解题思路：

依旧是利用next数组的性质，m % (m - next[m]) == 0;

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#define LL long long
#define FOR(i,x,y) for(int i=x;i<=y;i++)
using namespace std;
const int maxn = 1000000 + 10;
char s[maxn];
int next[maxn];
int main()
{
    while(scanf("%s", s)!=EOF)
    {
        if(strcmp(s,".") == 0)
            break;
        int m = strlen(s);
        next[0] = 0; next[1] = 0;
        for(int i=1;i<m;i++)
        {
            int j = next[i];
            while(j && s[i] != s[j]) j = next[j];
            next[i+1] = (s[i] == s[j]) ? j + 1 : 0;
        }
        int ans = 1;
        if(next[m] > 0 && m % (m - next[m]) == 0) ans = m / (m - next[m]);
        printf("%d\n", ans);
    }
    return 0;
}

POJ 2406 Power String(KMP)