poj--2406--Power Strings(KMP)

Time Limit: 3000MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u

Status

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


#include<stdio.h>
#include<string.h>
#include<math.h>
#define MAX 100000+10
using namespace std;
char p[MAX],t[MAX];
int f[MAX];
void get()
{
	int i=0,j=-1;
	f[0]=-1;
	int len=strlen(p);
	while(i<len)
	{
		if(j==-1||p[i]==p[j])
		{
			i++;
			j++;
			f[i]=j;
		}
		else j=f[j];
	}
}
int main()
{
	int len;
	while(scanf("%s",p),p[0]!='.')
	{
		get();
		len=strlen(p);
		if(len%(len-f[len]))
		printf("1
");
		else 
		printf("%d
",len/(len-f[len]));
	}
}


原文地址:https://www.cnblogs.com/playboy307/p/5273755.html