POJ——T 2406 Power Strings

Posted 2020-10-03

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

 1 #include <algorithm>
 2 #include <cstring>
 3 #include <cstdio>
 4 
 5 using namespace std;
 6 
 7 const int N(2333333);
 8 int ans,len,p[N];
 9 char s[N];
10 
11 inline void Get_next()
12 {
13     for(int i=2,j=0;i<=len;i++)
14     {
15         for(;s[i]!=s[j+1]&&j>0;) j=p[j];
16         if(s[i]==s[j+1]) j++;
17         p[i]=j;
18     }
19 }
20 
21 int main()
22 {
23     for(scanf("%s",s+1);s[1]!=‘.‘;scanf("%s",s+1))
24     {
25         memset(p,0,sizeof(p));
26         len=strlen(s+1);
27         Get_next();
28         int tmp=len-p[len];
29         if(len%tmp) puts("1");
30         else printf("%d\n",len/tmp);
31     }
32     return 0;
33 }