POJ——T 2406 Power Strings

http://poj.org/problem?id=2406

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 50627		Accepted: 21118

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

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

Waterloo local 2002.07.01

 

求最短循环节出现次数

 #include <algorithm>
 #include <cstring>
 #include <cstdio>

 using namespace std;

 const int N();
 int ans,len,p[N];
 char s[N];

 inline void Get_next()
 {
     for(int i=,j=;i<=len;i++)
     {
         for(;s[i]!=s[j+]&&j>;) j=p[j];
         if(s[i]==s[j+]) j++;
         p[i]=j;
     }
 }

 int main()
 {
     for(scanf("%s",s+);s[]!='.';scanf("%s",s+))
     {
         memset(p,,sizeof(p));
         len=strlen(s+);
         Get_next();
         int tmp=len-p[len];
         if(len%tmp) puts("");
         else printf("%d\n",len/tmp);
     }
     return ;
 }