How many（串的最小表示法）

Problem Description
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.
Input
The 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’).
Output
For 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

就一个字符串的最小表示套模板的，但是里面还用到了指针，我以前看到指针就蒙，这次也很蒙，但也多少知道了点。get(temp+k);是指向temp字符串中的第k个字符，相当于把首指针向后移动k位。

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<set>
using namespace std;
char s[250];
int len;
set<string>a;     //利用set去重 
int findMin(string s)    //字符串的最小表示 
{
    int len = s.length();
    int i=0,j=1,k=0;
    while(i<len&&k<len&&j<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);
}
void get(char *s)
{
	s[len/2]='\0';
	a.insert(s);
}

int main()
{
    int n;   
    while(scanf("%d",&n)!=EOF)
    {
    	a.clear();
        for(int i=0;i<n;i++)
        {
        	scanf("%s",s);
        	char temp[210];
        	strcpy(temp,s);    //字符串复制函数 
        	strcat(temp,s);     //字符串连接函数 
        	len=strlen(temp);
        	int k=findMin(temp);
        	get(temp+k);
		}
		cout<<a.size()<<endl;
    }
    return 0;
}