POJ 2406 Power Strings

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

分析：此题主要考察字符串模式匹配算法即KMP算法；

代码：

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std; 
char s[1000000];
int len,next[1000000];
void get_next()
{
     int i,j;
     i=1; next[1]=0; j=0;
     while(i<len)
     {
          if(j==0||s[i]==s[j])
          {
              ++i;
              ++j;
              next[i]=j;
          }
          else  j=next[j];
     }
}
int main()
{
    int k;
    while(cin>>s)
    {      
        if(s[0]=='.') break;
        len=strlen(s);
        get_next();
        if(len%(len-next[len])==0)
        {
   k=len/(len-next[len]);
   cout<<k<<endl;
  }
  else cout<<1<<endl;      
    }
    return 0;
}