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POJ 2406 Power Strings（kmp）

2024-07-31 03:20:48   220人阅读

Language:
Power Strings
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 33335 Accepted: 13852

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


求一个串的最小周期


对于一个串，如果abcdabc 那么next[len]=3,那么len-next【len】就大于len/2,那么len%(len-next[len])!=0;而对于一个周期串ababab next[len]=4,此时len-next[len]应该等于

最小串的长度，所以是不是有最小周期就可以用len%(len-next[len])是否为0来判断，（个人理解，如有错，请想告）



#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
using namespace std;
#define N 100005

char a[N];
int len,next[N];

void getfail(char *a)
{
	 int i,j;
	 len=strlen(a);
	 i=0;j=-1;
	 next[0]=-1;
	 while(i<len)
	 {
	 	if(j==-1||a[i]==a[j])
		{
			 i++;
			 j++;
			 next[i]=j;
		}
		else
			j=next[j];
	 }
}

int main()
{
	int i,j;
	while(scanf("%s",a))
	{
		if(a[0]=='.') break;
		getfail(a);
		int ans=len%(len-next[len]);
		if(ans==0)
			printf("%d\n",len/(len-next[len]));
		else
			printf("1\n");
	}
	return 0;
}






POJ 2406 Power Strings（kmp）


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