hdu 2609 How many 最小表示法

How many

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1248    Accepted Submission(s): 486

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

 #include <iostream>
 #include <stdio.h>
 #include <algorithm>
 #include <string.h>
 #include <set>
 using namespace std;
 set<string>ss;
 void make(char a[],int x,int l)
 {
     char b[];
     int i;
     for(i=; i<l; i++)
         b[i]=a[x+i>=l?x+i-l:x+i];
     b[i]='\0';
     ss.insert(b);
 }
 void moremin(char a[])
 {
     int len=strlen(a);
     int i,j,k,t;
     k=i=;
     j=;
     while(i<len&&j<len&&k<len)
     {
         int t=a[i+k>=len?i+k-len:i+k]-a[j+k>=len?j+k-len:j+k];
         if(!t)k++;
         else
         {
             if(t>) i+=k+;
             else j+=k+;
             if(i==j)j++;
             k=;
         }
     }
     make(a,(i>j?j:i),len);
 }
 int main()
 {
     int n;
     char a[];
     while(~scanf("%d",&n))
     {
         ss.clear();
         while(n--)
         {
             scanf("%s",a);
             moremin(a);
         }
         cout<<ss.size()<<endl;
     }
 }