hdu2609

How many

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2768    Accepted Submission(s): 1191

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

4011011001001001141010010110000001

 

Sample Output

1
2
题意分析：
有n个环形字符串，一个环形字符串移动会形成不同的字符串，我们把它们看作同一串字符串，求有多少个不同的字符串？
用最小表示法将一个环形串的最小字典序找出来，然后让这个环形串按照这个顺序来组成一个新的串，其他串都这样处理，然后去重
AC代码：
#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char s[10005][105];struct node{    char ch[105];}t[10005];int cmp(const node a,const node b){    if(strcmp(a.ch,b.ch)<0)    return 1;    else    return 0;}int work(int m,char str[]){    int i,j,l;    i=0; j=1;    while(i<m && j<m)    {        for(l=0;l<m;l++)             if(str[(i+l)%m]!=str[(j+l)%m]) break;        if(l>m) break;        if(str[(i+l)%m] > str[(j+l)%m])            i=i+l+1;        else            j=j+l+1;        if(i==j) j=i+1;    }    if(i<j) return i;    return j;}int main(){    int n;    while(scanf("%d",&n)>0)    {        for(int i=1;i<=n;i++)        scanf("%s",s[i]);        for(int i=1;i<=n;i++)        {            int len=strlen(s[i]);            int cnt=work(len,s[i]);            strcpy(t[i].ch,s[i]+cnt);            s[i][cnt]='\0';            strcpy(t[i].ch+len-cnt,s[i]);        }        sort(t+1,t+1+n,cmp);        int sum=1;        for(int i=2;i<=n;i++)        {            if(strcmp(t[i-1].ch,t[i].ch)!=0)            sum++;        }        printf("%d\n",sum);    }    return 0;}