poj 2406 Power Strings kmp算法

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 27368		Accepted: 11454

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.

kmp算法里的make next函数就可以了,不过我今天还是没搞明白这个算法的原理,只能说暂时能写对这个算法

#include<stdio.h>
#include<string.h>
char str[1000];
int next[1000];
void make_next()
{
	int i = 1, j = 0;
	next[0] = -1;
	while(str[i])
	{
		if(j == -1 || str[j] == str[i])
			next[++i] = ++j;
		else
			j = next[j];
	}
}
int main()
{
	while(scanf("%s", str), str[0] + str[1] != '.')
	{
		make_next();
		int i = strlen(str);
		int len = i - next[i];
		if(i % len == 0)
			printf("%d\n",i / len);
		else printf("1\n");
	}
	return 0;
}