Palindromes

要注意:单个字符是没有对应反序串的。

几个测试:

 1 NOTAPALINDROME -- is not a palindrome.
2
3 ISAPALINILAPASI -- is a regular palindrome.
4
5 2A3MEAS -- is a mirrored string.
6
7 ATOYOTA -- is a mirrored palindrome.
8
9 B -- is a regular palindrome.
10
11 E -- is a regular palindrome.
12
13 8 -- is a mirrored palindrome.

我用的 switch 结构:

/* UVa 401 */
# include <stdio.h>
# include <string.h>

# define MAXN 25

char s[MAXN], ss[MAXN];

char ch_map(char ch);
void s_map(char *s, char *ss);
int is_pal(char *s);

int main()
{
int isPal, isMir, i;

while (~scanf("%s", s))
{
for (i = 0; s[i] ; ++i)
printf("%c", s[i]);
s_map(s, ss);
isPal = is_pal(s);
isMir = is_pal(ss);
if (strlen(s)==1 && ch_map(s[0])!=s[0]) isMir = 0;
if (isPal && isMir) printf(" -- is a mirrored palindrome.\n");
else if (isMir) printf(" -- is a mirrored string.\n");
else if (isPal) printf(" -- is a regular palindrome.\n");
else printf(" -- is not a palindrome.\n");
printf("\n");
}
return 0;
}

char ch_map(char ch)
{
switch (ch)
{
case 'E': return '3';
case 'J': return 'L';
case 'L': return 'J';
case 'S': return '2';
case 'Z': return '5';
case '2': return 'S';
case '3': return 'E';
case '5': return 'Z';
case 'A': case 'H': case 'I':
case 'M': case 'O': case 'T':
case 'U': case 'V': case 'W':
case 'X': case 'Y': case '1':
case '8':return ch;
default : return '0';
}
}

void s_map(char *s, char *ss)
{
int i, len, L;
len = strlen(s);
L = len / 2;
for (i = 0; i < L; ++i)
ss[i] = s[i];
for ( ; s[i] ; ++i)
ss[i] = ch_map(s[i]);
ss[i] = '\0';
}

int is_pal(char *s)
{
int i, len, L;
len = strlen(s);
L = len / 2;

for (i = 0; i < L; ++i)
if (s[i] != s[len-1-i]) return 0;
return 1;
}

posted on 2012-03-15 10:07  getgoing  阅读(196)  评论(0编辑  收藏  举报

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