Password CodeForces - 126B（KMP）

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.

Example
Input

fixprefixsuffix

Output

fix

Input

abcdabc

Output

Just a legend

/****
题目要求求最大前缀和后缀，以及在字符串除掉头尾出现过一次。
很容易就想到kmp，但是我没有想到后一点。。。。。。
next[]数组求得是strlen（a）+1，保证是全部的字符串元素。
先说错误的解题方法：

****/

//WA原因是因为自己最初只是单纯的想着过样例，所以wa忘记一种特殊情况
//qwerqwerqwer
//原先错误的思路是求出next[]数组，对其进行sort排序，从后往前遍历寻找最大的切相等的两个next[]，值，极为正确答案。
//qwerqwerqwer救是一种特例
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn =1e6+10;
int next[maxn];
char a[maxn];

int m;

void get_next()
{
    int i,j;
    i=0;
    j=-1;
    next[0]=-1;
    while (i<=m)
    {
        if(j==-1||a[i]==a[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
        j=next[j];
    }

}
int main ()
{
    scanf("%s",a);
    m=strlen(a);
    get_next();
    int i,j,k;
    for (i=0;i<=m;i++)
    cout<<next[i];
    cout<<endl;

   if (m<3)
   cout<<"Just a legend"<<endl;
   sort (next,next+m+1);
//    for (i=0;i<=m;i++)
//    cout<<next[i];
//    cout<<endl;

    int result ;
   for (i=m;i>0;i--)
   {
    if(next[i]==next[i-1])
    {
        result=next[i];
        break;
    }
   }
   if(result)
   {
    for (i=0;i<result;i++)
        cout<<a[i];
        cout<<endl;
   }
   else
   cout<<"Just a legend"<<endl;

    return 0;
}

/***
正确的做题方式：
先求出全部的next[]数组（即strlen(m)+1），然后标记出现过的next[]数组，对其进行反向遍历，即从i=next[m]开始遍历，找到第一个在vis[i]中标记过得数据，对应的next[i]即为最大公共前缀的长度。
注意：标记是不要标记最后以为即stren（a）+1；
代码如下：
***/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn =1e6+10;
int next[maxn];
char a[maxn];
bool vis[maxn];
int m;

void get_next()
{
    int i,j;
    i=0;
    j=-1;
    next[0]=-1;
    while (i<=m)
    {
        if(j==-1||a[i]==a[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
        j=next[j];
    }

}
int main ()
{
    memset(vis,0,sizeof(vis));
    scanf("%s",a);
    m=strlen(a);
    get_next();
    int i,j,k;
    for (i=0;i<m;i++)
    vis[next[i]]=1;
    i=next[m];
    while (i)
    {
        if (vis[i])
        {
            for (j=0;j<i;j++)
            cout<<a[j];
            cout<<endl;
            return 0;
        }
        else
            i=next[i];
    }
    cout<<"Just a legend"<<endl;

    return 0;
}

PS：（如果不懂请参考：http://blog.youkuaiyun.com/chen_minghui/article/details/53339246）