POJ 2406 Power Strings

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 38400		Accepted: 15938

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 <cstdio>
#include <cstring>

const int N = 1000010;
char str[N];
int next[N];

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

int main() {

	while (~scanf("%s", str), str[0] != '.') {
		get_next();
		int len = strlen(str);
		int cir = len - next[len];
		if (len % cir == 0)
			printf("%d\n", len / cir);
		else
			puts("1");
	}

	return 0;
}