hdu2609---How many

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

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/*************************************************************************
    > File Name: hdu2609.cpp
    > Author: ALex
    > Mail: [email protected]
    > Created Time: 2015年02月17日 星期二 22时06分29秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

const int N = 10010;
const int M = 110;
char str[N][M];
char s[M];
set <string> st;

void minway (char str[])
{
    int len = strlen (str);
    int i = 0;
    int j = 1;
    int k = 0;
    while (i < len && j < len && k < len)
    {
        int t = str[(i + k) % len] - str[(j + k) % len];
        if (!t)
        {
            ++k;
        }
        else
        {
            if (t > 0)
            {
                i += k + 1;
            }
            else
            {
                j += k + 1;
            }
            k = 0;
            if (i == j)
            {
                ++j;
            }
        }
    }
    int start = min (i, j);
    for (int i = 0; i < len; ++i)
    {
        s[i] = str[(i + start) % len];
    }
    s[len] = ‘\0‘;
    st.insert(s);
}

int main ()
{
    int n;
    while (~scanf("%d", &n))
    {
        st.clear();
        for (int i = 1; i <= n; ++i)
        {
            scanf("%s", str[i]);
            minway(str[i]);
        }
        int size = st.size();
        printf("%d\n", size);
    }
    return 0;
}