POJ

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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

ps:一道KMP基础题，拿来练手。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int next[1000010],m;char s[1000010];void get_next(){     int i=0,j=-1;       next[0]=-1;     while(i<m)     {         if(j==-1||s[i]==s[j])              {                                  i+=1;                j+=1;               next[i]=j;          }        else   j=next[j];      }}int main(){      while(scanf("%s",s)!=EOF)      {            if(s[0]=='.')   break;            m=strlen(s);            get_next();                        if(m%(m-next[m])==0)    printf("%d\n",m/(m-next[m]));            else printf("1\n");    }    return 0;}