codeforces 126 B. Password - c++编程基础

题目链接：

huangjing

思路:灵活运用kmp的next值，首先求kmp的next值，首先如果长度小于3那么肯定是不符合条件的，然后对2――len进行标记，然后求next[len+1]，然后知道这个值出现过，说明中中间必然出现了前缀，那么就至少出现了3次了。。

题目：

B. Password time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

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 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 test(s) input

fixprefixsuffix

output

fix

input

abcdabc

output

Just a legend

代码：

#include
  
   
#include
   
     #include
    
      #include
     
       #include
       #include
       
         #include
        
          #include
         
           #include
          
            #define eps 1e-9 #define ll long long #define INF 0x3f3f3f3f using namespace std; const int maxn=1000000+10; char p[maxn]; int m,vis[maxn],next[maxn]; void Getnext() { int j,k; next[1]=0; k=0; j=1; while(j<=m) { if(k==0||p[k]==p[j]) { k++; j++; next[j]=k; } else k=next[k]; } } int main() { int k,ans,i; while(~scanf("%s",p+1)) { m=strlen(p+1); Getnext(); if(m<3) { printf("Just a legend\n"); continue; } memset(vis,false,sizeof(vis)); for(i=2;i<=m;i++) vis[next[i]]=true; i=next[m+1]; while(!vis[i]&&i) i=next[i]; if(i==1) printf("Just a legend\n"); else { for(int j=1;j