CodeForces - 126B Password

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

计算出next数组,再用贪心的方法匹配看是否前后缀相同且中间也出现过。
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 
 5 char s[1000005];
 6 int Next[1000005],c[1000005];
 7 
 8 void getnext(char* s,int m)
 9 {
10     Next[0]=0;
11     Next[1]=0;
12     for(int i=1;i<m;i++)
13     {
14         int j=Next[i];
15         while(j&&s[i]!=s[j])
16             j=Next[j];
17         if(s[i]==s[j])
18             Next[i+1]=j+1;
19             else
20                 Next[i+1]=0;
21     }
22 }
23 
24 int main()
25 {
26     scanf("%s",s);
27     int n=strlen(s);
28     getnext(s,n);
29     for(int i=0;i<n;i++)
30         c[Next[i]]=1;
31     c[0]=0;
32     for(int i=n;i;i=Next[i])
33         if(c[Next[i]]==1)
34         {
35             for(int j=0;j<Next[i];j++)
36                 printf("%c",s[j]);
37             return 0;
38         }
39     printf("Just a legend");
40     
41     return 0;
42 }

 

posted @ 2017-08-08 18:13  西北会法语  阅读(137)  评论(0编辑  收藏  举报