题目意思：都是给定m个颜色，然后去个长度为N的项链染色，无其他限制。
然后反转，顺时针旋转算是一种方案啊
思路:比较基础的polya题目，算是模板题吧。。
1）旋转 将项链旋转i格，循环个数为gcd（N，i）
2）反转：
A.n为奇数时，共有 n个循环节数为（n+1）/2的循环群
B、n为偶数，可以根据一条边及一个点对称，有n/2 个循环节数为（n+2）/2的循环群
还可以根据两个点对称，有n/2个循环节数为n/2的循环群
code：
1 #include <iostream> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cstdio> 5 #include <cmath> 6 #include <algorithm> 7 8 typedef __int64 LL; 9 using namespace std; 10 __int64 n, m, gcd[30][30]; 11 12 void init(){ 13 for (int i = 1;i <= 24; ++i) 14 for (int j = 1; j <= 24; ++ 15 j) 16 for (int k = 1; k <= min(i,j); ++k) 17 if (j % k == 0 && i % k ==0) gcd[i][j] = k; 18 19 } 20 21 void solve(){ 22 LL sum = 0; 23 if (n == 0){ 24 printf("%d\n", 0); 25 return ; 26 } 27 for (int i = 1; i <= n; ++i) 28 sum += (LL)( pow( m * 1.0, gcd[i][n] * 1.0)); 29 if (n&1){ 30 sum += (LL)( n* pow(m * 1.0, (n + 1)/ 2.0)); 31 } else { 32 sum += (LL)((n / 2) * pow(m *1.0, (n + 2) / 2 * 1.0)); 33 sum += (LL)((n / 2) * pow(m *1.0, n / 2 * 1.0)); 34 } 35 printf("%lld\n", sum / (2 * n)); 36 37 } 38 39 int main(){ 40 freopen("poj1286.in", "r", stdin); 41 freopen("poj1286.out","w", stdout); 42 init(); 43 m = 3; 44 while (scanf("%lld", &n) != EOF && n != -1) 45 solve(); 46 }
code2：
1 #include <iostream> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cstdio> 5 #include <cmath> 6 #include <algorithm> 7 8 typedef __int64 LL; 9 using namespace std; 10 __int64 n, m, gcd[34][34]; 11 12 void init(){ 13 for (int i = 1;i <= 33; ++i) 14 for (int j = 1; j <= 33; ++ 15 j) 16 for (int k = 1; k <= min(i,j); ++k) 17 if (j % k == 0 && i % k ==0) gcd[i][j] = k; 18 19 } 20 21 void solve(){ 22 LL sum = 0; 23 if (n == 0){ 24 printf("%d\n", 0); 25 return ; 26 } 27 for (int i = 1; i <= n; ++i) 28 sum += (LL)( pow( m * 1.0, gcd[i][n] * 1.0)); 29 if (n&1){ 30 sum += (LL)( n* pow(m * 1.0, (n + 1)/ 2.0)); 31 } else { 32 sum += (LL)((n / 2) * pow(m *1.0, (n + 2) / 2 * 1.0)); 33 sum += (LL)((n / 2) * pow(m *1.0, n / 2 * 1.0)); 34 } 35 printf("%I64d\n", sum / (2 * n)); 36 37 } 38 39 int main(){ 40 freopen("poj2409.in", "r", stdin); 41 freopen("poj2409.out","w", stdout); 42 init(); 43 while (scanf("%I64d%I64d", &m, &n) != EOF && n != 0 && m != 0) 44 solve(); 45 }