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. 

4

0110

1100

1001

0011

4

1010

0101

1000

0001

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=,j=,k=;

    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+;

            k = ;

        }

        else

        {

            j = j+k+;

            k = ;

        }

        if(i==j)

            j++;

    }

    return min(i,j);

}

int main()

{

    int n;

    string tmp;

    tmp.reserve();

    while(scanf("%d",&n)!=EOF)

    {

        for(int i=;i<n;i++)

        {

            tmp.clear();

            cin>>str;

            int pos = GetMin(str,str.size()),L = str.size();

            for(int j=pos,cnt=;cnt<L;j=(j+)%L,cnt++)

            {

                tmp.push_back(str[j]);

            }

            S.insert(tmp);

        }

        cout<<S.size()<<endl;

        S.clear();

    }

}