uva 401.Palindromes

Posted 2020-07-03 windysai

题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=342

题目意思：给出一段字符串（大写字母+数字组成）。判断是否为回文串 or 镜像串 or 回文镜像串 or 什么都不是。每个字母的镜像表格如下

 

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

 

注意是没有数字0的哦。（该题，数字 0 与字母 O 看成是一样的）

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <cstring>
 5 using namespace std;
 6 
 7 const int maxn = 1000 + 5;
 8 char mirror[] = "A   3  HIL JM O   2TUVWXY51SE Z  8 ";
 9 const char *msg[4] = {" -- is not a palindrome.", " -- is a regular palindrome.", " -- is a mirrored string.", " -- is a mirrored palindrome."};
10 char s[maxn];
11 
12 char change(char ch)
13 {
14     if (ch >= ‘A‘ && ch <= ‘Z‘)
15        return mirror[ch-‘A‘];
16     return mirror[ch-‘0‘+25];
17 }
18 
19 int main()
20 {
21     #ifndef ONLINE_JUDGE
22         freopen("in.txt", "r", stdin);
23     #endif // ONLINE_JUDGE
24 
25     while (scanf("%s", s) != EOF) {
26         int len = strlen(s);
27         int p = 1, m = 1;
28 
29         for (int i = 0; i < len; i++) {
30             if (s[i] != s[len-i-1]) p = 0;
31             if (change(s[i]) != s[len-i-1]) m = 0;
32         }
33         printf("%s%s\n\n", s, msg[p+m*2]);
34     }
35     return 0;
36 }