P

Give you n ( n < 10000) necklaces ,the length of necklace will not large than 100,tell me 
How many kinds of necklaces total have.(if two necklaces can equal by rotating ,we say the two necklaces are some). 
For example 0110 express a necklace, you can rotate it. 0110 -> 1100 -> 1001 -> 0011->0110. 

InputThe input contains multiple test cases. 
Each test case include: first one integers n. (2<=n<=10000) 
Next n lines follow. Each line has a equal length character string. (string only include '0','1'). 
OutputFor each test case output a integer , how many different necklaces.Sample Input

4
0110
1100
1001
0011
4
1010
0101
1000
0001

Sample Output

1
2

#include<iostream>
#include<set>
#include<map>
#include<vector>
#include<string>
#include<algorithm>
#include<cstring>
using namespace std;
#define MAXN 10002
/*
题目相当于求所有字符串中
能通过相互循环位移得到的字符串数目
对每个字符串求最小表示法，然后加入到set中
*/
string str;
set<string> S;
int GetMin(string s,int len)
{
    int i=0,j=1,k=0;
    while(i<len&&j<len&&k<len)
    {
        if(s[(i+k)%len]==s[(j+k)%len])
            k++;
        else if(s[(i+k)%len]>s[(j+k)%len])
        {
            i = i+k+1;
            k = 0;
        }
        else 
        {
            j = j+k+1;
            k = 0;
        }
        if(i==j)
            j++;
    }
    return min(i,j);
}
int main()
{
    int n;
    string tmp;
    tmp.reserve(101);
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
            tmp.clear();
            cin>>str;
            int pos = GetMin(str,str.size()),L = str.size();
            for(int j=pos,cnt=0;cnt<L;j=(j+1)%L,cnt++)
            {
                tmp.push_back(str[j]);
            }
            S.insert(tmp);
        }
        cout<<S.size()<<endl;
        S.clear();
    }
}