POJ 2406 Power Strings

B - Power Strings

Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2406

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.

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<queue>using namespace std;#define maxn 1000100#define inf 0x3f3f3f3fchar p[100005];int f[100005];void getfail(){  int m=strlen(p);  f[0]=f[1]=0;  for(int i=1;i<m;i++){    int j=f[i];    while(j&&p[j]!=p[i])j=f[j];    f[i+1]=p[j]==p[i]?j+1:0;  }}int main(){    //freopen("in.txt","r",stdin);    while(~scanf("%s",p)){        if(strlen(p)==1&&p[0]=='.')break;        getfail();        int n=strlen(p);        if(n%(n-f[n])==0)printf("%d\n",n/(n-f[n]));        else printf("%d\n",1);    }}