A. Mr. Kitayuta's Gift
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mr. Kitayuta has kindly given you a string s consisting of lowercase English letters. You are asked to insert exactly one lowercase English letter into s to make it a palindrome. A palindrome is a string that reads the same forward and backward. For example, "noon", "testset" and "a" are all palindromes, while "test" and "kitayuta" are not.

You can choose any lowercase English letter, and insert it to any position of s, possibly to the beginning or the end of s. You have to insert a letter even if the given string is already a palindrome.

If it is possible to insert one lowercase English letter into s so that the resulting string will be a palindrome, print the string after the insertion. Otherwise, print "NA" (without quotes, case-sensitive). In case there is more than one palindrome that can be obtained, you are allowed to print any of them.

Input

The only line of the input contains a string s (1 ≤ |s| ≤ 10). Each character in s is a lowercase English letter.

Output

If it is possible to turn s into a palindrome by inserting one lowercase English letter, print the resulting string in a single line. Otherwise, print "NA" (without quotes, case-sensitive). In case there is more than one solution, any of them will be accepted.

Sample test(s)
Input
revive
Output
reviver
Input
ee
Output
eye
Input
kitayuta
Output
NA
Note

For the first sample, insert 'r' to the end of "revive" to obtain a palindrome "reviver".

For the second sample, there is more than one solution. For example, "eve" will also be accepted.

For the third sample, it is not possible to turn "kitayuta" into a palindrome by just inserting one letter.

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 100001
const int inf=0x7fffffff; //无限大
int check(string s)
{
for(int i=;i<s.size();i++)
{
if(s[i]!=s[s.size()--i])
return ;
}
return ;
}
int main()
{
string s;
cin>>s ;
for(char c='a';c<='z';c++)
{
for(int i=;i<=s.size();i++)
{
string a = s;
string b=" ";
b[] = c;
a.insert(i,b);
if (check(a))
{
cout<<a<<endl;
return ;
}
}
}
cout<<"NA"<<endl;
return ;
}

Codeforces 505A Mr. Kitayuta's Gift 暴力的更多相关文章

  1. codeforces 505A. Mr. Kitayuta's Gift 解题报告

    题目链接:http://codeforces.com/problemset/problem/505/A 题目意思:给出一个长度不大于10的小写英文字符串 s,问是否能通过在字符串的某个位置插入一个字母 ...

  2. Codeforces Round #286 (Div. 2)A. Mr. Kitayuta's Gift(暴力,string的应用)

    由于字符串的长度很短,所以就暴力枚举每一个空每一个字母,出现行的就输出.这么简单的思路我居然没想到,临场想了很多,以为有什么技巧,越想越迷...是思维方式有问题,遇到问题先分析最简单粗暴的办法,然后一 ...

  3. Codeforces 506E - Mr. Kitayuta's Gift(神仙矩阵乘法)

    Codeforces 题目传送门 & 洛谷题目传送门 神仙题 %%%%%%%%%%%%% u1s1 感觉这道题风格很省选( 下记 \(m=|s|\),首先探讨 \(n+m\) 为偶数的情形. ...

  4. Codeforces 506E Mr. Kitayuta's Gift (矩阵乘法,动态规划)

    描述: 给出一个单词,在单词中插入若干字符使其为回文串,求回文串的个数(|s|<=200,n<=10^9) 这道题超神奇,不可多得的一道好题 首先可以搞出一个dp[l][r][i]表示回文 ...

  5. 水题 Codeforces Round #286 (Div. 2) A Mr. Kitayuta's Gift

    题目传送门 /* 水题:vector容器实现插入操作,暴力进行判断是否为回文串 */ #include <cstdio> #include <iostream> #includ ...

  6. 【CF506E】Mr. Kitayuta's Gift dp转有限状态自动机+矩阵乘法

    [CF506E]Mr. Kitayuta's Gift 题意:给你一个字符串s,你需要在s中插入n个字符(小写字母),每个字符可以被插在任意位置.问可以得到多少种本质不同的字符串,使得这个串是回文的. ...

  7. codeforces Round 286# problem A. Mr. Kitayuta's Gift

    Mr. Kitayuta has kindly given you a string s consisting of lowercase English letters. You are asked ...

  8. CodeForces 505B Mr. Kitayuta's Colorful Graph

    Mr. Kitayuta's Colorful Graph Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d ...

  9. codeforces 505B Mr. Kitayuta's Colorful Graph(水题)

    转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Mr. Kitayuta's Colorful Graph Mr. Kitayut ...

随机推荐

  1. aarch64_p1

    PEGTL-devel-1.3.1-2.fc26.aarch64.rpm 2017-02-14 08:00 63K fedora Mirroring Project PackageKit-1.1.6- ...

  2. js如何查看元素类型

    <script type="text/javascript"> //定义变量temp var temp = Object.prototype.toString.appl ...

  3. Android性能测试工具之APT

    1.APT工具简介: APT是一个eclipse插件,可以实时监控Android手机上多个应用的CPU.内存数据曲线,并保存数据:另外还支持自动获取内存快照.PMAP文件分析等,方便开发人员自测或者测 ...

  4. Angular CLI 命令行工具

    工欲善其事必先利其器.好的工具让开发更加简单便捷. 1.全局安装angular cli npm install -g @angular/cli 2.安装完成后就可以使用angular-cli命令行工具 ...

  5. Scrapy的【SitemapSpider】的【官网示例】没有name属性

    Windows 10家庭中文版,Python 3.6.4,Scrapy 1.5.0, 上午看了Scrapy的Spiders官文,并按照其中的SitemapSpider的示例练习,发现官文的示例存在问题 ...

  6. 夜神模拟器调试android studio项目

    这几天为了android studio也是醉了,先是R文件丢失忙活一下午,各种百度谷歌,最后终于解决这个小问题,没想到在启动avd这个问题上更是棘手,网上的方法试了,主要有三种,上篇博文http:// ...

  7. 洛谷P1972 HH的项链

    传送门啦 分析: 题目描述不说了,大意是,求一段区间内不同元素的种数. 看到区间,我们大概先想到的是暴力(然后炸掉).线段树.树状数组.分块. 下面给出的是一种树状数组的想法. 首先,对于每一段区间里 ...

  8. InnoDB 锁

    参看文章: innodb的意向锁有什么作用? 2.<MySQL技术内幕:InnoDB存储引擎> InnoDB存储引擎中的锁 InnoDB中的锁介绍 InnoDB存储引擎既支持行级锁,也支持 ...

  9. PL/SQL开发中动态SQL的使用方法

    一般的PL/SQL程序设计中,在DML和事务控制的语句中可以直接使用SQL,但是DDL语句及系统控制语句却不能在PL/SQL中直接使用,要想实现在PL/SQL中使用DDL语句及系统控制语句,可以通过使 ...

  10. elastucasearch基础理论以及安装

    一.elasticasearch核心概念 Near Realtime(NRT 近实时) Elasticsearch 是一个近实时的搜索平台.您索引一个文档开始直到它被查询时会有轻微的延迟时间(通常为1 ...