Power Strings
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

題意：
就是一道求s 最小迴圈節的題，輸出迴圈的最大次數。

分析：
KMP演算法中，f[i]陣列存的是[0,i)的最長真字首等於最長真字尾的長度。那麼若前i個字元組成的字首是迴圈的，那麼i%(i-f[i])==0 一定成立。 錯位部分就是迴圈節。

AC程式碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

const int maxn = 1000010;
char s[maxn];
int f[maxn];

void getfail(char* p)
{
    int n = strlen(p);
    f[0] = f[1] = 0;
    for(int i=1;i<n;i++)
    {
        int j = f[i];
        while(j && p[i]!=p[j]) j = f[j];
        f[i+1] = p[i]==p[j] ? j+1 : 0;
    }
}

int main()
{
    while(scanf("%s",s)!=EOF)
    {
        if(strcmp(s,".")==0) break;
        getfail(s);
        int n = strlen(s);
        if(f[n] > 0 && n%(n-f[n])==0) printf("%d\n",n/(n-f[n]));
        else printf("1\n");
    }
    return 0;
}