8.11 D - Period

最新推荐文章于 2023-04-07 20:39:25 发布

原创最新推荐文章于 2023-04-07 20:39:25 发布 · 409 阅读

0 ·

CC 4.0 BY-SA版权

kmp 专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一种用于检测字符串周期性的算法实现，该算法能够高效地找出字符串中所有可能的循环节及其长度。通过构建next数组来辅助计算，进而判断每个前缀是否具有周期性特征。

D - Period

For each prefix of a given string S with N characters(each character has an ASCII code between 97 and 126, inclusive), we want toknow 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 theprefix of S with length i can be written as A ^K,that is Aconcatenated K times, for some string A. Of course, we also want to know theperiod K.

Input

The input consists of several test cases. Each test caseconsists 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 fileends with a line, having the
number zero on it.

Output

For each test case, output "Test case #" andthe consecutive test case number on a single line; then, for each prefix withlength i that has a period K > 1, output the prefix size i and the period Kseparated 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

题意：输入一个数字n，然后输入长度为n的一个字符串，然后从前往后判断循环节，当某一位字符前含有2个循环节时，输出当前为第几个字符，并输出循环节的个数，例如aabaabaabaab，判断到第二个字符时，有两个循环节，单循环节为a，所以输出2 2，到第六个字符时，也有两个循环节，单循环节为aad，所以输出6 2，以此类推，输出全部符合条件的数据。输出的样例之间含有换行。

#include<stdio.h>
#include<string.h>
int next[1110000];
char s[1100000];
int main()
{
	int i,j,k,m,n,w,z=1;
	while(scanf("%d",&n)!=EOF)
	{
		if(n==0)
		break;
		getchar();
		for(i=0;i<n;i++)
			scanf("%c",&s[i]);
			
	//	puts(s);
		i=0;j=-1;
		next[0]=-1;
		while(i<n)
		{
			if(j==-1||s[i]==s[j])
			{
				i++;
				j++;
				next[i]=j;
			}
			else
			{
				j=next[j];
			}
		}
/*		for(i=0;i<=n;i++)
		{
			printf("%d ",next[i]);
		}
		printf("\n");
*/	//	if(z!=1)
		
		printf("Test case #%d\n",z++);
		for(i=2;i<=n;i++)
		{
			w=i-next[i];//判断单个循环节长度
			
			if(next[i]%w==0&&w!=0)//当存在符合条件的循环节时
			{
				if(next[i]/w+1!=1)//计算是否含有2个以上循环节
				printf("%d %d\n",i,next[i]/w+1);
			}
		}
		
		printf("\n");
		
	}
	
	return 0;
}