Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13519    Accepted Submission(s): 6330

 

Problem Description

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 AK , that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input

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.

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

問題描述：給定字串S（長度為N），問S的長度為 i （1<i<=n ）的字首是否為迴圈串，如果是列印字首長度 i ，以及有多少個迴圈節。比如樣例一aaa。長度為2的字首為aa，迴圈節長度為1（即a），有2個迴圈節，則輸出2 2。長度為3的字首為aaa，迴圈節長度為1（即a），有3個迴圈節，則輸出3 3。

解題思路：對KMP中next陣列的運用。要記住迴圈節的長度為 i-next[i]。如果i是迴圈節長度的倍數（不能為1倍），則可輸出，具體看程式。

AC的C++程式：

#include<iostream>
#include<cstring>

using namespace std;

const int N=1000010;

char p[N];
int Next[N];

void GetNext(char p[],int next[])
{
	next[0]=-1;
	int k=-1,j=0,plen=strlen(p);
	while(j<plen)
	{
		//p[k]表示字首 p[j]表示字尾
		if(k==-1||p[k]==p[j])
		{
			k++;
			j++;
			next[j]=k;
		}
		else
		  k=next[k];
	}
}

int main()
{
	int n,cnt=1;
	while(~scanf("%d",&n)&&n)
	{
		scanf("%s",p);
		GetNext(p,Next);
		printf("Test case #%d\n",cnt++);
		for(int i=2;i<=n;i++)//列舉字首長度 
		{
			int l=i-Next[i];//迴圈節的長度 
			if(i%l==0&&i/l>1)
			  printf("%d %d\n",i,i/l); 
		}
		printf("\n"); 
	}
	return 0;
 }