【链接】 我是链接,点我呀:)
【题意】
【题解】
KMP的应用 f[i]就是以i为结尾的后缀能匹配的最长前缀的长度 因此只要知道f[n]的值。 然后在做kmp的时候,看看中间有没有哪个时刻能匹配到长度为f[n]的前缀就好(开个数组标记一下就好); 如果没有就让j = f[f[n]] 直到匹配不到为止.【代码】
import java.io.*;
import java.util.*;
public class Main {
static int N = (int)1e6;
static InputReader in;
static PrintWriter out;
public static void main(String[] args) throws IOException{
//InputStream ins = new FileInputStream("E:\rush.txt");
InputStream ins = System.in;
in = new InputReader(ins);
out = new PrintWriter(System.out);
//code start from here
new Task().solve(in, out);
out.close();
}
static class Task{
public void solve(InputReader in,PrintWriter out) {
int [] f = new int[N+10];
int [] ok = new int[N+10];
int n;
String s = in.next();
n = s.length();
f[0] = 0;f[1] = 0;
for (int i = 1;i < n;i++) {
int j = f[i];
while (j>0 && s.charAt(i)!=s.charAt(j)) j = f[j];
if (s.charAt(i)==s.charAt(j)) {
f[i+1] = j+1;
if (i<n-1) {
ok[j+1] = 1;
}
}
else
f[i+1] = 0;
}
int j = f[n];
int temp = 0;
while (j>0) {
if (ok[j]==1) {
temp = Math.max(temp, j);
}
j = f[j];
}
if (s.charAt(n-1)==s.charAt(0) && ok[1]==1) {
temp = Math.max(1, temp);
}
if (temp==0) {
out.print("Just a legend");
}else {
for (int i = 0;i < temp;i++) {
out.print(s.charAt(i));
}
}
}
}
static class InputReader{
public BufferedReader br;
public StringTokenizer tokenizer;
public InputReader(InputStream ins) {
br = new BufferedReader(new InputStreamReader(ins));
tokenizer = null;
}
public String next(){
while (tokenizer==null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(br.readLine());
}catch(IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}