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[];
 int main()
 {
     while(scanf("%s",s)==&&strcmp(s,".")!=)
     {
         int len=strlen(s);

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

         }
     }
     return ;
 }