Title Link: palindromes
UVA-401
Palindromes
Time Limit:3000MS |
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Memory Limit: Unknown |
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64bit IO Format:%lld &%llu |
Submit Status
Description
A regular palindrome is a string of numbers or letters so is the same forward as backward. For example, the string"abcdedcba" was a palindrome because it was the same when the string was read from L EFT to right as when the string was read from right to left.
A mirrored string is a string for which when each of the elements of the string was changed to its reverse (if it had a rev Erse) and the string is read backwards the result is the same as the original string. For example, the string"3AIAE" was a mirrored string because "a" and "I" are their own Reverses, and"3" and "E" are each others ' reverses.
A mirrored palindrome is a string that meets the criteria of a regular palindrome and the criteria of a mirrored string. The string"Atoyota" is a mirrored palindrome because if the string was read backwards, the string is the Same as the original and because if each of the characters are replaced by their reverse and the result is read backwards, th E result is the same as the original string.
Of course, "A", "T" , "O" , and "Y" Is all their own reverses.
A List of all valid characters and their reverses are as follows.
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 |
|
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Note that O (zero) and 0 (the letter) is considered the same character and therefore only the "0" is a valid character.
Input
Input consists of strings (one per line) each of the which would consist of one to twenty valid characters. There'll be is no invalid characters in any of the strings. Your program should read to the end of file.
Output
For each input string, you should print the string starting in column 1 immediately followed by exactly one of the FOLLOWI ng strings.
STRING |
CRITERIA |
" --is not a palindrome." |
If the string is not a palindrome and is not a mirrored string |
" --is a regular palindrome." |
If the string is a palindrome and was not a mirrored string |
" --is a mirrored string." |
If the string is not a palindrome and is a mirrored string |
" --is a mirrored palindrome." |
If the string is a palindrome and is a mirrored string |
Note The output line was to include the -' s and spacing exactly as shown in the table above and demonstrated in the Sample Output below.
In addition, after the output line, you must print a empty line.
Sample Input
Notapalindrome Isapalinilapasi 2a3meas Atoyota
Sample Output
Notapalindrome Isapalinilapasi 2a3meas Atoyota --is a mirrored palindrome.
Hint
Use the C + + 's class of string would be convenient, but not a must
Do a string trick.
Test instructions: Judging is not a palindrome string and symmetric string
Thinking: Write two for the function of judgment OK ~ But that is the valid characters all to list, a little trouble.
AC code (PE up to two times.) ):
/*************************************************************************> File name:d.cpp> Author:zzuspy > Mail: [email protected] > Created time:2014 December 01 Monday 16:42 35 seconds **************************************** /#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <cstdlib> #include <cmath> #include <stack> #include <queue>using namespace Std;int a[100];void init () {a[' a '] = ' a '; a[' E '] = ' 3 '; a[' h '] = ' h '; a[' i '] = ' I '; a[' J '] = ' l '; a[' l '] = ' J '; a[' M '] = ' M '; a[' o '] = ' o '; a[' S '] = ' 2 '; a[' t '] = ' t '; a[' u '] = ' u '; a[' V '] = ' V '; a[' w '] = ' w '; a[' x '] = ' x '; a[' y '] = ' y '; a[' Z '] = ' 5 ' ; a[' 1 '] = ' 1 '; a[' 2 '] = ' S '; a[' 3 '] = ' E '; a[' 5 '] = ' Z '; a[' 8 '] = ' 8 ';} int Judge1 (char b[], int n) {for (int i=0; i<=n/2; i++) {if (b[i]!=b[n-i-1]) return 0;} return 1;} int Judge2 (char b[], int n) {for (int i=0; i<n; i++) {if (b[i]==0) return 0;} for (int i=0; i<=n/2; i++) {if (A[b[i]! = b[n-i-1]) return 0;} Return 1;} int main () {init (); Char Str[25];while (scanf ("%s", str)!=eof) {int len = strlen (str); int a = Judge1 (str, len), B = Judge2 (str , Len), if (a==1&&b==1) printf ("%s--is a mirrored palindrome.\n\n", str); else if (a==0&&b==1) printf ("%s- -is a mirrored string.\n\n ", str); else if (a==1&&b==0) printf ('%s--is a regular palindrome.\n\n ', str); else if (a= =0&&b==0) printf ("%s--is not a palindrome.\n\n", str);} return 0;}
Uva-401-palindromes