POJ2406 Power Strings 【KMP】

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 31388		Accepted: 13074

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.

题意：给定一个串，它一定是某个前缀反复K次组成的。求K的最大值。

题解：利用next数组求最小循环节点len-next[len]，注意还需推断len是否能整除这个节点，若不能，那么k==1，否则k==len/最小循环节点。

#include <stdio.h>
#define maxn 1000002

char str[maxn];
int next[maxn], len, cir;

void getNext()
{
	int i = 0, j = -1;
	next[0] = -1;
	while(str[i]){
		if(j == -1 || str[i] == str[j]){
			++i; ++j;
			next[i] = j; //mode 1
		}else j = next[j];
	}
	len = i;
}

int main()
{
	//freopen("stdin.txt", "r", stdin);
	while(scanf("%s", str) == 1){
		if(str[0] == '.') break;
		getNext();
		cir = len - next[len];
		if(len % cir != 0) printf("1
");
		else printf("%d
", len / cir);
	}
	return 0;
}