[KMP nx数组循环节] E - Period HDU - 1358

最新推荐文章于 2025-06-03 20:14:20 发布

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最新推荐文章于 2025-06-03 20:14:20 发布

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分类专栏：字符串 KMP

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本文介绍了一种用于检测字符串前缀是否为周期性字符串的有效算法。对于给定的字符串S，算法可以找出所有可能的前缀及其周期K，即前缀能够被长度为K的子串重复构成的情况。通过构建next数组，算法能够在O(N)的时间复杂度内完成这一任务。

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Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12695 Accepted Submission(s): 5948

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

#include <bits/stdc++.h>
using namespace std;

const int mn = 1000010;

int n;
char a[mn];
int nx[mn];

void cal_next(char b[])
{
	nx[0] = -1;
	int k = -1;
	for (int i = 1; i <= n - 1; i++)
	{
		while (k > -1 && b[k + 1] != b[i])
			k = nx[k];
		if (b[k + 1] == b[i])
			k++;
		nx[i] = k;
	}
}

int main()
{
	#ifndef ONLINE_JUDGE
		freopen("D:\\in.txt", "r", stdin);
	#endif // ONLINE_JUDGE
	
	int cnt = 0;
	while (~scanf("%d", &n))
	{
		if (n == 0)
			break;
		scanf("%s", a);
		printf("Test case #%d\n", ++cnt);
		
		cal_next(a);
		
		for (int i = 0; i < n; i++)
		{
			if (nx[i] == -1)
				continue;
				
			int len = i + 1 - (nx[i] + 1);
			if ((i + 1) % len == 0) // 可由循环节组成
				printf("%d %d\n", i + 1, (i + 1) / len);
		}
		
		printf("\n");
	}
	
	return 0;
}