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
fix
abcdabc
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;
}
codeforces 126B Password,布布扣,bubuko.com
原文：http://blog.csdn.net/kimi_r_17/article/details/38238667