codeforce D. Password

最新推荐文章于 2022-05-26 19:50:53 发布

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最新推荐文章于 2022-05-26 19:50:53 发布

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分类专栏： KMP

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本文介绍了一个算法问题，即在一个字符串中找到同时作为开头、中间部分和结尾的最长子串。通过KMP算法中的next数组来解决该问题，并提供了一段实现代码。

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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[N];
char str[N];
int next[N];
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;
}