題目描述
輸入格式
輸出格式
題意翻譯
輸入是一個字符串,判斷它是否爲迴文串以及鏡像串。輸入字符串保證不含數字00 。所謂迴文串,就是反轉以後和原串相同,如abbaabba 和madammadam 。所謂鏡像串,就是左右鏡像之後和原串相同,如2S2S 和3AIAE3AIAE 。注意,並不是每個字符在鏡像之後都能得到一個合法字符。在本題中,每個合法字符的鏡像如下表所示:
Character Reverse
A A
E 3
H H
I I
J L
L J
M M
O O
S 2
T T
U U
V V
W W
X X
Y Y
Z 5
1 1
2 S
5 Z
8 8
輸入輸出樣例
輸入 #1
NOTAPALINDROME
ISAPALINILAPASI
2A3MEAS
ATOYOTA
輸出 #1
NOTAPALINDROME -- is not a palindrome.
ISAPALINILAPASI -- is a regular palindrome.
2A3MEAS -- is a mirrored string.
ATOYOTA -- is a mirrored palindrome.
代碼如下:
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<cstdio>
using namespace std;
#pragma warning(disable:4996)
int n, l, i, a[10000];
const char* rev = "A 3 HIL JM O 2TUVWXY51SE Z 8 ";
const char* msg[] = {"not a palindrome","a regular palindrome","a mirrored string","a mirrored palindrome"};
char r(char ch) {
if (isalpha(ch))return rev[ch-'A'];//是字母,返回反轉的字母/數字
return rev[ch-'0'+25];//如果是數字,返回字母(字母26+數字9)
}
int main()
{
char s[30];
while (scanf("%s",s)==1)
{
int len = strlen(s);//計算長度,二分,左邊右邊相比較
int p = 1,m = 1;//標誌
for (int i=0;i<(len+1)/2;i++)//左右相比
{
if (s[i] != s[len - 1 - i])p = 0;//不是迴文串
if (r(s[i]) != s[len - 1 - i])m = 0;//不是鏡像串
}
printf("%s -- is %s.\n\n",s,msg[m*2+p]);//按格式打印
}
return 0;
}