How many
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5865 Accepted Submission(s): 2751
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
板子题：套一个字符串的最小表示法 的板子，构造最小字符串，然后用set去重，最后得到的set.size()即要求的答案。
#include <iostream> #include<cstdio> #include<cstring> #include<string> #include<set> using namespace std; const int maxn = 1e4 + 100; int minString(string s) { int i = 0, j = 1, k = 0; int len = s.size(); while(i < len && j < len && k < len) { if(s[(i + k) % len] == s[(j + k) % len]) k++; else { if(s[(i + k) % len] > s[(j + k) % len]) i = i + k + 1; else j = j + k + 1; if(i == j) j++; k = 0; } } return i < j ? i : j; } string s[maxn]; string s2[maxn]; int visited[maxn]; int main() { int n; while(~scanf("%d", &n)) { for(int i = 0; i < n; i++) { cin>>s[i]; } int len = s[0].size(); set<string>myset; for(int i = 0; i < n; i++){ int index = minString(s[i]); string str = s[i].substr(index, len - index) + s[i].substr(0,index); // cout<<str<<endl; myset.insert(str); } cout<<myset.size()<<endl; } return 0; } /* ababab ba */