hdu1817 Necklace of Beads（Polya 计数原理）

最新推荐文章于 2025-02-01 23:34:35 发布

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最新推荐文章于 2025-02-01 23:34:35 发布

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分类专栏：组合数学 Polya 计数原理

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Polya 计数原理

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本文探讨了一道经典的组合数学问题：如何计算不同颜色珠子组成圆形项链的不重复排列数量，忽略旋转和翻转产生的相同形态。文章通过具体示例输入输出说明问题背景，并提供了一段使用 C++ 实现的代码，该算法利用了旋转群和反射操作来减少重复计数。

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Necklace of Beads

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1005 Accepted Submission(s): 360

Problem Description

Beads of red, blue or green colors are connected together into a circular necklace of n beads ( n < 40 ). If the repetitions that are produced by rotation around the center of the circular necklace or reflection to the axis of symmetry are all neglected, how many different forms of the necklace are there?

Input

The input has several lines, and each line contains the input data n.
-1 denotes the end of the input file.

Output

The output should contain the output data: Number of different forms, in each line correspondent to the input data.

Sample Input

4
5
-1

Sample Output

21
39

#include <bits/stdc++.h>
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000007
#define ll long long
#define i128 __int128
#define N 1100000


std::ostream& operator<<(std::ostream& os, __int128 T)
{
    if (T<0)
        os<<"-";
    if (T>=10 )
        os<<T/10;
    if (T<=-10)
        os<< -(T/10);
    return os<<( (int) (T%10) >0 ? (int) (T%10) : -(int) (T%10) ) ;
}
i128 sum, m = 3;
i128 quick_pow(i128 a, i128 b)
{
    i128 ans = 1;
    while(b) {
        if(b&1)
            ans = ans * a;
        a *= a;
        b >>= 1;
    }
    return ans;
}
int main()
{
    int n;
    while(cin>> n, n!=-1) {
        sum  = 0;
        if(n==0) {
            cout << 0 << endl;
            continue;
        }
        for(int i = 1; i <= n; ++i) {//旋转
            sum += quick_pow(m, __gcd(i, n));
        }
        if(n & 1) {//翻转
            sum += n*quick_pow(m, (n+1)/2);
        } else {
            sum += n/2*quick_pow(m, (n)/2);
            sum += n/2*quick_pow(m, (n)/2+1);
        }
        sum = sum/2/n;
        cout << sum << endl;
    }
    return 0;
}