POJ 2406 Power Strings

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 26177		Accepted: 10976

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.

Source

Waterloo local 2002.07.01

   其实这道题目的思路来自POJ 1961，建议先做做1961，做完之后这道题目就会很容易的解决了

/***************************************************************
 > File Name: E:\我的程序\C语言\power.c
 > Author: SDUT_GYX
 > Mail: [email protected]
 > Created Time: 2013/5/9 10:14:47
 **************************************************************/

#include
#include 
#include 
char s1[1100000];
int next[1100000],dp[1100000];
int main()
{
	void get_next(int l);
	int i,j,n,m,s,t,l;
	while(gets(s1))
	{
        if(strcmp(s1,".")==0)
		{
			break;
		}
		l=strlen(s1);
		get_next(l);
		dp[0]=0;
		for(i=2;i<=l;i++)
		{
		   n=next[i];
		   if(n+n==i)
		   {
			   dp[i-1]=n;
		   }else if(n+n