ,BEL},
                      {'\0','\0',SYN ,'\0','\0','\0','\0',EOT ,'\0','\0','\0','\0',DC4 ,NAK ,'\0',SUB},
                      { SP ,'\0','\0','\0','\0','\0','\0','\0','\0','\0','\0','.' ,'<' ,'(' ,'+' ,'|'},
                      {'&' ,'\0','\0','\0','\0','\0','\0','\0','\0','\0','!' ,'$' ,'*' ,')' ,';' ,'\0'},
                      {'-' ,'/' ,'\0','\0','\0','\0','\0','\0','\0','\0','\0',',' ,'%' ,'_' ,'>' ,'\?'},
                      {'\0','\0','\0','\0','\0','\0','\0','\0','\0','`' ,':' ,'#' ,'@' ,'\'','=' ,'\"'},
                      {'\0','a' ,'b' ,'c' ,'d' ,'e' ,'f' ,'g' ,'h' ,'i' ,'\0','\0','\0','\0','\0','\0'},
                      {'\0','j' ,'k' ,'l' ,'m' ,'n' ,'o' ,'p' ,'q' ,'r' ,'\0','\0','\0','\0','\0','\0'},
                      {'\0','~' ,'s' ,'t' ,'u' ,'v' ,'w' ,'x' ,'y' ,'z' ,'\0','\0','\0','\0','\0','\0'},
                      {'^' ,'\0','\0','\0','\0','\0','\0','\0','\0','\0','[' ,']' ,'\0','\0','\0','\0'},
                      {'{' ,'A' ,'B' ,'C' ,'D' ,'E' ,'F' ,'G' ,'H' ,'I' ,'\0','\0','\0','\0','\0','\0'},
                      {'}' ,'J' ,'K' ,'L' ,'M' ,'N' ,'O' ,'P' ,'Q' ,'R' ,'\0','\0','\0','\0','\0','\0'},
                      {'\\','\0','S' ,'T' ,'U' ,'V' ,'W' ,'X' ,'Y' ,'Z' ,'\0','\0','\0','\0','\0','\0'},
                      {'0' ,'1' ,'2' ,'3' ,'4' ,'5' ,'6' ,'7' ,'8' ,'9' ,'\0','\0','\0','\0','\0','\0'}};


int main()
{
    char a,b;

    while(1)
    {
        a=getchar();
        if(a=='\n')
            break;
        b=getchar();
        if(a>='0'&&a<='9')
        {
            if(b>='0'&&b<='9')
                printf("%02X",c[a-'0'][b-'0']);
            else if(b>='A'&&b<='F')
                printf("%02X",c[a-'0'][b-'A'+10]);
        }
        else if(a>='A'&&a<='F')
        {
            if(b>='0'&&b<='9')
                printf("%02X",c[a-'A'+10][b-'0']);
            else if(b>='A'&&b<='F')
                printf("%02X",c[a-'A'+10][b-'A'+10]);
        }
    }
    printf("\n");

    return 0;
}