cf 126B-Password(KMP)

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: 要求找出字符串中前缀和后缀相同的且出现在字符串中的小串..

思路: kmp 的next 数组的使用...不可使用优化后的next 数组...原始的next 数组才能够记录相同的前缀和后缀是否在中间出现...

         使用vis 数组记录后缀长度是否在字符串中间出现...

CODE

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
char t[1000005];
int next[1000005],vis[1000005];
int len;

void getnext()
{
int i=0,j=-1;
next[0]=-1;
while(i<len)
{
if(j==-1||t[i]==t[j])
++i,++j,next[i]=j;
else j=next[j];
}
for(int i=0;i<len;i++)
vis[next[i]]++;
}
int main()
{
//freopen("P.txt","r",stdin);
while(~scanf("%s",t))
{
memset(vis,0,sizeof(vis));
len=strlen(t);
getnext();
int j=len;
int ok=0;
while(next[j]!=0)
{
if(vis[next[j]])
{
for(int i=0;i<next[j];i++)
printf("%c",t[i]);
ok=1;
break;
}
j=next[j];
}
if(ok==0) printf("Just a legend");
printf("\n");
}
return 0;
}