How many

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3272 Accepted Submission(s): 1457

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

Author

yifenfei

Source

奋斗的年代

Recommend

yifenfei

题解：

1.求出每个字符串的最小表示法。

2.对所有字符串的最小表示进行排序，然后统计。

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <string>
 6 #include <vector>
 7 #include <map>
 8 #include <set>
 9 #include <queue>
10 #include <sstream>
11 #include <algorithm>
12 using namespace std;
13 typedef long long LL;
14 const double eps = 1e-6;
15 const int INF = 2e9;
16 const LL LNF = 9e18;
17 const int MAXN = 1e4+10;
18 
19 char s[MAXN][110], tmp[110];
20 
21 int cmp(const void *a,const void *b)
22 {
23      return (strcmp((char*)a,(char*)b));
24 }
25 
26 int getmin(char *s, int len) //返回最小表示法的始端
27 {
28     int i = 0, j = 1, k = 0;
29     while(i<len && j<len && k<len)
30     {
31         int t = s[(i+k)%len]-s[(j+k)%len];
32         if (!t) k++;
33         else
34         {
35             if (t>0) i += k+1;
36             else j += k+1;
37             if (i==j) j++;
38             k = 0;
39         }
40     }
41     return i<j?i:j;
42 }
43 
44 int main()
45 {
46     int n;
47     while(scanf("%d", &n)!=EOF)
48     {
49         for(int i = 1; i<=n; i++)
50         {
51             scanf("%s", tmp);
52             int len = strlen(tmp);
53             int k = getmin(tmp, len);
54             for(int j = 0; j<len; j++)
55                 s[i][j] = tmp[(k+j)%len];
56             s[i][len] = 0; //!!
57         }
58         qsort(s+1, n, sizeof(s[1]), cmp);
59 
60         int ans = 0;
61         for(int i = 1; i<=n; i++)
62             if(i==1 || strcmp(s[i], s[i-1]))
63                 ans++;
64 
65         printf("%d
", ans);
66     }
67 }

View Code