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.
Examples
input
fixprefixsuffix
output
fix
input
abcdabc
output
Just a legend
题目大意：
在一个字符串里找一个子串，它在头，尾，中间都出现一次。
思路：
遍历一遍字符串与头部子串进行匹配。
方法：
跑单for进行匹配，把最后的子串标记为不可取。
AC代码：
#include<bits/stdc++.h> using namespace std; typedef long long ll; #define endl "\n" const ll N = 1e6+7; ll dp[N]; bool vis[N]; string s; void solve(){ cin >> s; ll len=s.size(),cnt=0; for(ll i = 1 ; i < len ; i ++){ //如果当前的字母不匹配了,往前匹配 while(cnt && s[i] != s[cnt])cnt=dp[cnt-1]; //如果当前字符匹配,长度++ if(s[i] == s[cnt])cnt++; //末尾与头相等的字符串不进行匹配 if(i < len-1)vis[cnt]=1; //记录此点可取子串的长度 dp[i]=cnt; } vis[0]=1,cnt=dp[len-1]; //除去末尾与头相等的部分 while(cnt && !vis[cnt])cnt=dp[cnt-1]; cnt > 0 ? cout << s.substr(0,cnt) << endl : cout << "Just a legend" << endl; return; } int main() { ll t=1;//cin >> t; while(t--)solve(); return 0; }