Poj2406【KMP】

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 33595		Accepted: 13956

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>
int next[1000001], l;
char s[1000001];
void getnext()
{
	next[0] = -1;
	int i = 0, j = -1;
	while(i < l)
	{
		if(j == -1 || s[i] == s[j])
		{
			i++;
			j++;
			next[i] = j;
		}
		else
		j = next[j];
	}
}
int main()
{
	int  i, j, k;
	while(scanf("%s", s) != EOF)
	{
		if(s[0] == '.')
		break;
		l = strlen(s);
		getnext();
        if(l%(l-next[l]) == 0 && l/(l-next[l]) > 1)
		printf("%d\n", l/(l-next[l]));
        else
        printf("1\n");
	}
	return 0;
}

题意：求出字符串的字串最大循环周期。

思路：用KMP算法求出next数组的值，接着就是判断其循环字串的长度是否能被l整除且结果不等于1.

体会：运用的是对KMPnext数组的理解，即i-next[i]即为循环字串的长度。