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}}

 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 }

E - Power Strings,求最小周期串