POJ - 2406 - Power Strings

Power Strings Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 51263 Accepted: 21405

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. 问 把所给字符串写成  s^n 的模式 ，n最大是多少 #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> using namespace std; const int len=1e6+8; char a[len]; int kmp_next[len]; void KMP_next() { int i=0,j=-1; kmp_next[0]=-1; while(a[i]) if(j==-1||a[i]==a[j]) kmp_next[++i]=++j; else j=kmp_next[j]; } int main() { while(scanf("%s",a)&&a[0]!='.') { KMP_next(); int n=strlen(a); if(n%(n-kmp_next[n])) printf("1\n"); else printf("%d\n",n/(n-kmp_next[n])); } return 0; }