po2406Power Strings

最新推荐文章于 2025-12-05 07:17:18 发布

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本文介绍了一道关于字符串周期性的编程题目，通过使用KMP算法的next数组来高效解决输入字符串的最大周期问题。

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 49274		Accepted: 20527

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.

题目意思：一道求周期的题目，让你找出一个前缀在串中周期出现的次数最多，例如abababab就是4

解题思路：一开始用kmp去做没想太多直接一个一个找，最后超时了。然后仔细想了下，发现直接用kmp的next数组做就行。

#include<iostream>
#include<cstring>

using namespace std;

int const maxn=1000005;

char str[maxn];
int next1[maxn];

int len;

void sign()
{
	int i=-1,j=0;
	next1[0]=-1;
	int len=strlen(str);
	while(j<len)
	{
		if(i==-1||str[i]==str[j])
		{
			i++;
			j++;
			next1[j]=i;
		}
		else
		{
			i=next1[i];
		}
	}
}



int main()
{
	int num;
	while(cin>>str&&str[0]!='.')
	{
		memset(next1,0,sizeof(next1));
		len=strlen(str);
	    sign();
	    int i=len-next1[len];
	    if(len%i==0)    //一定要判断整除，只有能整除的才有周期 
	    {
	    	num=len/i;
		}
		else
		   num=1;
		cout<<num<<endl;
		
	}
	return 0;
}