codeforces 126B Password

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本文介绍了一种使用KMP算法解决字符串中同时作为前缀、内嵌和后缀的最长子串的问题。通过巧妙地利用KMP算法的特性，文章提供了一个高效的解决方案，并附带了完整的AC代码。

Description

Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.

A little later they found a string s, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring t of the string s.

Prefix supposed that the substring t is the beginning of the string s; Suffix supposed that the substring t should be the end of the string s; and Obelix supposed that t should be located somewhere inside the string s, that is, t is neither its beginning, nor its end.

Asterix chose the substring t so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring t aloud, the temple doors opened.

You know the string s. Find the substring t or determine that such substring does not exist and all that's been written above is just a nice legend.

Input

You are given the string s whose length can vary from 1 to 10⁶ (inclusive), consisting of small Latin letters.

Output

Print the string t. If a suitable t string does not exist, then print "Just a legend" without the quotes.

Sample Input

Input

fixprefixsuffix

Output

fix

Input

abcdabc

Output

Just a legend

题意：找前面中间和最后都出现的子串。

思路：KMP，就是注意abcdabcdabcd的这种情况，所以要i=f[i]；

可以自己看着代码理解一下。

AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <stdlib.h>

using namespace std;

int f[1000006];
char s[1000006];
int len;
int vis[1000006];

void init(char s[]){
    f[0]=0;f[1]=0;
    for(int i=1;i<len;i++){
        int j=f[i];
        while(j&&s[j]!=s[i]) j=f[j];
        f[i+1]=(s[i] == s[j] ? j+1 : 0);
    }
}

int main(){
    scanf("%s",&s);
    len=strlen(s);
    init(s);
    memset(vis,false,sizeof(vis));
    for(int i=0;i<len;i++){
        vis[f[i]]=true;
    }
    int i=len,j=0;
    bool flag=false;
    while(f[i]!=0){
        if(vis[f[i]]){
            for(int j=0;j<f[i];j++){
                printf("%c",s[j]);
            }
            flag=true;
            break;
        }
        i=f[i];
    }
    if(!flag){
        printf("Just a legend");
    }
    printf("\n");
    return 0;
}