Arif in Dhaka(First Love Part 2)

Input: standard input

Output: standard output

Time Limit: 2 seconds

Our hero Arif is now in Dhaka (Look at problem 10244 – First Love if you want to know more about Arif, but that information is not necessary for this problem. In short, Arif is a brilliant programmer working at IBM) and he is looking for his first love. Days pass by but his destiny theory is not working anymore, which means that he is yet to meet his first love. He then decides to roam around Dhaka on a rickshaw (A slow vehicle pulled by human power), running DFS (by physical movement) and BFS (with his eyes) on every corner of the street and market places to increase his probability of reaching his goal. While roaming around Dhaka he discovers an interesting necklace shop. There he finds some interesting necklace/bracelet construction sets. He decides to buy some of them, but his programmer mind starts looking for other problems. He wants to find out how many different necklace/bracelet can be made with a certain construction set. You are requested to help him again. The following things are true for a necklace/bracelet construction set.

a)      All necklace/bracelet construction sets has a frame, which has N slots to place N beads.

b)      All the slots must be filled to make a necklace/bracelet.

c)      There are t types of beads in a set. N beads of each type are there in the box. So the total number of beads is tN (t multiplied by N), of which exactly N can be used at a time.

Fig: Different types of necklace for t=2 and different value of N

The figure above shows necklaces for some different values of N (Here, t is always 2). Now let’s turn out attentions to bracelets. A bracelet is a necklace that can be turned over (A junior programmer in Bangladesh says that wrist watch is a necklace (Boys!!! Don’t mind :-))). So for abracelet the following two arrangements are equivalent. Similarly, all other opposite orientation or mirror images are equivalent.

 

So, given the description of a necklace/bracelet construction set you will have to determine how many different necklace and bracelet can be formed with made with that set

 

Input

The input file contains several lines of input. Each line contains  two positive integers N(0<N<51) and t(0<t<11) as described in the problem statement. Also note that within this input range inputs will be such that no final result will exceed 11 digits. Input is terminated by end of file.

Output

For each line of input produce one line of output which contains two round numbers NN and NB separated by a single space, where NN is the number of total possible necklaces and NB is the number of total possible bracelets for the corresponding input set.

Sample Input

5 2

5 3

5 4

5 5

Sample Output

8 8

51 39

208 136

629 377

旋转:每移动i颗珠子,循环节为gcd(i,n),故ans1=sum(t^gcd(i,n)),t为颜色种数;

翻转:奇:有n条对称轴,共(n+1)/2个循环节!  ans2=n*t^((n+1)/2)

   偶:1)穿过珠子有n/2条对称轴,有(n/2+1)个循环节!

     2)不穿过珠子也有n/2条对称轴,有n/2个循环节!  ans2=n/2*(t^(n/2+1)+t^(n/2))

根据polya定理:项链ans=ans/n;手镯ans=(ans1+ans2)/2n;

 #include<stdio.h>
#include<string.h>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll; int gcd(int a,int b)
{
return b?gcd(b,a%b):a;
}
ll ans1,ans2,pow[];
int main()
{
//printf("%d***\n",gcd(6,9));
int i,n,t;
while(~scanf("%d%d",&n,&t))
{
pow[]=;
ans1=;
for(i=;i<=n;i++)pow[i]=pow[i-]*t;
for(i=;i<n;i++)ans1+=pow[gcd(i,n)];//旋转
if(n&)ans2=n*pow[(n+)/];
else ans2=n/*(pow[n/+]+pow[n/]);//翻转
printf("%lld %lld\n",ans1/n,(ans1+ans2)//n);
}
}

UVa 10294(polya 翻转与旋转)的更多相关文章

  1. 项链与手镯Uva 10294——Polya定理

    题意 项链和手镯都是由若干珠子串成的环形首饰,区别在于手环可以翻转,但项链不可以. 输入整数 $n$ 和 $t$,输出用 $t$ 中颜色 $n$ 颗珠子能制作成的项链和手镯的个数.($1\leq n ...

  2. Uva 10294 Polya

    #include <bits/stdc++.h> using namespace std; typedef long long LL; int gcd(int a,int b) { ? a ...

