POJ 2406 Power Strings

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 50920		Accepted: 21254

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

题目链接：http://poj.org/problem?id=2406

题意：输入字符串，问其中最短循环出现的次数。

解题思路：最短循环节的长度cir其实就是字符串的长度len减去最后一个字符的next值，也就是cir=len-next[len]。如果cir能够被len整除，那么循环节出现的次数就是len/cir。如果不能整除那么说明该循环节是第一次出现。

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 1e6 + 10; //最大的字符串长度 
char str[MAXN];
int fail[MAXN];
void get_fail(int n) //失配数组 
{
	memset(fail,0,sizeof(fail));
	fail[0]=0; 
	fail[1]=0; 
	for(int i=1;i<n;i++)
	{
		int j=fail[i];
		while(j && str[i]!=str[j]) j=fail[j];
		fail[i+1]=str[i]==str[j]?j+1:0;
	}
}
void solve()
{
	int len=strlen(str); //字符串的长度 
	get_fail(len); //得到失陪数组 
	int cir=len-fail[len]; //最短的循环节长度 
	if(len%cir!=0) printf("1\n"); //第一次出现 
	else printf("%d\n",len/cir); //出现次数 
}
int main(void)
{
	while(gets(str)) //输入字符串 
	{
		if(str[0]=='.') break; //不满足要求，退出循环 
		solve(); //处理函数 
	}
	return 0;
}