E-power strings: returns the minimum periodic string.

E-power strings Time limit:3000 Ms Memory limit:65536kb 64bit Io format:% I64d & % i64usubmit 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 shoshould print the largest N such that S = a ^ N for some string.

Sample Input

abcdaaaaababab.

Sample output

143


After talking about this, it is actually a problem of finding the minimum cycle in a string. The time limit is 3 s. We can use enumeration.
Note the following:Cycle stringBasic algorithm:
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; // remember to use break
}
}

 1 #include<cstdio> 2 #include<string.h> 3 using namespace std; 4 char s[1000100]; 5 int main() 6 { 7     while(scanf("%s",s)==1&&strcmp(s,".")!=0) 8     { 9         int len=strlen(s);10 11         for(int i=1;i<=len;i++)12             if(len%i==0)13         {14              int ok=1;15             for(int j=i;j<len;j++)16             {17                 if(s[j]!=s[j%i])18                 {19                     ok=0;20                     break;21                 }22             }23             if(ok)24             {25                  printf("%d\n",len/i);26                  break;27             }28 29         }30     }31     return 0;32 }