HDU 1358 Period（Kmp 循环节）

Period
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13517 Accepted Submission(s): 6328

Problem Description
For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as AK , that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input
The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.

Output
For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

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
About next 数组,Kmp循环节问题,
给定一个字符串，所有前缀串可以由某个串重复k次组成，并且这个k (k>1) 最大是多少.
如果存在k（k>1）输出前缀长度及其中相同的子字符串数（K）。

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;
int nex[1110000];
void Get_Next(string s,int nex[])
{
  int len=s.length();
  int i=0,j=-1;
  nex[0]=-1;
  while(i<len)
  {
    if(j==-1||s[i]==s[j])
    {
      i++;
      j++;
      nex[i]=j;
    }else j=nex[j];
  }
}
int main()
{
  ios::sync_with_stdio(false);
  int n,cnt=0;
  string s;
  while(cin>>n,n)
  {
    cnt++;
    memset(nex,0,sizeof(nex));
    cin>>s;
    cout<<"Test case #"<<cnt<<endl;
    Get_Next(s,nex);
    for(int i=0;i<=s.length();i++)
    {
      int j=i-nex[i];//最小循环节 
      if(i%j==0&&i/j>1)
        cout<<i<<" "<<i/j<<endl;
    }
    cout<<endl;
  }
  return 0;
}