E - Power Strings,求最小周期串

E - 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

abcd
aaaa
ababab
.

Sample Output

1
4
3


讲了这么多，其实就是一个求一个字符串里最小周期的问题。限时有3S，我们可以用枚举。
这里需要注意的是一个：求周期串的基本算法：
for(i=1;i<=len;i++)
{
  ok=1;
  if(len%i==0)
    {
     for(j=i;j<len;j++)
      {
        if(s[j]!=s[j%i]){ok=0;break;}
      }
    }
if(ok==1)
{
 printf("%d\n",len/i);
  break;//记得要break
}
}

#include<cstdio>
#include<string.h>
using namespace std;
char s[1000100];
int main()
{
    while(scanf("%s",s)==1&&strcmp(s,".")!=0)
    {
        int len=strlen(s);

        for(int i=1;i<=len;i++)
            if(len%i==0)
        {
             int ok=1;
            for(int j=i;j<len;j++)
            {
                if(s[j]!=s[j%i])
                {
                    ok=0;
                    break;
                }
            }
            if(ok)
            {
                 printf("%d\n",len/i);
                 break;
            }

        }
    }
    return 0;
}