Power Strings poj2406（神代码）

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 29402		Accepted: 12296

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
 
两种做法：

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 using namespace std;
 char a[];
 int t;
 int main()
 {
     char x;
     while()
     {
         t=;
         x=getchar();
         if(x=='.')break;
         while()
         {
             a[t++]=x;
             if(x=='\n')break;
             x=getchar();
         }
         int p=,k=;
         t--;
         for(int i=; i<t; i++)
         {
              if(a[i]==a[i%p])
              k++;
              else
              {
                  if(p==k)k++;
                  p=k;
              }
         }
         if(t%p==)
         printf("%d\n",t/p);
         else printf("1\n");
     }
 }

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 using namespace std;
 char a[];
 int t,next[];
 void fun()
 {
     next[]=;
     int i,j=;
     for(i=;i<t;i++)
     {
         while(j>&&a[i]!=a[j])j=next[j-];
         if(a[i]==a[j])j++;
         next[i]=j;
     }
 }
 int main()
 {
     char x;
     while()
     {
         t=;
         x=getchar();
         if(x=='.')break;
         while()
         {
             a[t++]=x;
             if(x=='\n')break;
             x=getchar();
         }
         t--;
       fun();
       if(t%(t-next[t-])==)
       printf("%d\n",t/(t-next[t-]));
       else printf("1\n");
     }
 }