How many

题目描述：

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.

输入：

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’).

输出：

For each test case output a integer , how many different necklaces.

样例输入：

4
0110
1100
1001
0011
4
1010
0101
1000
0001

样例输出：

1
2

code：

给出n个01字符串表示链，如果两个01字符串可以通过旋转得到就说这两个串是同一个，最后求有多少个不同的链。

解题思路：用最小表示法

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<set>
using namespace std;
char a[100050];
char b[100050];
set<string> jiang;
void findd()
{
	int n=strlen(a);
	int i=0,j=1,k=0;
	while(i<n&&j<n&&k<n)
	{
		int tot=a[(i+k)%n]-a[(j+k)%n];
		if(tot==0)k++;
		else
		{
			if(tot>0)
			{
				i+=k+1;
			} 
			else j+=k+1;
			if(i==j)j++;
			k=0;
		}
	}
	int tt;
	if(i>j)tt=j;
	else tt=i;
	for(int i=0;i<n;i++)
	{
		b[i]=a[(tt+i)%n];
	}
	b[n]='\0';
	jiang.insert(b);
}
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		for(int i=0;i<n;i++)
		{
			scanf("%s",a);
			findd();
		}
		printf("%d\n",jiang.size());
		jiang.clear();
	}
	return 0;
}