HDU 1817Necklace of Beads(置换+Polya计数)

Posted 2020-07-07 zhaop

Necklace of Beads

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1817

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

题目大意：

n个珠子串成一个圆，用三种颜色去涂色。问一共有多少种不同的涂色方法。

不同的涂色方法被定义为：如果这种涂色情况翻转，旋转不与其他情况相同就为不同。

 

解题思路：

Polya定理模版题。

对于顺时针长度为i的旋转，为pow(3, gcd(n,i)；

对于翻转，当为奇数时，有：n*pow(3, n/2+1); 

   当为偶数时，有：n/2*pow(3.0,n/2)+n/2*pow(3.0,n/2+1);

一共有2*n种情况，最后要除以2*n

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <cmath>
 6 using namespace std;
 7 typedef long long LL;
 8 int n;
 9 LL Pow(LL a, LL b) // 手写long long
10 {
11     LL res = 1;
12     for (int i = 0; i < b; i++)
13         res *= a;
14     return res;
15 }
16 int gcd(int a, int b)
17 {
18     if (a == 0)
19         return b;
20     return gcd(b % a, a);
21 }
22 int main()
23 {
24     while (scanf("%d", &n) != EOF && n != -1)
25     {
26         if (n == 0)
27         {
28             printf("0\\n");
29             continue;
30         }
31         LL ans = 0;
32         for (int i = 1; i <= n; i++)
33             ans += Pow(3, gcd(i, n));
34 
35         if (n & 1)
36         {
37             ans += n * Pow(3, (n + 1) / 2);
38         }
39         else
40         {
41             ans += n / 2 * Pow(3.0, n / 2 + 1);
42             ans += n / 2 * Pow(3.0, n / 2);
43             //ans += n / 2 * (pow(3.0, n / 2 + 1) + pow(3.0, n / 2));
44         }
45         printf("%I64d\\n", ans / 2 / n);
46     }
47     return 0;
48 }

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