算法学习之路|回文镜像串

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：
NOTAPALINDROME
ISAPALINILAPASI
2A3MEAS
ATOYOTA

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<stdio.h>
#include<string.h>
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;
}