Period （next與循環串）

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.

Input

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.

Output

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.

Sample Input

3
aaa
12
aabaabaabaab
0      

Sample Output

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3

12 4







題目大意：給出一個長度為n的字元串，從長度為2的字首開始，一一判斷 字首 是否為 循環串 構成，并輸出 字首長度 和 循環串的個數 。

e.g Test case #2中 aabaabaab 是長度為9的字首，其循環串為aab，個數為3，故輸出 9 3 。




知識點：如果i，滿足i%（i-next[i]）=0則長度為i的字首，是由循環串構成， 其個數為i/（i-next[i]）。

Tip：但如此會出現循環串個數為1的情況，與題意不符，故排除掉。




抄襲加原創代碼（抄襲部分出處未知）：





       
#include<stdio.h>

int next[1000005];
char a[1000005];

void getnext(char *a,int n)
{
	int i,j;
	i=0,j=next[0]=-1;
	while(i<n)
	{
		if(j==-1||a[i]==a[j])
		{
			i++,j++;
			next[i]=j;
		}
		else
		{
			j=next[j];
		}
	}
}

int main()
{
	int n,i,t=1;
	while(~scanf("%d",&n)&&n!=0)
	{
		scanf("%s",a);
		printf("Test case #%d\n",t++);
		
		getnext(a,n);
		for(i=2;i<=n;i++)
		{
			if(i%(i-next[i])==0&&i/(i-next[i])>1)
			{
				printf("%d %d\n",i,i/(i-next[i]));
			}
		}
		printf("\n");
	}
}