[KMP][字符串Hash] #93 div1 cf 126B Password

[KMP][Hash] #93 div1 cf 126B Password

@(ACM题目)[KMP,字符串Hash]

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 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

Input

fixprefixsuffix

Output

fix

Input

abcdabc

Output

Just a legend

题目分析

本题给定一个字符串s，求最长的子串，使得其同时为原字符串的前缀、后缀，并且去原字符串掉第一个和最后一个字母后，该子串仍然存在。无解输出“Just a legend”。
此题利用KMP中的失配函数，通过可以得到以当前位置的“前缀与后缀相同”的最大长度，见下图：

字符串开始的下标定为1。以最后一个字母d为例，当d失配后，其前方的后缀“ab”是整个字符串的前缀，故失配函数lost6（即字母d失配后要匹配的位置）为3（即字母c）。故此时“前缀与后缀相同”的最大长度为“ab”的长度2。
遍历字符串的每个位置i，通过失配函数losti找到当前最长公共前、后缀s[1,losti−1]，在用字符串Hash来check一下该子串是否是原字符串整个串的后缀即可。

代码

#include<bits/stdc++.h>typedef long long LL;typedef unsigned long long ull;using namespace std;const int maxn = 1e6+7;const ull base = 163;char s[maxn];ull hah[maxn];ull pw[maxn];int n, lost[maxn];void getFail(char* P){    int m = strlen(P+1);    lost[1] =  lost[2] = 1;    for(int i = 2; i <= m; i++)    {        int j = lost[i];        while(j > 1 && P[i] != P[j]) j = lost[j];        lost[i+1] = P[i] == P[j] ? j+1 : 1;    }}void calcHash(char* s){    pw[0]  = 1;    hah[0] = 0;    for(int i = 1; i <= n; i++)    {        pw[i] = pw[i-1] * base;        hah[i] = hah[i-1] * base + s[i];    }}ull getHash(int l, int r){    return hah[r] - hah[l-1] * pw[r-l+1];}int main(){    scanf("%s", s+1);    n = strlen(s+1);    calcHash(s);    getFail(s);    int p = 0;    for(int i = 1; i <= n; ++i)    {        int l = lost[i] - 1;        if(l > p && getHash(1, l) == getHash(n-l+1, n)) p = l;    }    if(p > 0)    {        s[p+1] = '\0';        puts(s+1);    }    else        puts("Just a legend");    return 0;}