poj 2406 Power Strings

题目链接

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.

 

#include
#include
#define MAX 1000005
using namespace std;
int next[MAX];
char s[MAX];
int t;
void getNext()
{
    int i=0;
    int j=-1;
    next[0]=-1;
    while(i>s)
    {
        if(s[0]=='.')
        break;
        t=strlen(s);
        getNext();
        if(t%(t-next[t])==0)
            cout<