首页 > 代码库 > codeforces 126B Password

codeforces 126B Password

2024-07-14 14:33:23   216人阅读

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 strings.

Prefix supposed that the substring t is the beginning of the strings; Suffix supposed that the substring t should be the end of the string s; and Obelix supposed thatt should be located somewhere inside the strings, 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 substringt aloud, the temple doors opened.

You know the string s. Find the substringt 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 from1 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：找前面中间和最后都出现的子串。

思路：KMP，就是注意abcdabcdabcd的这种情况，所以要i=f[i]；

可以自己看着代码理解一下。

AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <stdlib.h>

using namespace std;

int f[1000006];
char s[1000006];
int len;
int vis[1000006];

void init(char s[]){
    f[0]=0;f[1]=0;
    for(int i=1;i<len;i++){
        int j=f[i];
        while(j&&s[j]!=s[i]) j=f[j];
        f[i+1]=(s[i] == s[j] ? j+1 : 0);
    }
}

int main(){
    scanf("%s",&s);
    len=strlen(s);
    init(s);
    memset(vis,false,sizeof(vis));
    for(int i=0;i<len;i++){
        vis[f[i]]=true;
    }
    int i=len,j=0;
    bool flag=false;
    while(f[i]!=0){
        if(vis[f[i]]){
            for(int j=0;j<f[i];j++){
                printf("%c",s[j]);
            }
            flag=true;
            break;
        }
        i=f[i];
    }
    if(!flag){
        printf("Just a legend");
    }
    printf("\n");
    return 0;
}





联系
我们