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.
-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 "); 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 ", ans / 2 / n); 46 } 47 return 0; 48 }