HDU ——1358 Period （kmp+循环节）

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#kmp #循环节

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本文探讨了如何识别字符串的所有前缀是否为周期性字符串，即能否由更小的子串重复构成。通过分析nexts[]数组，我们能确定每个前缀的最小周期及其重复次数。文章提供了一个C++实现示例，包括输入字符串、计算nexts数组、输出结果的完整流程。

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

题意：给出一个字符串s，问在[0, i]区间是否有完整的循环节，看样例知道是要求最小循环节，若有，输出i并输出循环次数(循环次数要>=2)；

题解：求循环节的问题，，主要还是nexts[]数组，如果nexts[i]==0说明在0——(i-1)中就一个循环节，如果nexts[i]!=0 ，那么0——(i-1)的最小循环节，就是i-nexts[i],那循环次数就是i/(i-nexts[i]). 上代码：

#include <iostream>
#include <cstring>
using namespace std;
typedef long long ll;
const int MAX = 1e6+520;
char str[MAX];
int nexts[MAX];
int l;
void getnexts(){//nexts[]模板
	//memset(nexts,0,sizeof(nexts));//初始，不初始化都对，还没碰上有错的
	int i,j;
	i=j=0;
	nexts[0]=-1;
	j=-1;
	while(i<l){
		if(j==-1||str[i]==str[j]) nexts[++i]=++j;
		else j=nexts[j];
	}
}
int main(){
	int cas=1;
	while(cin >> l,l){
		scanf("%s",str);
		cout << "Test case #" << cas++ << endl;
		getnexts();
		for (int i = 2; i <= l;i++){//注意加等号
			if(i%(i-nexts[i])||!nexts[i]) continue;//如果不是正好循环完或者只有一个循环节，跳过
		    else cout << i << " " << i/(i-nexts[i]) << endl;
		}
		cout << endl;
	}
	return 0;
}

这个题会了，可以去做一下HDU——3746 、 POJ——2406