HDU-1358 Period(kmp)

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本文介绍了一种使用next[]数组特性来判断字符串是否为循环字符串的方法，并提供了完整的代码实现。通过检查next[]数组，可以确定字符串的循环部分及其重复次数。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1358

Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13560 Accepted Submission(s): 6341

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

aaa

aabaabaabaab

Sample Output

Test case #1

2 2

3 3

Test case #2

2 2

6 2

9 3

12 4

题意：判断给出的字符串是否是循环字符串，要求输出循环串的长度及循环次数。

解题思路：利用next[ ]数组的性质即可，如果一个字符串为循环串时，（例如adcabcabc）那么它的next[ ]数组满足下面性质：
1: len%(len-next[len])==0; 2: len/(len-next[len])就是循环的次数。

这里题目中要求循环的长度必须大于1，所以再满足第一个条件下要确保次数是大于1的，即len/(len-next[len])!=1即可。

代码：

#include <stdio.h>
#include <string.h>
#define N 1000000  
int next[N];

void getnext(char s[])
{
	int k=-1;
	int j=0;
	int len=strlen(s);
	next[0]=-1;
	while(j<len)
	{
		if(k==-1||s[j]==s[k])
		{
			j++;
			k++;
			next[j]=k;
		}
		else
			k=next[k];
	}
}
int main()
{
	
	int n,i;
	char s[N];
	int c=1;
	while(scanf("%d",&n)!=EOF)
	{
		if(n==0)
			break;
		scanf("%s",s);
		getnext(s);
		/*for(i=0;i<n;i++)
			printf("%d ",next[i]);*/
		//printf("%d\n",next[n]);
		printf("Test case #%d\n",c++);
		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");
	}
	return 0;
}