Hexadecimal View （2011ACM亚洲大连赛区现场赛D题）

Hexadecimal View

---

Time Limit: 2 Seconds      Memory Limit: 65536 KB

---

Hexadecimal is very important and useful for computer programmers. You are requested to provide a hexadecimal view for given data. The hexadecimal view is made up of one or more rows. Every row except the last one represents 16 characters. Each row consists of three columns separated by a space:

• addr: the 4-digit hexadecimal beginning address of this row.
• dump: the hexadecimal representation of this row, separating every two characters by a whitespace. If there are less than 16 characters in the last row, pad it with spaces.
• text: the ASCII translation of this row, with uppercase characters converted to lowercase and lowercase characters converted to uppercase.

Use lowercase for the letter digits. See sample for more details.

Input

There are multiple test cases. Each line is a test case. The line is made up of no less than 1 and no more than 4096 printable characters including spaces.

Output

For each test case, output its hexadecimal view. Do not output any extra spaces after the last character of text.

Sample Input

Hex Dump
#include <cstdio>
printf("Hello, World!\n");
main = do getLine >>= print . sum . map read . words

Sample Output

0000: 4865 7820 4475 6d70                     hEX dUMP
0000: 2369 6e63 6c75 6465 203c 6373 7464 696f #INCLUDE <CSTDIO
0010: 3e                                      >
0000: 7072 696e 7466 2822 4865 6c6c 6f2c 2057 PRINTF("hELLO, w
0010: 6f72 6c64 215c 6e22 293b                ORLD!\N");
0000: 6d61 696e 203d 2064 6f20 6765 744c 696e MAIN = DO GETlIN
0010: 6520 3e3e 3d20 7072 696e 7420 2e20 7375 E >>= PRINT . SU
0020: 6d20 2e20 6d61 7020 7265 6164 202e 2077 M . MAP READ . W
0030: 6f72 6473                               ORDS

---

Author: WU, Zejun

题目比较简单，输入输出注意点。。。

看来要去复习下C++的输入输出的函数了。。。。。

WR了好多次，原来数组太小。题目给的是4096的，我卡的太紧了，，，下次开大点了。。。

C语言的函数轻松过掉

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<iomanip>
using namespace std;
char str[5000];
int main()
{
    int t=0;
    int len;
    while(cin.getline(str,4100))//这里要开大点，错了好几次了 
    {
        t=0;
        len=strlen(str);
        for(t=0;t<len;t+=16)
        {
             printf("%04x: ",t);
             for(int i=t;i<t+16;i+=2)
             {
                 if(i<len)  printf("%02x",str[i]);
                 else  printf("  ");
                 if(i+1<len) printf("%02x ",str[i+1]);
                 else  printf("   ");
             }    
             for(int i=t;i<t+16&&i<len;i++)
             {
                 if(str[i]>='a'&&str[i]<='z') printf("%c",str[i]-'a'+'A');
                 else if(str[i]>='A'&&str[i]<='Z')  printf("%c",str[i]-'A'+'a');
                 else printf("%c",str[i]);
             }    
            printf("\n");
         }    
    }  
    return 0;  
}