  3. 【uva 10294】 Arif in Dhaka (First Love Part 2) (置换,burnside引理|polya定理)

    题目来源:UVa 10294 Arif in Dhaka (First Love Part 2) 题意:n颗珠子t种颜色 求有多少种项链和手镯 项链不可以翻转 手镯可以翻转 [分析] 要开始学置换了. ...

  4. Arif in Dhaka (First Love Part 2) UVA - 10294(Polya定理)

    这题和POJ-1286一样 题意: 给出t种颜色的n颗珠子 (每种颜色的珠子个数无限制,但总数必须是n), 求能制作出项链和手镯的个数 注意手镯可以翻转和旋转  而 项练只能旋转 解析: 注意Poly ...

  5. UVA 10294 项链与手镯 (置换)

    Burnside引理:对于一个置换\(f\), 若一个着色方案\(s\)经过置换后不变,称\(s\)为\(f\)的不动点.将\(f\)的不动点数目记为\(C(f)\), 则可以证明等价类数目为\(C( ...

  6. UVa 10294 Arif in Dhaka (First Love Part 2)(置换)

    题目链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=35397 [思路] Polya定理. 旋转:循环节为gcd(i,n) ...

  7. AS3.0 位图翻转、旋转

    /* * * *-------------------------* * | *** 位图翻转.旋转 *** | * *-------------------------* * * 作 者:fengz ...

  8. 项目实战:Qt+Ffmpeg+OpenCV相机程序(打开摄像头、支持多种摄像头、分辨率调整、翻转、旋转、亮度调整、拍照、录像、回放图片、回放录像)

    若该文为原创文章,未经允许不得转载原博主博客地址:https://blog.csdn.net/qq21497936原博主博客导航:https://blog.csdn.net/qq21497936/ar ...

  9. UVa 10294 Arif in Dhaka (First Love Part 2) (Polya定理)

    题意:给定 n 和 m 表示要制作一个项链和手镯,项链和手镯的区别就是手镯旋转和翻转都是相同的,而项链旋转都是相同的,而翻转是不同的,问你使用 n 个珠子和 m 种颜色可以制作多少种项链和手镯. 析: ...

随机推荐

  1. python之random随机函数

    random.random()用于生成 用于生成一个指定范围内的随机符点数,两个参数其中一个是上限,一个是下限.如果a > b,则生成随机数 1 n: a <= n <= b.如果 ...

  2. python连接字符串的几种方法--转子(香草拿铁的园子)

    一. str1+str2 string类型 ‘+’号连接 >>> str1="Good" >>> str2="Luck" & ...

  3. HyperV - glossary

    Root Partition - sometimes called partition. Manages machine-level functions such as device drivers, ...

  4. CentOS7 如何挂载网络设备

    CentOS 自动挂载网络设备的方法 手动挂载: [root@mysql ~]# mount -o username=USER,password=PASSWORD //192.168.10.212/z ...

  5. Git-Runoob:Git 工作区、暂存区和版本库

    ylbtech-Git-Runoob:Git 工作区.暂存区和版本库 1.返回顶部 1. Git 工作区.暂存区和版本库 基本概念 我们先来理解下Git 工作区.暂存区和版本库概念 工作区:就是你在电 ...

  6. Quartz安装包中的15个example

    Welcome======= Welcome to the Quartz Examples directory. This directory contains 15 examples that sh ...

  7. python-接口开发flask模块(一)工具类准备

    我们常常听说测试http接口.测试java接口,测试socket接口等等:那么python这么强大的语言当然也可以用来开发接口了. flask模块介绍: python中用来开发接口的模块:flask, ...

  8. WPF 带有提示文本的透明文本框

    <TextBox Text="{Binding SearchInfo, UpdateSourceTrigger=PropertyChanged}" Grid.Row=&quo ...

  9. Windows 2008任务计划执行bat脚本失败返回0x1

    测试环境: C:\>systeminfo | findstr /c:"OS Name"OS Name:                   Microsoft Windows ...

  10. 程序的内存分布 - 以 Linux 为例,基于 C 语言分析

    这里以 Linux 为例,用 C 语言进行演示. 内存模型 - 内存空间名称 内容 读写操作 分配时机 高地址 kernel 内核空间 命令行参数.环境变量等 不可读写 程序运行时 - stack 栈 ...