poj2406 Power Strings

 

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 16472		Accepted: 6863

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.

#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <string> using namespace std; char s[1111111]; int next[1111111]; void getnext(int x){ int i=0,j=-1; next[0]=-1; while(i<x){ if(j==-1||s[i]==s[j]){ i++;j++; if(s[i]==s[j]) next[i]=next[j]; else next[i]=j; } else j=next[j]; } } int main() { int x; while(scanf("%s",s),s[0]!='.'){ x=strlen(s); int len; getnext(x); //能整除就重复了 x/(x-next[x]) 次 //不能整除就是它本身 if(x%(x-next[x])!=0) len=1; else len=x/(x-next[x]); printf("%d/n",len); } }