题目描述:
输入一个字符串,判断它是否为回文以及镜像串。输入字符串保证不含数字0.所谓回文串,就是反转之后原串相同,如abba和madam。所谓镜像串,就是左右镜像之后和原串相同,如2S和3AIAE。注意,并不是每个字符在镜像之后都能得到一个合法字符,本题中,每个字符的镜像如下所示,(空白项表示该字符镜像后不能得到一个合法的字符)。
Character | Reverse | Character | Reverse | Character | Reverse |
A | A | M | M | Y | Y |
B | N | Z | 5 | ||
C | O | O | 1 | 1 | |
D | P | 2 | S | ||
E | 3 | Q | 3 | E | |
F | R | 4 | |||
G | S | 2 | 5 | Z | |
H | H | T | T | 6 | |
I | I | U | U | 7 | |
J | L | V | V | 8 | 8 |
K | W | W | 9 | ||
L | J | X | X |
样例输入:
NOTAPALINDROME ISAPALINILAPASI 2A3MEAS ATOYOTA 样例输出:
NOTAPALINDROME -- is not a palindrome. ISAPALINILAPASI -- is a regular palindrome. 2A3MEAS -- is a mirrored string. ATOYOTA -- is a mirrored palindrome.
#include<stdio.h>
#include<string.h>
char r(char ch)
{
char s[]="A 3 HIL JM O 2TUVWXY51SE Z 8 ";
if(ch>='A'&&ch<='Z')
return s[ch-'A'];
else if(ch>='1'&&ch<='9')
return s[ch-'0'+25];
else return ' ';
}
int main()
{
int i,n;
char s[50];
char m[4][100]={"not a palindrome","a regular palindrome","a mirrored string","a mirrored palindrome"};
while(scanf("%s",s)==1)
{
int flag1=1,flag2=1;
n=strlen(s);
for(i=0; i < (n+1)/2;i++)
{
if(s[i]!=s[n-i-1]) flag1=0; //判断是否为回文串
if(r(s[i])!=s[n-i-1]) flag2=0; //判断是否为镜像串
}
printf("%s -- is %s\n",s,m[flag2*2+flag1]);
}
return 0;
}