1. Problem Description:

Title: a string with length of n (2 < = n < = 1 000 000), ask whether the prefix string with length of K (k > 1) is a periodic string, i.e. k = A...A; if so, output K in the order of K from small to large, i.e. the number of cycles;

Sample Input
3 aaa
12 aabaabaabaab
0
 
Sample Output
Test case #1
2 2
3 3
 
Test case #2
 2   2
 6   2
 9   3
12  4

2. In fact, it is the use of next array in KMP algorithm, assuming that next[j] = k

Then if the periodic string has: J% (J - K) remainder of 0, then the string is considered to be periodic, and the number of repetitions: j / (j - k)

The code is as follows:

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class next Array example_Period string{
	public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
	    List<String> list = new ArrayList<String>();
	    while (true) {
	      int n = sc.nextInt();
	      if (n == 0) {
	        break;
	      }
	      String s = sc.next();
	      list.add(s);
	    }
	    for(int j = 0; j<list.size();j++) {
	      String s = list.get(j);
	      int[] next = next(s);
	      System.out.println("Test case #" + (j + 1));
	      boolean flag = false;
	      for(int i = 2; i < next.length; i++) {
	        int k = next[i];
	        int t = i - k;
	        if (i%t==0&&i/t>1){
	          System.out.println(i + " " + i / t);
	        }
	      }
	      System.out.println();
	    }
	  }
	    
	  private static int[] next(String s) {
	    if (s == null || s.length() == 0) return null;
	    int[] next = new int[s.length()+1];
	    next[0] = -1;
	    if (s.length() == 1)
	      return next;
	    next[1] = 0;
	    int j = 1;
	    int k = next[j];
	    while (j < s.length()) {
	      if (k == -1 || s.charAt(j) == s.charAt(k)) {
	        next[++j] = ++k;
	      } else {
	        k = next[k];
	      }
	    }
	    return next;
	  }
	}