Power Strings
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.
题目大意：求字符串的循环节数
KMP求串的前后缀相同长度，然后求循环节数
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int len;
int p[1000010];
char s[1000010];
int main()
{
for (;;) {
scanf("%s", s + 1);
if (s[1] == '.') break;
len = strlen(s + 1);
int j = 0;
p[1] = 0;
for (int i = 2; i <= len; ++i) {
while (j && s[j + 1] != s [i]) j = p[j];
if (s[j + 1] == s[i]) ++j;
p[i] = j;
}
if (len % (len - p[len]) == 0) printf("%d\n", len / (len - p[len]));
else printf("1\n");
}
}
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