Poj2406-Brute Force Search, KMP

Power strings

Time limit:3000 Ms		Memory limit:65536 K
Total submissions:25491		Accepted:10694

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 = "" (
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

Hint

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

The question is to give you a string, and then you can find the smallest loop section in it and output the number of loops. For example, for ABCD, the loop section is itself, repeating once. While the AAAA loop section is a and has four cycles. The abababab loop section is AB and has three loops.

In fact, the brute-force search has passed.

11371876

Tserrof

2406

Accepted

4784 K

1891 Ms

C ++

427b

2013-03-20 14:07:58

#include<iostream>#include <algorithm>#include <string>using namespace std;int main(){string s;while (cin>>s && s!="."){unsigned int n=1;for(n=1;n<=s.size();++n){if( s.size()%n ) continue;string substr=s.substr(0,n);string temp=substr;for(unsigned t=1;t<s.size()/n;++t)substr.append(temp);if(substr==s)break;}cout<<s.size()/n<<endl;}return 0;}

It seems to be slow, but it can be done using the KMP idea.