ZOJ 3542 Hexadecimal View [2011大连现场赛]

题意：

分析：

//AC CODE：

#include<iostream>
#include<cmath>
#include<algorithm>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>

using namespace std;

char str[4100];
void print_num(int a)
{
if(a<16)
{
printf("000%x: ",a);
}
else if(a<0x100)
{
printf("00%x: ",a);
}
else if(a<0x1000)
{
printf("0%x: ",a);
}
else
{
printf("%x: ",a);
}
}
int main()
{
while(gets(str))
{
int len=strlen(str);
int countn=(int)(len/16.0+0.999999);
for(int i=0;i<countn;i++)
{
print_num(16*i);

for(int j=i*16;j<(i+1)*16;j++)
{
if(j<len)
{
printf("%x",str[j]);
}
else
{
printf("  ");
}
if((j+1)%2==0)
printf(" ");
}

for(int j=i*16;j<(i+1)*16&&j<len;j++)
{
if(str[j]>='A'&&str[j]<='Z')
{
printf("%c",str[j]+'a'-'A');
}
else if(str[j]>='a'&&str[j]<='z')
{
printf("%c",str[j]-'a'+'A');
}
else
printf("%c",str[j]);
}

printf("\n");
}
}
return 0;
}

附上大牛的代码，转自http://watashi.ws/blog/2102/icpc-2011-dalian-dlut-cont/comment-page-1/#comment-20486

#include <cctype>
#include <cstdio>
#include <cassert>
#include <cstring>

int main()
{
int len;
char buf[1 << 20];
while (fgets(buf, sizeof(buf), stdin) != NULL)
{
len = strlen(buf) - 1;
assert(0 < len && len <= 4096);
for (int off = 0; off < len; off += 16)
{
printf("%04x: ", off);
for (int pos = off; pos < off + 16; ++pos)
{
if (pos < len)
{
printf("%02x", buf[pos]);
}
else
{
printf("  ");
}
if (pos % 2 != 0)
{
putchar(' ');
}
}
for (int pos = off; pos < off + 16 && pos < len; ++pos)
{
assert(isprint(buf[pos]));
putchar(isupper(buf[pos]) ? tolower(buf[pos]) : toupper(buf[pos]));
}
putchar('\n');
}
}

return 0;
}

仔细研读，可以学到不少东西！

printf("%04x: ", off);

putchar(isupper(buf[pos]) ? tolower(buf[pos]) : toupper(buf[pos]));