Password
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
Input
fixprefixsuffix
Output
fix
Input
abcdabc
Output
Just a legend
Sample Output
Case 1: 13 Case 2: 6
Hint
找字符串中是否存在前后缀相同的字符串，KMP pre数组应用
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cstring>
4 #include <algorithm>
5 #include <iostream>
6 #include <cmath>
7 #include <queue>
8 #include <map>
9 #include <stack>
10 #include <list>
11 #include <vector>
12
13 using namespace std;
14
15 const int maxn = 1000010;
16 int nb;
17 char b[maxn];
18 int pre[maxn];
19 int vis[maxn];
20
21 //b是模式串，a是目标串
22 void getpre(char *b, int *pre) {
23 int j, k;
24 pre[0] = -1;
25 j = 0;
26 k = -1;
27 while(j < nb) {
28 if(k == -1 || b[j] == b[k]) {
29 j++;
30 k++;
31 pre[j] = k;
32 }
33 else {
34 k = pre[k];
35 }
36 }
37 }
38
39 int main() {
40 // freopen("in", "r", stdin);
41 while(~scanf("%s", b)) {
42 memset(pre, 0, sizeof(pre));
43 memset(vis, 0, sizeof(vis));
44 nb = strlen(b);
45 if(nb < 3) {
46 printf("Just a legend\n");
47 continue;
48 }
49 getpre(b, pre);
50 for(int i = 0; i < nb; i++) {
51 vis[pre[i]] = 1;
52 }
53 int flag = 0;
54 while(pre[nb]) {
55 if(vis[pre[nb]]) {
56 printf("%.*s\n", pre[nb], b);
57 flag = 1;
58 break;
59 }
60 nb = pre[nb];
61 }
62 if(!flag) {
63 printf("Just a legend\n");
64 }
65 }
66 return 0;
67 }