hdu1358——经典循环节问题

最新推荐文章于 2024-10-28 14:12:21 发布

mumei314

最新推荐文章于 2024-10-28 14:12:21 发布

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分类专栏： kmp 文章标签：循环节

本文链接：https://blog.youkuaiyun.com/qq_43472263/article/details/98762091

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本文探讨了字符串前缀的周期性问题，通过分析每个前缀是否可以视为某个基本字符串的重复，介绍了如何判断和计算字符串的循环节。利用KMP算法预处理next数组，有效地解决了这一经典问题。

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

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

题目翻译：

给出一个字符串s,问在[0, i]区间是否有完整的循环节,若有,输出i并输出循环次数

经典的循环节问题，求循环次数并且判断是否存在循环节。

关于循环节的问题可以参考我的这篇文章：https://blog.youkuaiyun.com/qq_43472263/article/details/98759664

#include <iostream>
using namespace std;
const int maxn=1e6+7;
int n;
char str[maxn];
int nxt[maxn];
void getnext(){
	int j=0,k=-1;
	nxt[0]=-1;
	while(j<n){
		if(k==-1||str[j]==str[k]){
			j++;
			k++;
			nxt[j]=k;
		}
		else k=nxt[k];
	}
}
int main(int argc, char** argv) {
	ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	int ncase=1;
	while(cin>>n&&n){
		scanf("%s",str);
		getnext();
		printf("Test case #%d\n",ncase++);
		for(int i = 1;i<=n;++i)
			if((i)%(i-nxt[i])==0&&nxt[i]!=0)
				printf("%d %d\n",i,i/(i-nxt[i])); 
		printf("\n");
	}
	return 0;
}