UVA

Description

Problem D: Power Strings

Given two strings a and b we define a*b to be their concatenation. For example, ifa = "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).

Each test case is a line of input representing s, a string of printable characters. For eachs you should print the largest n such that s = a^n for some stringa. 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.

Sample Input

abcd
aaaa
ababab
.

Output for Sample Input

1
4
3

题意：求循环节

思路：KMP模板求循环节

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN = 1000000;

char str[MAXN];
int next[MAXN];

int main() {
	while (scanf("%s", str) != EOF && str[0] != '.') {
		int len = strlen(str);
		int i = 0, j = -1;
		next[0] = -1;
		while (i < len) {
			if (j == -1 || str[i] == str[j]) {
				i++, j++;
				next[i] = j;
			}
			else j = next[j];
		}
		printf("%d
", len/(len-next[len]));
	}
	return 0;
}