Password Uva1262 KMP

D - Password

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  CodeForces 126B

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：给出一个字符串，求出最大的前缀和后缀且它能在串的中部找到，若存在则输出，否则输出

Just a legend

思路：KMPnext数组的应用，一些细节要非常注意！

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

int N;
int nextval[1000010];
char str[1000010],pstr[1000010];

int get_nextval()
{
    int i=0;
    int len=strlen(str);
    int j=-1;
    nextval[i]=-1;
    while (i<len)
    {
        if (j==-1||str[i]==str[j])
        {
            i++;
            j++;
            nextval[i]=j;
        }
        else
            j=nextval[j];
    }
//    for (int i=0;i<=len;i++)
//        printf("%d ",nextval[i]);
//    printf("\n");
//    printf("%d\n",nextval[len]);
    return nextval[len];
}

int kmp_search()
{
    int ans=0;
    int plen=strlen(pstr);
    int slen=strlen(str);
    int i=0;
    int j=0;
    while (i<slen)
    {
        if (j==-1||pstr[j]==str[i])
        {
            i++;
            j++;
            if (j==plen)
            {
                ans++;
                j=nextval[j];
            }
        }
        else
            j=nextval[j];
    }
    return ans;
}

int main()
{
    while (~scanf("%s",str))
    {
        int len=strlen(str);
        int ma=get_nextval();
        for (int i=0;i<ma;i++)
            pstr[i]=str[i];
        pstr[ma]='\0';
        if (ma==0)
        {
            printf("Just a legend\n");
            continue;
        }
//        printf("ma=%d\n",ma);
        if (ma>=2&&ma==len-1)
        {
            for (int i=0;i<len-2;i++)
                printf("%c",str[i]);
            printf("\n");
            continue;
        }
        int j=ma;
        int ok=0;
        while (j!=-1)
        {
            int ans=kmp_search();
//            printf("ans=%d\n",ans);
            if (ans>2)
            {
                for (int t=0;t<j;t++)
                    printf("%c",str[t]);
                printf("\n");
                ok=1;
                break;
            }
//            printf("j=%d\n",j);
            j=nextval[j];
            if (j==0)
                break;
            pstr[j]='\0';
        }
//        printf("ok=%d\n",ok);
        if (!ok)
            printf("Just a legend\n");
    }
    return 0;
}