poj 2406 kmp算法巩固之next数组的再理解

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

143

Hint

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

我学习的人家的，感觉还好，似乎很容易懂

#include <stdio.h>#include <string.h>#include <iostream>const int N=1000005;int next[N],len;char a[N];void get_next(){    int i=0,j=-1;    next[0]=-1;    while(i<len)    {        if(j==-1||a[i]==a[j])            next[++i]=++j;        else            j=next[j];    }}int main(){    while(~scanf("%s",a))    {        if(a[0]=='.')           break;        len=strlen(a);        get_next();        if(len%(len-next[len])!=0)            printf("1\n");        else            printf("%d\n",len/(len-next[len]));    }    return 0;}