地址：http://acm.hdu.edu.cn/showproblem.php?pid=2609
题目：
How many
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2625 Accepted Submission(s): 1135
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.
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').
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
Author
yifenfei
Source
Recommend
yifenfei
思路：最大最小表示法+set
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <set>
5 #include <string>
6 using namespace std;
7
8 #define MP make_pair
9 #define PB push_back
10 typedef long long LL;
11 const double eps=1e-8;
12 const int K=1e6+7;
13 const int mod=1e9+7;
14
15 char sb[205];
16 //ff为真表示最小，为假表示最大
17 //S串应该为原串复制两次后的字符串
18 int mx_mi_express(char *S,bool ff,int len)
19 {
20 int i=0,j=1,k;
21 while(i<len&&j<len)
22 {
23 k=0;
24 while(k<len&&S[i+k]==S[j+k]) k++;
25 if(k==len) return i<=j?i:j;
26 if((ff&&S[i+k]>S[j+k]) || (!ff&&S[i+k]<S[j+k]))
27 {
28 if(i+k+1>j) i=i+k+1;
29 else i=j+1;
30 }
31 else if((ff&&S[i+k]<S[j+k]) || (!ff&&S[i+k]>S[j+k]))
32 {
33 if(j+k+1>i) j=j+k+1;
34 else j=i+1;
35 }
36 }
37 return i<=j?i:j;
38 }
39 string tmp;
40 set<string>st;
41 int main(void)
42 {
43 int t,n,len;
44 while(scanf("%d",&n)==1&&n)
45 {
46 st.clear(),tmp.clear(),len=0;
47 for(int i=1,be;i<=n;i++)
48 {
49 scanf("%s",sb);
50 if(!len)
51 {
52 len=strlen(sb);
53 for(int j=0;j<len;j++)
54 tmp+='0';
55 }
56 for(int j=0;j<len;j++)
57 sb[j+len]=sb[j];
58 be=mx_mi_express(sb,1,len);
59 for(int j=0;j<len;j++)
60 tmp[j]=sb[j+be];
61 st.insert(tmp);
62 }
63 printf("%d
",st.size());
64 }
65 return 0;
66 }