How many

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 974    Accepted Submission(s): 377

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

4011011001001001141010010110000001

 

Sample Output

12
 
先用最小表示法将字符串还原到以最小字典序循环的串，然后把它放到set中进行判重就可以了。。。
# include<iostream># include<stdio.h># include<string># include<string.h># include<set>using namespace std;int min(int i,int j){return i>j?j:i;}inline int getmin(string &s){   int i=0,j=1,k=0;   int len = s.length();   while(i<len && j<len && k<len)   {       int t = s[(i+k)%len] - s[(j+k)%len];       if(!t) k++;       else       {           if(t>0) i = i+k+1;           else j = j+k+1;           if(i == j) j++;           k = 0;       }   }   return min(i,j);}int main(){   int n;   string s,s1;   while(~scanf("%d",&n))   {       int num = 0;       set<string> set1;       set<string>::iterator it;       for(int i=0;i<n;++i)       {           cin>>s;           int j = getmin(s);           s1.assign(s,j,s.length()-j);           if(j) s1.append(s,0,j);           if(!set1.count(s1))           {               num++;               set1.insert(s1);           }       }       cout<<num<<endl;   }   return 0;}