How many HDU - 2609 (最小表示法)

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本文介绍了一种算法，用于计算给定多个二进制串（代表项链）时，通过旋转来确定不同项链的数量。该算法首先找到每个字符串的最小表示形式，将其标准化，然后将这些标准化后的字符串排序并计数，最终输出不同的项链总数。

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

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;
}