判断一个字符串是不是镜像串和回文串

下表是题目所给的对称字符。

输入格式：输入多组数据，每行一个字符串

输出格式：”字符串—是否是回文串，镜像串，回文镜像串."

输入样例：

NOTAPALINDROMEISAPALINILAPASI2A3MEASATOYOTA

输出样例：

NOTAPALINDROME -- is not a palindrome.ISAPALINILAPASI -- is a regular palindrome.2A3MEAS -- is a mirrored string.ATOYOTA -- is a mirrored palindrome.

直接打表，列举情况就可以

#include
    
     #include
     
      int main(){    char save[200],a[200];    int i,j;    int flag1=1,flag2=1;    memset(save,0,sizeof(save));    save['A']='A';    save['E']='3';    save['H']='H';    save['I']='I';    save['J']='L';    save['L']='J';    save['M']='M';    save['O']='O';    save['S']='2';    save['T']='T';    save['U']='U';    save['V']='V';    save['W']='W';    save['X']='X';    save['Y']='Y';    save['Z']='5';    save['2']='S';    save['1']='1';    save['3']='E';    save['5']='Z';    save['8']='8';    while(scanf("%s",a)!=EOF)    {        flag1=1;        flag2=1;        i=strlen(a);        for(j=0;j<=(i+1)/2-1;j++)        {            if(a[j]!=a[i-j-1])                flag1=0;            if(a[j]!=save[a[i-j-1]])                flag2=0;        }        if(flag1)        {            if(flag2)                printf("%s -- is a mirrored palindrome.\n",a);            else                printf("%s -- is a regular palindrome.\n",a);        }        else if(flag2)            printf("%s -- is a mirrored string.\n",a);        else            printf("%s -- is not a palindrome.\n",a);        printf("\n");    }    return 0;}