POJ 2406 Power Strings KMP 求最小循环节

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#POJ 2406 #Power Strings #KMP #求最小循环节

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本文探讨了如何通过KMP算法中的next数组找到给定字符串的最小循环节，即求解给定字符串能由某个最短子串重复多次组成的情况。通过分析next数组的特性，我们能够高效地确定该最小子串。

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 29027		Accepted: 12124

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

Waterloo local 2002.07.01

给定一个字符串m，求一个最短的子串n，m是n重复多次组成的（求m的最小循环节）。

利用kmp中next（我的数组名是p）数组的性质。

表达能力拙计。。。分析样例的next数组找数学规律吧。

还有,用cin是可以AC的。

#include <iostream>
using namespace std;
string s;
int l,i,j,p[1000010];
int main()
{
	ios::sync_with_stdio(false);
	while (cin>>s && s!="."){
		l=s.size();
		j=-1; p[0]=-1;
		for (i=1;i<l;++i){
			while (j>-1 && s[i]!=s[j+1]) j=p[j];
			if (s[i]==s[j+1]) ++j;
			p[i]=j;
		}
		i=l-p[l-1]-1;
		if (l%i==0) cout<<l/i<<endl; 
		else cout<<"1\n";
	}
	return 0;
}

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