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CF 505A(Mr. Kitayuta's Gift-回文串)

霸姨 2022-10-25 阅读 88



A. Mr. Kitayuta's Gift



time limit per test



memory limit per test



input



output



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.

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.

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



s (1 ≤ |s| ≤ 10). Each character in s



Output



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



r' to the end of "revive" to obtain a palindrome "reviver".

eve" will also be accepted.

kitayuta" into a palindrome by just inserting one letter.




给一个小写字母字符串,必须添一个字符,使添后的字符串回文,求添后的字符串。

贪心,

如果字符串已经回文,在中间塞一个字符

abc->abbc

abba->abcba

如果不回文,从外向内找到第一个不回文字符

eg:

ad...la 

显然只存在2种添法:ad...lda   ald...la。一一验证即可。


#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
char s[100];
bool check(int i,int j)
{
while(i<=j)
{
if (s[i]!=s[j]) return 0;
i++,j--;
}
return 1;
}
int main()
{
// freopen("palindromes.in","r",stdin);
// freopen(".out","w",stdout);
scanf("%s",s+1);
int n=strlen(s+1);
For(i,n/2)
{
if (s[i]!=s[n-i+1])
{
if (check(i,n-i))
{
For(j,i-1) printf("%c",s[j]);
printf("%c",s[n-i+1]);
Fork(j,i,n) printf("%c",s[j]);
printf("\n");
return 0;
}
else if (check(i+1,n-i+1))
{
For(j,n-i+1) printf("%c",s[j]);
printf("%c",s[i]);
Fork(j,n-i+2,n) printf("%c",s[j]);
printf("\n");
return 0;
}
else break;
}
}
if (check(1,n))
{
int t=(1+n)/2;
For(j,t) printf("%c",s[j]);
printf("%c",s[t]);
Fork(j,t+1,n) printf("%c",s[j]);
printf("\n");
return 0;


}

printf("NA\n");

return 0;
}




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