How many HDU - 2609 (最小表示法)

How many

HDU - 2609

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

思路：把所有字符串用最小表示法表示出来，然后全部放在vector中，然后从小到大排个序

前n个有多少个不同字符就有多少个不同串，如果前n个字符全一样说明这n个都可以表示成1个

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
int n;
char s[101],ss[101];
int getmin(){
    int i = 0,j = 1, k = 0;
    int len = strlen(s);
    while(i < len && j < len && k < len){
        int t = s[(j+k)%len]-s[(i+k)%len];
        if(t == 0) k++;
        else{
            if(t > 0) j = j+k+1;
            else i = i+k+1;
            if(i == j) j++;
            k = 0;
        }
    }
    int ans = min(i,j);
    return ans;
}
int main(){
    int i,j,k;
    while(~scanf("%d",&n)){
        vector<string>sn;
        for(i = 0; i < n; i++){
            scanf("%s",s);
            int len = strlen(s);
            k = getmin();
            for(j = 0; j < len; j++){
                ss[j] = s[(k+j)%len];
            }
            string st(ss);
            sn.push_back(st);
        }
        sort(sn.begin(),sn.end());
        int ans = 1;
        for(i = 1; i < n; i++){
            if(sn[i] != sn[i-1])
                ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}