Necklace of Beads
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 7874   Accepted: 3290

Description

Beads of red, blue or green colors are connected together into a circular necklace of n beads ( n < 24 ). 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
/*Polya原理*/
#include <cstdio>
#include <iostream>
using namespace std;
long long aaaa(long long a,long long b)
{
long long res=,tmp=a;
while(b)
{
if (b&)
res*=tmp;
tmp*=tmp;
b>>=;
}
return res;
}
int gcd(int a,int b)
{
return b==?a:gcd(b,a%b);
}
int main()
{
int n;
while ( ~scanf("%d", &n) && n != -)
{
long long a = , b = ;
for (int i=;i<n;i++)
{
a+=aaaa(,gcd(i,n));
}
if (n&)
b=(long long)n*aaaa(,(n+)/);
else
b=(long long)n/*(aaaa(,n/+)+aaaa(,n/));
if (n==)
printf("0\n");
else
printf("%lld\n",(a+b)//n);
}
return ;
}

POJ 1286 Necklace of Beads(项链的珠子)的更多相关文章

  1. poj 1286 Necklace of Beads &amp; poj 2409 Let it Bead(初涉polya定理)

    http://poj.org/problem?id=1286 题意:有红.绿.蓝三种颜色的n个珠子.要把它们构成一个项链,问有多少种不同的方法.旋转和翻转后同样的属于同一种方法. polya计数. 搜 ...

  2. POJ 1286 Necklace of Beads(Polya简单应用)

    Necklace of Beads 大意:3种颜色的珠子,n个串在一起,旋转变换跟反转变换假设同样就算是同一种,问会有多少种不同的组合. 思路:正规学Polya的第一道题,在楠神的带领下,理解的还算挺 ...

  3. 数学计数原理(Pólya):POJ 1286 Necklace of Beads

    Necklace of Beads Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7763   Accepted: 3247 ...

  4. poj 1286 Necklace of Beads poj 2409 Let it Bead HDU 3923 Invoker <组合数学>

    链接:http://poj.org/problem?id=1286 http://poj.org/problem?id=2409 #include <cstdio> #include &l ...

  5. POJ 1286 Necklace of Beads(Polya原理)

    Description Beads of red, blue or green colors are connected together into a circular necklace of n ...

  6. poj 2409 Let it Bead && poj 1286 Necklace of Beads(Polya定理)

    题目:http://poj.org/problem?id=2409 题意:用k种不同的颜色给长度为n的项链染色 网上大神的题解: 1.旋转置换:一个有n个旋转置换,依次为旋转0,1,2,```n-1. ...

  7. POJ 1286 Necklace of Beads(Polya定理)

    点我看题目 题意 :给你3个颜色的n个珠子,能组成多少不同形式的项链. 思路 :这个题分类就是polya定理,这个定理看起来真的是很麻烦啊T_T.......看了有个人写的不错: Polya定理: ( ...

  8. poj 1286 Necklace of Beads (polya(旋转+翻转)+模板)

      Description Beads of red, blue or green colors are connected together into a circular necklace of ...

  9. poj 1286 Necklace of Beads【polya定理+burnside引理】

    和poj 2409差不多,就是k变成3了,详见 还有不一样的地方是记得特判n==0的情况不然会RE #include<iostream> #include<cstdio> us ...

随机推荐

  1. 一个oracle存储过程

    打开plsql,在packages文件夹里新建存储过程 在sql窗口中运行如下语句 create or replace package SY_USER_PKG1 is TYPE MYCURSOR IS ...

  2. ios如何判断键盘是否已经显示

    ios如何判断键盘是否已经显示   在群里看到有人问:ios如何判断键盘已经显示在界面上. 其实这个解决很简单: 写一个单例来管理键盘的状态. 这个单例在初始化方法init种监听2个事件,分别是 UI ...

  3. 自制单片机之四……LCD1602的驱动

    LCD1602已很普遍了,具体介绍我就不多说了,市面上字符液晶绝大多数是基于HD44780液晶芯片的,控制原理是完全相同的,因此HD44780写的控制程序可以很方便地应用于市面上大部分的字符型液晶.字 ...

  4. Linq to object 技巧、用法集锦

    一.一个字符串,一个字符串数组.判断字符串数组里的元素出现在字符串中的有几个. class Program { static void Main(string[] args) { string str ...

  5. Qt编程之在QGraphics scene中使用图片

    http://stackoverflow.com/questions/5960074/qimage-in-a-qgraphics-scene http://stackoverflow.com/ques ...

  6. 文本编辑BOM标记(Byte Order Mark)

    微软的自带记事本程序notepad.exe会给UTF-8编码的文件头加入三个隐藏的字节(即BOM).这是一种很愚蠢的做法.就是为了让编辑器不去猜测文件本身是ASCII码还是UTF-8. 什么是BOM ...

  7. logstash 发送zabbix告警

    <pre name="code" class="html">[elk@dr-mysql01 test]$ cat t1.conf input { s ...

  8. About Adultism and why things ar the way they are

    About - Adultism About Adultism and why things ar the way they are In this page we will try to clari ...

  9. paip.索引优化---sql distict—order by 法

    paip.索引优化---sql distict—order by 法 作者Attilax ,  EMAIL:1466519819@qq.com 来源:attilax的专栏 地址:http://blog ...

  10. jquery.tochart.js

    var _jq, _hc; var jqsrc = "http://code.jquery.com/jquery-1.7.min.js"; var hcsrc = "ht ...