Period

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 14653		Accepted: 6965

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.

The input 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.

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.

3

aaa

12

aabaabaabaab

0

Test case #1

2 2

3 3

Test case #2

2 2

6 2

9 3

12 4
题意：求前缀的长度，以及其中最小循环节的循环次数  但是循环次数必须大于1

#include<stdio.h>

#include<string.h>

#define MAX 1000100

int next[MAX];

char str[MAX];

int a[MAX],b[MAX];

int n,m,k;

void getfail()

{

	int i,j;

	next[0]=next[1]=0;

	for(i=1;i<n;i++)

	{

		j=next[i];

		while(j&&str[i]!=str[j])

		    j=next[j];

		next[i+1]=str[i]==str[j]?j+1:0;

	}

}

void kmp()

{

	int i,j;

	for(i=1;i<=n;i++)

	{

		k=i;

		if(k==(k-next[k]))//k-next[k]最小循环节长度

		continue;

		if(k%(k-next[k])==0)

		printf("%d %d\n",i,k/(k-next[k]));

	}

	printf("\n");

}

int main()

{

	int i,t=1;

	while(scanf("%d",&n),n)

	{

		getchar();

		for(i=0;i<n;i++)

		scanf("%c",&str[i]);

		getfail();

		printf("Test case #%d\n",t++);

		kmp();

	}

	return 0;

}