Power Strings

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链接 http://poj.org/problem?id=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

abcdaaaaababab.

Sample Output

Hint

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

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
注意s数组的初始化，next【】数组的求法比较简单；
主要是分析next数组求出后的三步；
next数组的含义是这样的，举个例子：
比如：ababababab，初始的时候j=-1,i=0,next[0]=-1;设置i的主要用途是判断串的长度是否到头；

next数组记录的是字符串匹配过程中失配的情况下最多可以向前跳几个字符，next数组也描述子串的匹配程度，next的值越大，子串的匹配程度越高。匹配的速度也越快。(跳的字符多嘛!)next也可以是指前i个字符串对称程度。

#include<stdio.h>

#include<iostream>
#include<cstring>
using namespace std;
char s[1000050];
int next[1000050];
int main()
{
int i,j,k,m,n;
memset(s,'\0',sizeof(s));
while(scanf("%s",s)!=EOF)
{
if(s[0]=='.')
   break;
   int len;
len=strlen(s);
i=0;
j=-1;
next[0]=-1;
while(i<len)
{
if(j==-1||s[i]==s[j])
{
i++;
j++;
next[i]=j;
}
else
{
j=next[j];
}
}
int t=i-next[i];
if(t==0)
    k=i;
   else if(i%t==0)
      k=i/t;
   else
   k=1;
printf("%d\n",k);
memset(s,'\0',sizeof(s));
}
return 0;
}