Palindrome. Again Palindrome

1354. Palindrome. Again Palindrome

Time limit: 1.0 second
Memory limit: 64 MB

A word is the nonempty sequence of symbols a 1 a 2… an. A palindrome is the word a 1 a 2… an that is read from the left to the right and from the right to the left the same way ( a 1 a 2… an = ana n−1… a 1). If S 1 = a 1 a 2… an and S 2 = b 1 b 2… bm, then S 1 S 2 = a 1 a 2… anb 1 b 2… bm. The input contains some word S 1. You are to find a nonempty word S 2 of the minimal length that S 1 S 2 is a palindrome.

Input

The first input line contains S 1 (it may consist only of the Latin letters). It’s guaranteed that the length of S 1 doesn’t exceed 10000 symbols.

Output

S 1 S 2.

Samples

input	output
No	NoN
OnLine	OnLineniLnO
AbabaAab	AbabaAababA

/**使用求回文串的算法：Manacher算法**/
/** 要添加最短的串,使得新形成的串是回文串,首先,要考虑原串的回文属性,寻找包含末尾的最大回文后缀串,但是,要注意,当原串全部回文的时候,需要找最大回文后缀,但不包括开头~最后,要注意,当原串长度为1时,直接复制一遍输出即可.**/
#include <iostream>
#include<cstdio>
#include<cstring>using namespace std;
#define maxn 50000char s[maxn],p[maxn];
int slen[maxn],len;void read()
{scanf("%s",s);len=strlen(s);for(int i=1; i<=len; i++)//Manacher的预处理{p[(i<<1)-1]='#';p[i<<1]=s[i-1];}p[0]='?';
}void Manacher()
{int j,k,maxid=0,id=0;for(int i=1; p[i]!='\0'; i++){if(maxid>i)slen[i]=min(maxid-i,slen[2*id-i]);//实质为动态规划elseslen[i]=1;while(p[i+slen[i]]==p[i-slen[i]])slen[i]++;if(slen[i]+i>maxid){maxid=slen[i]+i;id=i;}}
}int deal()
{int maxi=0;for(int i=strlen(p)-1; i>(strlen(p)+1)/2; i--)//寻找最长的回文后缀,注意中间的判断条件,如果是大于等于那就会找出包含开头的最大回文后缀,会导致错误{if(p[i+slen[i]]=='\0')maxi=i;}return maxi;
}int main()
{read();if(len==1)//原串长度为1时,需要特判{printf("%s%s",s,s);return 0;}Manacher();int j=deal();printf("%s",s);for(int i=j-slen[j]; i>=1; i--)//计算出要添加的那一段是从哪里开始的if(p[i]!='#')printf("%c",p[i]);printf("\n");return 0;
}