poj-2406 Power Strings

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 37660		Accepted: 15576

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
# include<stdio.h>
# include<string.h>
# include<algorithm>

using namespace std;

char s[1001000];
int p[1001000];

void get()
{
    int i = 0, j = -1;
    p[0] = -1;
    int len = strlen(s); 
    while(i < len)
    {
   	    if(s[i] == s[j] || j == -1)
   	    {
   	  	   i++;
   	  	   j++;
   	  	   p[i] = j;
   	    }
   	    else
   	       j = p[j];
    }	
}

int main()
{
	int len;
	while(scanf("%s",s), s[0] != '.')
	{
	      get();
	      len = strlen(s);  //这个有点不清楚 就是上下两个语句不能调换位置 否则在oj过不去。
	      if(len % (len-p[len]))  如果结果不为零则说明有不属于循环体的内容。
              printf("1\n");  
              else  
              printf("%d\n", len/(len-p[len]));//<span style="font-family: 'Courier New', Courier, monospace;">len-p[len]是每个循环体的长度，用整个字节数除就是重复的次数。</span><span style="font-family: 'Courier New', Courier, monospace;">  </span>
	}
	return 0;
}