POJ 2406 (KMP)

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 41709 Accepted: 17343

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

abcdaaaaababab.

Sample Output

143

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

Waterloo local 2002.07.01

题意：求串的最小循环节的循环次数。

n-next[n]求出最小循环节，然后判断下是不是被长度整除。

#include <cstring>#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <cmath>using namespace std;#define maxn 1111111char T[maxn];int n;#define next Nextint next[maxn];void get_next (char *p) {      int m = strlen (p);    int t;      t = next[0] = -1;      int j = 0;      while (j < m) {          if (t < 0 || p[j] == p[t]) {//匹配              j++, t++;              next[j] = t;        }          else //失配              t = next[t];      } }  int main () {    while (scanf ("%s", T) == 1) {        n = strlen (T);        if (n == 1 && T[0] == '.')            break;        get_next (T);         int len = n-next[n];        if (n%len == 0)             printf ("%d\n", n/len);        else             printf ("1\n");    }    return 0;}