HDU 2609 How many

How many

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2870    Accepted Submission(s): 1241

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

最小表示法，每个都弄到字典序最小，存到map中，输出map的大小就可以了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;
string a;
int len,n;
string check(string a,int len)
{
  int i=0,j=1,k=0;
  while(i<len && j<len && k<len)
  {
    int t=a[(j+k)%len]-a[(i+k)%len];
    if(t==0) k++;
    else
    {
      if(t<0) i+=k+1;
      else j+=k+1;
      if(i==j) j++;
      k=0;
    } 
  }
  int minn=min(i,j);
  int cnt=0;
  string ans;
  while(cnt<len)
  {
    ans+=a[(cnt+minn)%len];
    cnt++;
  }
  return ans;
}
int main()
{
  map<string,int>m;
  while(cin>>n)
  {
    while(n--)
    {
      cin>>a;
      len=a.size();
      string s=check(a,len);
      m[s]=1;
    }
    printf("%d
",m.size());
    m.clear();
  }
  return 0;
}