Uva401Palindromes
| Palindromes |
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.
| 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 |
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.
| 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 is 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 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.
Miguel Revilla 2001-04-16
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int namx = ; bool is_mirrored(char a, char b)
{
if((a == 'E' && b == '') || (a == '' && b == 'E'))
return true;
else if((a == 'L' && b == 'J') || (a == 'J' && b == 'L'))
return true;
else if((a == 'S' && b == '') || (a == '' && b == 'S'))
return true;
else if((a == 'Z' && b == '') || (a == '' && b == 'Z'))
return true;
else if(a == b)
{
if(a == 'A' || a == 'H' || a == 'I' || a == 'M' || a == 'O'
|| a == 'T' || a == 'U' || a == 'V' || a == 'W' || a == 'X'
|| a == 'Y' || a == '' || a == '')
{
return true;
}
}
else
return false;
} int main()
{
char str[namx]; while(scanf("%s%*c", str) != EOF)
{
int len = strlen(str);
//printf("%d\n", len);
bool mark = true;
bool tag = true;
for(int i = ; i <= len / ; ++i)
{
if(str[i] != str[len-i-])
{
mark = false;
//printf("%c == %c %d\n", str[i], str[len-i-1], len-i-1);
break;
}
} for(int i = ; i <= len / ; ++i)
{
if(!is_mirrored(str[i], str[len-i-]))
{
tag = false;
//printf("%c == %c %d\n", str[i], str[len-i-1], len-i-1);
break;
}
}
if(mark == false && tag == false)
printf("%s -- is not a palindrome.\n", str); else if(mark == false && tag == true)
printf("%s -- is a mirrored string.\n", str); else if(mark == true && tag == false)
printf("%s -- is a regular palindrome.\n", str); else
printf("%s -- is a mirrored palindrome.\n", str);
printf("\n"); }
return ;
}
一个空格, PE一晚上,淡定淡定,不要太大压力
在14/11/2队内赛时的代码
#include <iostream>
#include <cstdio>
#include <cstring> using namespace std;
const int MAXN = ;
char str[MAXN]; bool CharTest(char c)
{
if(c == 'A' || c == 'H' || c == 'I' || c == 'M' || c == 'O' || c == 'T' || c == 'U'
|| c == 'V' || c == 'W' || c == 'X' || c == 'Y' || c == '' || c == '')
{
return true;
}
return false;
} bool StrTest(char *s)
{
int sum = ;
int len = strlen(s);
for(int i = ; i < len; ++i)
{
sum += (CharTest(s[i]) == true) ? : ;
}
if(sum == len)
return true;
return false;
} char transfer(char s)
{
if(s == '')
return 'E';
else if(s == 'J')
return 'L';
else if(s == '')
return 'S';
else if(s == '')
return 'Z';
} int test(char *s, int len)
{
int sum = ;
for(int i = ; i < len/; ++i)
{
if(str[i] == str[len-i-])
{
sum++;
}
}
return sum;
} int main()
{
while(scanf("%s", str) != EOF)
{
int len = strlen(str); if(test(str, len) == len/)
{
if(StrTest(str))
{
printf("%s -- is a mirrored palindrome.\n\n", str);
}
else
{
printf("%s -- is a regular palindrome.\n\n", str);
}
}
else
{
char s1[];
for(int i = ; i < len; ++i)
{
s1[i] = str[i];
}
s1[len] = '\0';
for(int i = ; i < len; ++i)
{
if(str[i] == '' || str[i] == 'J' || str[i] == '' || str[i] == '')
{
str[i] = transfer(str[i]);
}
} if(test(str, len) == len/)
{
printf("%s -- is a mirrored string.\n\n", s1);
}
else
{
printf("%s -- is not a palindrome.\n\n", s1);
}
} }
return ;
}
Uva401Palindromes的更多相关文章
- [算法练习] UVA-401-Palindromes
UVA Online Judge 题目401 Palindromes 回文串 问题描述: 回文串(Palindromes)就是正着读和反着读完全一样的字符串,例如"ABCDEDCBA&qu ...
- UVa-401-Palindromes(回文)
这一题的话我们可以把映像字符的内容给放入一个字符串常量里面,然后开辟一个二维的字符串常量数组,里面放置答案. 对于回文实际上是很好求的,对于镜像的话,我们写一个返回char的函数,让它接收一个char ...
- UVA401-Palindromes(紫书例题3.3)
A regular palindrome is a string of numbers or letters that is the same forward as backward. For exa ...
随机推荐
- oracle 的PACKAGE恢复过程
SELECT obj# FROM obj$ AS OF TIMESTAMP TO_TIMESTAMP('2016-06-30', 'YYYY-MM-DD') WHERE NAME = 'PFWZ_AP ...
- 第K 小数
[问题描述]有两个正整数数列,元素个数分别为N和M.从两个数列中分别任取一个数相乘,这样一共可以得到N*M个数,询问这N*M个数中第K小数是多少.[输入格式]输入文件名为number.in.输入文件包 ...
- 词频统计_输入到文件_update
/* 输入文件见337.in.txt 输出文件见338.out.txt */ #include <iostream> #include <cctype> #include &l ...
- Retrofit与RXJava整合
Retrofit 除了提供了传统的 Callback 形式的 API,还有 RxJava 版本的 Observable 形式 API.下面我用对比的方式来介绍 Retrofit 的 RxJava 版 ...
- NotePad ++的妙用:添加代码行数和格式不变复制代码
NotePad ++ 不仅安装包小而且功能强大,可以支持很多语言.这里简单阐述下两个功能: 一.在代码前添加行数: 1.用NotePad ++打开一个文件,一般NotePad ++会自动识别这是什么语 ...
- linux cpuInfo
转自:http://blog.csdn.net/lgstudyvc/article/details/7889364 /proc/cpuinfo文件分析 在Linux系统中,提供了proc文件系统显 ...
- .NET生成带Logo的二维码
使用ThoughtWorks.QRCode生成,利用这个库来生成带Logo的二维码(就是中间嵌了一个图片的二维码),直接见代码: HttpContext context = HttpContext.C ...
- 一致性hash算法简介与代码实现
一.简介: 一致性hash算法提出了在动态变化的Cache环境中,判定哈希算法好坏的四个定义: 1.平衡性(Balance) 2.单调性(Monotonicity) 3.分散性(Spread) 4.负 ...
- 1080P、720P、4CIF、CIF所需要的理论带宽
转自:http://blog.sina.com.cn/s/blog_64684bf30101hdl7.html 在视频监控系统中,对存储空间容量的大小需求是与画面质量的高低.及视频线路等都有很大关系. ...
- Unity3D打Box游戏
先学习一些基本的脚本实现: 1.动态创建物体.默认位置是(0,0)位置 GameObject goNew = GameObject.CreatePrimitive(PrimitiveType.Cube ...