POJ 2406 Power Strings

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal 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

abcdaaaaababab.

Sample Output

143

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: [e-mail] > Created Time: 2013/5/9 10:14:47 **************************************************************/#include<stdio.h>#include <string.h>#include <math.h>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<i)   {   dp[i-1]=0;   }else   {   dp[i-1]=dp[n-1];   }}if(dp[l-1]==0){printf("1\n");}else{printf("%d\n",l/dp[l-1]);}}return 0;}void get_next(int l){int i,j;next[0]=-1;next[1]=0;for(i=2,j=0;i<=l;){if(j==-1||s1[i-1]==s1[j]){i++;j++;next[i-1]=j;}else{j=next[j];}}}