【Codeforces 126B】Password

【链接】 我是链接,点我呀:)
【题意】

给你一个字符串s 让你从中选出来一个字符串t 这个字符串t是s的前缀和后缀 且在除了前缀和后缀之外的中间部位出现过。 且要求t的长度最长。 让你输出这个字符串t

【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());
		}
	}
}