UVa 401 - Palindromes -AC
这道题看完题思路大致清晰，总体分开两个函数写，一个判断回文串，一个判断镜像串。
回文串判断好写，循环判断前半截与后半截是否一致即可。
关键在判断镜像串，一开始就也没多想，简单的判断呗~虽然知道要写好多，但也懒得细想。先判断字符串中有没有不是镜像字符的，有就直接返回0，没有再依次判断对称位置上的镜像字符是否对应，有不对应的就直接返回0，全都对应最后返回1.
虽然直接AC了，但这样写完了190行，突然感觉我好傻~
先贴出来吧~
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_SIZE 10005
int pal(char str[MAX_SIZE])//palindrome
{
int i,j,slen;
slen=strlen(str);
for (i=0,j=slen-1;i<slen/2;i++,j--)
if (str[i]!=str[j])
break;
if (i==slen/2)
return 1;
return 0;
}
int mir (char str[MAX_SIZE])//mirrored
{
int i,j,slen;
slen=strlen(str);
for (i=0;i<slen;i++)
{
switch (str[i])
{
case 'B':
return 0;
case 'C':
return 0;
case 'D':
return 0;
case 'F':
return 0;
case 'G':
return 0;
case 'K':
return 0;
case 'N':
return 0;
case 'P':
return 0;
case 'Q':
return 0;
case 'R':
return 0;
case '4':
return 0;
case '6':
return 0;
case '7':
return 0;
case '9':
return 0;
}
}
for (i=0;i<slen/2;i++)
{
switch(str[i])
{
case 'A':
if (str[slen-i-1]=='A')
continue;
else
return 0;
case 'E':
if (str[slen-i-1]=='3')
continue;
else
return 0;
case 'H':
if (str[slen-i-1]=='H')
continue;
else
return 0;
case 'I':
if (str[slen-i-1]=='I')
continue;
else
return 0;
case 'J':
if (str[slen-i-1]=='L')
continue;
else
return 0;
case 'L':
if (str[slen-i-1]=='J')
continue;
else
return 0;
case 'M':
if (str[slen-i-1]=='M')
continue;
else
return 0;
case 'O':
if (str[slen-i-1]=='O')
continue;
else
return 0;
case 'S':
if (str[slen-i-1]=='2')
continue;
else
return 0;
case 'T':
if (str[slen-i-1]=='T')
continue;
else
return 0;
case 'U':
if (str[slen-i-1]=='U')
continue;
else
return 0;
case 'V':
if (str[slen-i-1]=='V')
continue;
else
return 0;
case 'W':
if (str[slen-i-1]=='W')
continue;
else
return 0;
case 'X':
if (str[slen-i-1]=='X')
continue;
else
return 0;
case 'Y':
if (str[slen-i-1]=='Y')
continue;
else
return 0;
case 'Z':
if (str[slen-i-1]=='5')
continue;
else
return 0;
case '1':
if (str[slen-i-1]=='1')
continue;
else
return 0;
case '2':
if (str[slen-i-1]=='S')
continue;
else
return 0;
case '3':
if (str[slen-i-1]=='E')
continue;
else
return 0;
case '5':
if (str[slen-i-1]=='Z')
continue;
else
return 0;
case '8':
if (str[slen-i-1]=='8')
continue;
else
return 0;
}
}
return 1;
}
int main()
{
char str[MAX_SIZE];
while (gets(str)!=NULL)
{
int pa,mi;
pa=pal(str);
mi=mir(str);
if (pa==0&&mi==0)
printf("%s -- is not a palindrome.\n",str);
else if (pa==1&&mi==0)
printf("%s -- is a regular palindrome.\n",str);
else if (pa==0&&mi==1)
printf("%s -- is a mirrored string.\n",str);
else if (pa&&mi)
printf("%s -- is a mirrored palindrome.\n",str);
printf("\n");
}
return 0;
}
AC之后感觉这题这么写实在太狠了，然后又看了一下网上代码，看到
首先可以用两个数组把题目中的表格存起来
char *ch = "AEHIJLMOSTUVWXYZ12358";
char *re = "A3HILJMO2TUVWXY51SEZ8";
这才受到启发，把原来代码进行了改进，但这样要注意判断镜像串的时候要讨论字符串长度是否为1，否则会WA
修改后AC的代码：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_SIZE 10005
int pal(char str[MAX_SIZE])
{
int i,j,slen;
slen=strlen(str);
for (i=0,j=slen-1;i<slen/2;i++,j--)
if (str[i]!=str[j])
break;
if (i==slen/2)
return 1;
return 0;
}
int mir (char str[MAX_SIZE])
{
char *ch="AEHIJLMOSTUVWXYZ12358";
char *re="A3HILJMO2TUVWXY51SEZ8";
int i,j,slen;
slen=strlen(str);
if (slen==1)
{
for (j=0;j<strlen(ch);j++)
if (str[0]==ch[j]&&str[0]==re[j])
return 1;
return 0;
}
else
{
for (i=0;i<slen/2;i++)
{
for (j=0;j<strlen(ch);j++)
if (str[i]==ch[j]&&str[slen-i-1]==re[j])
{
j=-1;
break;
}
if (j!=-1)
return 0;
}
return 1;
}
}
int main()
{
char str[MAX_SIZE];
while (gets(str)!=NULL)
{
int pa,mi;
pa=pal(str);
mi=mir(str);
if (pa==0&&mi==0)
printf("%s -- is not a palindrome.\n",str);
else if (pa==1&&mi==0)
printf("%s -- is a regular palindrome.\n",str);
else if (pa==0&&mi==1)
printf("%s -- is a mirrored string.\n",str);
else if (pa&&mi)
printf("%s -- is a mirrored palindrome.\n",str);
printf("\n");
}
return 0;
}