hdu 2609 How many

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How many

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3251    Accepted Submission(s): 1446

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

12

 

题不是很难，怪我没想到。借口，借口，还是能力不够。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <set>using namespace std;char str[10005][105];char st[10005][105];int mi_ma(char *s,int len){    int i = 0, j = 1, k = 0;    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 += k + 1;            else                i += k + 1;            if(i == j)                j++;            k = 0;        }    }    return min(i, j);}int main(){    int n;    while(cin >> n)    {        for(int i = 0; i < n; i++)        {            scanf("%s", str[i]);            int len=strlen(str[i]);            int t = mi_ma(str[i],len);            int j=0,k;            for(k=t;k<len;k++)            st[i][j++]=str[i][k];            for(k=0;k<t;k++)            st[i][j++]=str[i][k];            st[i][j]='\0';        }        set<string>ff;        for(int i=0;i<n;i++)        ff.insert(st[i]);        cout<<ff.size()<<endl;    }    return 0;}

参考博客

http://blog.csdn.net/piaocoder/article/details/48447193