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最新推荐文章于 2020-02-07 11:05:28 发布

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本文介绍了一种用于检测字符串前缀周期性的算法。通过计算每个前缀的最大周期K，该算法能够确定字符串是否由某个基串重复组成。利用next数组进行高效匹配，适用于多种编程竞赛及实际应用场景。

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Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10787 Accepted Submission(s): 5080

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 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 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
3

Test case #2
2
2
3
4

思路：

利用next求出当前最小循环节，公式len%(len-next[i]),若为0，则循环节长度为len/(len-next[i]),否贼为len

code:

#include <cstdio>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include<iostream>
#include<vector>
#include<string>
#define Max 1000000
using namespace std;
typedef long long LL;
int ls, ts, n, nexttt[Max];
char lss[Max], tss[Max];
void gget_next()
{
	nexttt[0] = -1;
	int k = 0, j = -1;
	while (k < ts)
	{
		if (j == -1 || tss[k] == tss[j])
		{
			nexttt[++k] = ++j;
		}
		else
			j = nexttt[j];
	}
}
int get_first()
{
	gget_next();
	int i = 0, j = 0;
	while (i < ls&&j < ts)
	{
		if (lss[i] == tss[j]||j==-1)
		{
			i++;
			j++;
		}
		else
		{
			j = nexttt[j];
		}
	}
	if (j == ts)
	{
		return i - ts;
	}
	else
		return -1;
}
int count()
{
	int ans = 0;
	int j = 0, i = 0,qian=-1;
	gget_next();
	if (ls == 1 && ts == 1)
	{
		if (lss[i] == tss[j])
			return 1;
		else
			return 0;
	}
	for (i = 0; i < ls; i++)
	{
		while (j > 0 && lss[i] != tss[j])
		{
			j = nexttt[j];
		}
		if (tss[j] == lss[i])
		{
			j++;
		}
		if (j == ts)
		{
			if((qian == -1 || i - qian >= ts))
				ans++, qian = i;;
		}
	}
	return ans;
}
int main()
{
	int t = 0;
	while (scanf("%d", &ts) && ts)
	{
		scanf("%s", tss);
		printf("Test case #%d\n", ++t);
		gget_next();
		for (int i = 1; i <= ts; i++)
		{
			int temp = i - nexttt[i];
			if (i%temp == 0 && i / temp > 1)
			{
				printf("%d %d\n", i, i / temp);
			}
		}
		printf("\n");
	}
	return 0;
}