Period

题目描述：

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A K , that is A concatenated K times, for some string A. Of course, we also want to know the period K.

输入：

The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.

输出：

For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

样例输入：

3
aaa
12
aabaabaabaab
0

样例输出：

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

code：

题意：
求字符串的最小周期
额，机翻的结果是这样的：

对于具有N个字符的给定字符串的每个前缀(每个字符的ASCII码在97到126之间，包括)，我们想知道该前缀是否是一个周期
字符串。即为每个我(2 < =我< = N)我们想知道最大的K > 1(如果有的话),这样的前缀长度为我可以写成一个K,这是一个
连接K倍,对于一些字符串A。当然,我们也想知道K。

输入

输入文件由几个测试用例组成。每个测试用例由两行组成。第一行包含N (2 <= N <= 1 000 000)—字符串s的大小。第二
行包含字符串s。

输出

对于每个测试用例，在一行上输出“测试用例#”和连续的测试用例号;然后，对于每个长度为i，周期为K > 1的前缀，输出
前缀大小i和周期K之间用一个空格隔开;前缀大小必须按递增顺序排列。在每个测试用例之后打印一个空行。

思路：
求出所有前缀(满足是两个或两个以上的循环节组成的字符串)
输出前缀的位置[0-i-1]的i，以及循环节的个数
KMP求解fail数组，i-f[i]就是循环节长度

一开始数组开小了，，，结果告诉我超时？？？？
改了很久才改对。。。。看来有时候不要太相信评测机给的反馈啊
又是一道KMP题，，，，

#include<stdio.h>
#include<string.h>
const int maxn=1e6+10;
char a[maxn];
int nextt[maxn]; 
int n;
void init()
{
	int i=0,j=-1;
	n=strlen(a);
	nextt[0]=-1;
	while(i<n)
	{
		if(j==-1||a[i]==a[j])
		{
			i++;
			j++;
			if((i%(i-j)==0)&&(i/(i-j)>1))
			{
				printf("%d %d\n",i,i/(i-j));
			}//这里是重点
			nextt[i]=j;
		}
		else j=nextt[j];
	}
}
int main()
{
	int n,kk=1;
	while(scanf("%d",&n)&&n)
	{
		scanf("%s",a);
		printf("Test case #%d\n",kk++);
		init();
		printf("\n");
	}
	return 0;
}