回文数判断
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Description
A regular palindrome is a string of numbers or letters that is the same forward as backward. For example, the string "ABCDEDCBA" is a palindrome because it is the same when the string is read from left to right as when the string is read from right to left.
A mirrored string is a string for which when each of the elements of the string is changed to its reverse (if it has a reverse) and the string is read backwards the result is the same as the original string. For example, the string "3AIAE" is 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 is read backwards, the string is the same as the original and because if each of the characters is replaced by its reverse and the result is read backwards, the result is the same as the original string.
Of course,"A","T", "O", and "Y" are all their own reverses.
A list of all valid characters and their reverses is as follows.
Note that O (zero) and 0 (the letter) are considered the same character and therefore ONLY the letter "0" is a valid character.
Input
Input consists of strings (one per line) each of which will consist of one to twenty valid characters. There will be 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 following strings.
Note that the output line is to include the -'s and spacing exactly as shown in the table above and demonstrated in the Sample Output below.
In addition, after each output line, you must print an empty line.
Sample Input
NOTAPALINDROME ISAPALINILAPASI 2A3MEAS ATOYOTA
Sample Output
NOTAPALINDROME -- is not a palindrome. ISAPALINILAPASI -- is a regular palindrome. 2A3MEAS -- is a mirrored string. ATOYOTA -- is a mirrored palindrome.
Hint
use the C++'s class of string will be convenient, but not a must
- #include<iostream>
- #include<cctype>
- #include<cstring>
- using namespace std;
- char zmrev[99] = "A***3**HIL*JM*O***2TUVWXY5"; //常数数组有时可以极大的简化代码。
- char surev[99] = "1SE*Z**8*";
- char s[1000];
- char diu(char c)
- {
- if (c >= 'A'&&c <= 'Z') return zmrev[c - 'A']; //diu函数返回处理过后的字符串,便于后面判断。
- if (c >= '1'&&c <= '9') return surev[c - '1'];
- }
- int main()
- {
- while (cin >> s)
- {
- int len = strlen(s);
- int ispalindrome = 1;
- int ismirrored = 1;
- for (int i = 0; i<=(len/ 2); i++) //要注意是等于号,因为开始没注意wrong了几次。
- {
- if (s[i] != s[len - i - 1]) ispalindrome = 0;
- if (s[i] != diu(s[len - i - 1])) ismirrored = 0;
- }
- if (ispalindrome == 0 && ismirrored == 0) cout << s << " -- is not a palindrome." << endl;
- else if (ispalindrome == 1 && ismirrored == 0) cout << s << " -- is a regular palindrome." << endl;
- else if (ispalindrome == 0 && ismirrored == 1) cout << s << " -- is a mirrored string." << endl;
- else if (ispalindrome == 1 && ismirrored == 1) cout << s << " -- is a mirrored palindrome." << endl;
- cout << endl; //格式。
- }
- }
这里的处理方法比较巧妙,主要是运用了常数组来提高效率,还有就是运用标志变量来衡量它们的类别,这都是ACM里常用的一些技巧。其中一个比较巧妙的设置就是数组的设定,通过观察可得出原来的字符与镜像字符之间的关系,当然不是我观察得出的,一开始我是用了暴力破解法,这篇博文是我在网上看到的。
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