poj------2406 Power Strings

A - Power Strings 难度：☆☆

Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2406

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

 from http://poj.org/problem?id=2406

考察kmp算法中的next数组....

next的两种表示方法，第一种是前缀next[]数组...

while(i<len)
{
    if(-1==j||ss[i]==ss[j])
    {
        i++;
        j++;
        next[i]=j; 
    }
    else  j=next[j];
}

 

另种表示方法文为：

while(i<len)
{
   if(j==-1||ss[i]==s[j])
 {
     i++:
     j++;
     if(ss[i]==ss[j])
    {
         next[i]=next[j];
    } 
   else  next[i]=j;
 }
  else  j=next[j];
}

 

 这道题是考察有多少个重复的最大n所以我们不妨看他的回溯长度len_D=len(已经匹配的位置) --next[len](对应部分的匹配值);

 

Java代码：

  

//package dek0;

import java.util.Scanner;

public class Main {    
    
    public static void main(String args[])
    {
        Scanner  reader = new Scanner(System.in);
        String ss="";
        while(reader.hasNext())    
        {
           ss=reader.next();
           if(ss.charAt(0)=='.') break;
           int len=ss.length();
           int next[]=new int [len+1];
               next[0]=-1;
           int i=0,j=-1;
           while(i<len)
           {
               if(j==-1||ss.charAt(i)==ss.charAt(j))
               {
                 i++;
                 j++;
            /*     if(ss.charAt(i)==ss.charAt(j))
                         next[i]=next[j];
                 else      next[i]=j;*/
                 next[i]=j;
               }
               else  j=next[j];
           }
           if(len%(len-next[len])==0)
             System.out.println(len/(len-next[len]));
           else
               System.out.println(1);
        }
    }
}