HDU 2609 How many

How many

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1005    Accepted Submission(s): 392

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

题目大意：一串由0和1组成的环形珠子，问共有多少串不同的珠子。

解题方法:先用最小表示法求出每个字符串最小字典序的字符串，然后统计不同字符串的数量。

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
using namespace std;

char str[10001][105];

int cmp(const void *p1, const void *p2)
{
    return strcmp((char *)p1, (char *)p2);
}

int Getmin(char str[])
{
    int nLen = strlen(str);
    int i = 0, j = 1, k;
    while (i < nLen && j < nLen)
    {
        for (k = 0; k < nLen; k++)
        {
            if (str[(i + k) % nLen] != str[(j + k) % nLen])
            {
                break;
            }
        }
        if (str[(i + k) % nLen] > str[(j + k) % nLen])
        {
            i += k + 1;
        }
        else
        {
            j += k + 1;
        }
        if (i == j)
        {
            j++;
        }
    }
    return min(i, j);
}

int main()
{
    int n;
    char strtemp[105];
    int ans = 1;
    while(scanf("%d", &n) != EOF)
    {
        ans = 1;
        for (int i = 0; i < n; i++)
        {
            scanf("%s", strtemp);
            int nLen = strlen(strtemp);
            int index = Getmin(strtemp);
            int nCount = 0;
            for (int j = index; j < nLen; j++)
            {
                str[i][nCount++] = strtemp[j];
            }
            for (int j = 0; j < index; j++)
            {
                str[i][nCount++] = strtemp[j];
            }
            str[i][nCount] = '\0';
            nCount = 0;
        }
        qsort(str, n, sizeof(str[0]), cmp);
        for (int i = 0; i < n - 1; i++)
        {
            if (strcmp(str[i], str[i + 1]) != 0)
            {
                ans++;
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}