poj 2406 kmpアルゴリズムの強固なnext配列の再理解

http://poj.org/problem?id=2406
Description
Given two stinings a and b we define a*b to be their concateation.For example、if="abc"andb="def"the n a*b="abcdef".If wethininininininininininininininininamltipcation、exponentiation=="aaaaatttttttttinininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininininin)
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 milion characters.A line containing a period follows the last test case.
Output
For each s you shoul d print the larget n such that=a^n for some string a.
Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

ベント
This problem has huge input、use scanf instead of cin to avoid time limit exced.
私が勉強している人の感じは大丈夫です。分かりやすいようです。

#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
");
        else
            printf("%d
",len/(len-next[len]));
    }
    return 0;
}