hdu 1358 Period

最新推荐文章于 2021-03-07 19:35:22 发布

原创最新推荐文章于 2021-03-07 19:35:22 发布 · 253 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#KMP算法

ACM算法_KMP 专栏收录该内容

17 篇文章

订阅专栏

本文介绍了一种用于检测字符串前缀是否具有周期性的算法。该算法通过构建字符串的next数组来高效判断每个前缀是否可以表示为某个字符串的重复。文章提供了一个完整的C++实现示例，并展示了如何通过输入输出样例进行验证。

点击打开链接

Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9581 Accepted Submission(s): 4597

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 2
3 3

Test case #2
2 2
6 2
9 3
12 4

好像做过。。。

#include <iostream>
#include <cstring>
#include <cstdio>
const int N=1e6;
char mo[N+10];
int Next[N+10];
int len;
using namespace std;
void getNext()
{
	Next[0]=-1;
	int i=0,j=-1;
	while(i<len)
	{
		if(j==-1||mo[i]==mo[j])
		{
			i++;
			j++;
			Next[i]=j;
		}
		else
			j=Next[j];
	}
}
int main()
{
	int k=1;
	while(scanf("%d",&len),len)
	{
		scanf("%s",mo);
		getNext();
		printf("Test case #%d\n",k++);
		for(int i=1;i<=len;i++)
		{
			if(i%(i-Next[i])==0&&i!=i-Next[i])
				printf("%d %d\n",i,i/(i-Next[i]));
		}
		printf("\n");
	}
	return 0;
}