hdu1358 Period--KMP-白红宇的个人博客

hdu1358 Period--KMP

发布日期：2021-10-03 20:31:47 浏览次数：6 分类：技术文章

本文共 2316 字，大约阅读时间需要 7 分钟。

原题链接：

一：原题内容

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

3aaa12aabaabaabaab0

Sample Output

Test case #12 23 3Test case #22 26 29 312 4

二：分析理解

一个字符串，从头到某个位置，字符串的前缀最多重复了多少次，要求输出次数，注意题意要求次数至少2次

三：AC代码

#include
    
     #include
     
      using namespace std;char ch[1000010];int nextval[1000010];int len;void GetNextval(){	nextval[0] = -1;	int i = 0, j = -1;	while (i < len)	{		if (j == -1 || ch[i] == ch[j])		{			i++;			j++;			//if (ch[i] != ch[j])				nextval[i] = j;			//else				//nextval[i] = nextval[j];		}		else			j = nextval[j];	}}int main(){	int num = 1;	while (scanf("%d",&len) && len != 0)	{		scanf("%s", ch);		GetNextval();		printf("Test case #%d\n", num++);		for (int i = 2; i <= len; i++)		{			int mix = 2 * nextval[i] - i;     //重叠部分			int cycleLen = nextval[i] - mix;  //循环节长度			if (mix >= 0 && i%cycleLen == 0)				printf("%d %d\n", i, i / cycleLen);		}		printf("\n");	}	return 0;}

在这里解释一下：

int mix = 2 * nextval[i] - i; //重叠部分

int cycleLen = nextval[i] - mix; //循环节长度

这两句代码。

在这个循环里，我们假设ch数组的前i个字符满足题意，即它们是周期性至少为2的周期字符数组。我们来看看周期字符数组的特点，假如是ababab#，周期3，由ab构成，此时i（这里的i==6）指向#（只是一个代表符号，代表末尾的下一个字符），我们知道未优化的next数组next[i]代表什么意思呢，代表着前i个字符的前缀后缀最大公共元素长度。也就是说在这里next[6]=4，前缀是abab，后缀是abab。所以int mix = 2 * nextval[i] - i;就代表前缀后缀的交集长度。

那么int cycleLen = nextval[i] - mix;自然很好理解，就是获得循环节长度，在这里，就是获得ab的长度，即2。

转载地址：https://blog.csdn.net/LaoJiu_/article/details/50926919 如侵犯您的版权，请留言回复原文章的地址，我们会给您删除此文章，给您带来不便请您谅解！

上一篇：hdu2087 剪花布条--KMP

下一篇：hdu3746 Cyclic Nacklace--KMP

发表评论

关于作者

喝酒易醉，品茶养心，人生如梦，品茶悟道，何以解忧？唯有杜康！

-- 愿君每日到此一游！

发表评论

最新留言

关于作者

推荐文章