zoukankan      html  css  js  c++  java
  • hdu 1817 Necklace of Beads (polya)

    Necklace of Beads

    Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1049    Accepted Submission(s): 378


    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

    help

    C/C++:

     1 #include <map>
     2 #include <queue>
     3 #include <cmath>
     4 #include <vector>
     5 #include <string>
     6 #include <cstdio>
     7 #include <cstring>
     8 #include <climits>
     9 #include <iostream>
    10 #include <algorithm>
    11 #define INF 0x3f3f3f3f
    12 #define LL long long
    13 using namespace std;
    14 const int MAX = 1e5 + 10;
    15 
    16 __int64 sum, n;
    17 
    18 __int64 gcd(__int64 a, __int64 b)
    19 {
    20     if (b == 0) return a;
    21     return gcd(b, a%b);
    22 }
    23 
    24 __int64 my_pow(__int64 a, __int64 m)
    25 {
    26     __int64 ans = 1;
    27     while (m)
    28     {
    29         if (m & 1) ans *= a;
    30         a *= a;
    31         m >>= 1;
    32     }
    33     return ans;
    34 }
    35 
    36 int main()
    37 {
    38     while (scanf("%I64d", &n), n != -1)
    39     {
    40         sum = 0;
    41         if (n <= 0)
    42         {
    43             printf("0
    ");
    44             continue;
    45         }
    46         for (__int64 i = 1; i <= n; ++ i)
    47         {
    48             __int64 temp = gcd(i, n);
    49             sum += my_pow(3, temp);
    50         }
    51         if (n & 1)
    52             sum += n * my_pow(3, (n + 1) >> 1);
    53         else
    54         {
    55             sum += (n >> 1) * my_pow(3, (n + 2) >> 1);
    56             sum += (n >> 1) * my_pow(3, n >> 1);
    57         }
    58         printf("%I64d
    ", sum / 2 / n);
    59     }
    60     return 0;
    61 }
  • 相关阅读:
    【AS3代码】小游戏打飞机源代码
    【AS3代码】键盘的输入和输出
    PHP多维数组排序
    【AS3代码】一个完整的游戏框架
    【AS3代码】小画板升级版(带重绘回放和清空功能)
    Liunx命令工作总结(1)
    Java NIO基础 我们到底能走多远系列(17)
    ibatis 一对多 解决方案
    图片上传+预览+剪切解决方案我们到底能走多远系列(20)
    Liunx命令工作总结(2)
  • 原文地址:https://www.cnblogs.com/GetcharZp/p/9557245.html
Copyright © 2011-2022 走看看