POJ Power Strings 2406【KMP】

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 37379 Accepted: 15443

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

 

#include <stdio.h>#include <string.h>#include <algorithm>#define maxn 1000000+10using namespace std;char P[maxn];int pre[maxn];void getnext(int plen){pre[0]=pre[1]=0;for(int i=1;i<plen;i++){int k=pre[i];while(k&&P[i]!=P[k])k=pre[k];pre[i+1]= P[k]==P[i]?k+1:0;}}int main(){while(scanf("%s",P)!=EOF){if(P[0]=='.')break;int plen=strlen(P);getnext(plen);if(plen%(plen-pre[plen])==0)printf("%d\n",plen/(plen-pre[plen]));else printf("1\n");}return 0;}