poj 2406 最短循环周期

最新推荐文章于 2019-08-03 10:14:31 发布

原创最新推荐文章于 2019-08-03 10:14:31 发布 · 1.4k 阅读

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本文介绍了一个编程挑战，任务是确定给定字符串是否可以表示为某个更短字符串的幂次重复。通过KMP算法预处理得到最大重复次数，适用于大规模字符串输入。

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 27099		Accepted: 11345

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

code:

#include<iostream>
using namespace std;
char s[1000005];
int next[1000005];
int len;
int get_next()
{
//	cout<<1<<endl;
	int i=0,j=-1;
	next[0]=-1;
	int lent=strlen(s);
	while(i<lent)
	{

		if(j==-1||s[i]==s[j])
		{
		//	cout<<1<<endl;
			++i,++j;
			next[i]=j;
		}
		else
		j=next[j];
				
	}
//	cout<<next<<endl;
	int cir=lent-next[lent];
		if(lent%cir==0)
	   return lent/cir;    
	   return 1;
}

int main()
{
	while(cin>>s)
	{
		getchar();
		memset(next,0,sizeof(next));
		
		if(s[0]=='.')
		 break;	
	    cout<<get_next()<<endl;	
	}
	return 0;
}