codeforce D. Password

D. Password

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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 test(s)

Input

fixprefixsuffix

Output

fix

Input

abcdabc

Output

Just a legend

题意：在一个字符串中寻找一个子串，要求子串出现在原串的开头，中间和结尾

思路：KMP中next数组的灵活运用，只要判断最后next [len]的数字是否在之前出现过，即是否在中间出现过即可

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 1000009
int vis
;
char str
;
int next
;
int len;

void getnext()
{
    int i,j;
    i=0;
    j=-1;
    next[0]=-1;

    while(i<len)
    {
        if(j==-1||str[i]==str[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
        j=next[j];
    }
}

int main()
{
    while(~scanf("%s",str))
    {
        len=strlen(str);
        getnext();

        memset(vis,0,sizeof vis);

        for(int i=1;i<len;i++)
        {
            vis[next[i]]=1;
        }

        int t=next[len];

        int flag=0;

        while(t)
        {
            if(vis[t])
            {
                flag=1;
                str[t]='\0';
                break;
            }
            else
            t=next[t];
        }
        
        if(flag)
        printf("%s\n",str);
        else
        cout<<"Just a legend"<<endl;

    }
    return 0;
}