1. 问题描述：

题目：一个长为N (2 <= N <= 1 000 000) 的字符串，问前缀串长度为k(k > 1)是否是一个周期串，即k = A…A;若是则按k从小到大的顺序输出k即周期数；

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. 其实就是对KMP算法中的next数组的使用，假设next[j] = k

那么如果是周期字符串有：j % (j – k)余数为0那么认为这个字符串是周期性的，而且重复的次数：j / (j – k)

代码如下：

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Main {
    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);
          for(int i = 0;i<next.length;i++){
        	  System.out.print(next[i]+" ");
          }
          System.out.print("\n");
          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);
            }
          }
          // if (!flag) System.out.println(0+" "+0);
          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;
      }
    }