题目描述：
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.
题意翻译:
Asterix，Obelix和他们的临时伙伴Suffix、Prefix已经最终找到了和谐寺。然而和谐寺大门紧闭，就连Obelix的运气也没好到能打开它。
不久他们发现了一个字符串S（|S|<=1000000），刻在和谐寺大门下面的岩石上。Asterix猜想那一定是打开寺庙大门的密码，于是就大声将字符串朗读了出来，然而并没有什么事发生。于是Asterix又猜想密码一定是字符串S的子串T。
Prefix认为T是S的前缀，Suffix认为T是S的后缀，Obelix却认为T应该是S中的某一部分，也就是说，T既不是S的前缀，也不是S的后缀。
Asterix选择子串T来满足所有伙伴们的想法。同时，在所有可以被接受的子串变形中，Asterix选择了最长的一个（因为Asterix喜欢长的字符串）当Asterix大声读出子串T时，寺庙的大门开了。 （也就是说，你需要找到既是S的前缀又是S的后缀同时又在S中间出现过的最长子串）
现在给你字符串S，你需要找到满足上述要求的子串T。
输入格式：
You are given the string ss whose length can vary from 11 to 10^{6}106 (inclusive), consisting of small Latin letters.
一个长度在[1,1000000]间的只包含小写字母的字符串S。
输出格式：
Print the string tt . If a suitable tt string does not exist, then print "Just a legend" without the quotes.
输出子串T，如果T不存在，输出 "Just a legend"，不包含引号。
题解:
首先一看到这道题，要判断字符串是否出现，最先想到kmp,但是单纯用kmp去匹配，是O(N^2)的时间复杂度，但是数据量却有1000000，肯定不行。
那么就得启发我们换换思路，是否一定要匹配呢？
换个角度思考，抛开字符串要在中间出现不谈，既然题目要求最长长度的前后缀，kmp的next数组不就是如此吗？
所以如果next[n]对应的字符串中间出现过，答案肯定就是。那么如何判断呢？如图所示：
如果2~n-1（为何不是1-n,因为题目中说了此串的第三次出现不能在前后缀位置）当中有串和1-next[n]相同，则意味着如果缩短1-i的字符串，把后几位扔掉，总会存在2~n-1的一个i，它的后缀与1~next[n]相同，但是1~next[n]又是1~i的前缀，所以这就是前后缀相同了，只要判断next[i]==next[n]就行了
如果next[n]没有出现过3次，那么1~next[n]的最长相同前后缀一定是答案（因为它长度最大，而且分别在1-next[n]、n-next[n]+1~next[n]出现了2次，总共4次，满足条件)
29行代码搞定
1 #include<bits/stdc++.h> 2 using namespace std; 3 int n; 4 char s[1000001],s1[1000001]; 5 int p[1000001],len,p1[1000001]; 6 int main(){ 7 int i; 8 scanf("%s",s+1); 9 n=strlen(s+1); 10 int j=0; 11 for(i=2;i<=n;i++){ 12 while(j&&s[i]!=s[j+1])j=p[j]; 13 if(s[i]==s[j+1])j++; 14 p[i]=j; 15 } 16 int x=p[n],y=n-p[n]+1; 17 for(i=2;i<=n-1;i++)if(p[i]==p[n])len=max(len,p[n]); 18 for(i=y;i<=n;i++)s1[i-y+1]=s[i]; 19 j=0; 20 for(i=2;i<=p[n];i++){ 21 while(j&&s1[i]!=s1[j+1])j=p1[j]; 22 if(s1[i]==s1[j+1])j++; 23 p1[i]=j; 24 } 25 if(p1[p[n]])len=max(len,p1[p[n]]); 26 if(len)for(i=1;i<=len;i++)putchar(s[i]); 27 else puts("Just a legend"); 28 return 0; 29 }