【Burnside引理】【Polya定理】Arif in Dhaka

Problem L

Arif in Dhaka (First LovePart 2)

Input: standard input

Output: standard output

Time Limit: 2 seconds

 

Our hero Arif is now in Dhaka (Lookat problem 10244– First Love if you want to know more about Arif, but thatinformation is not necessary for this problem. In short, Arif is a brilliantprogrammer working atIBM) and he islooking for his first love. Days pass by but his destiny theory is not workinganymore, which means that he is yet to meet his first love. He then decides toroam around Dhaka on arickshaw(A slow vehicle pulled by human power), runningDFS (by physical movement) and BFS(with his eyes) on every corner of the street and market places to increase hisprobability of reaching his goal. While roaming around Dhaka he discovers an interestingnecklace shop. There he finds someinteresting necklace/braceletconstruction sets. He decides to buy some of them, but his programmer mindstarts looking for other problems. He wants to find out how many differentnecklace/bracelet can be made with acertain construction set. You are requested to help him again. The followingthings are true for anecklace/braceletconstruction set.

 

a)      All necklace/braceletconstruction sets has a frame, which hasNslots to placeN beads.

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

c)      There are t types of beads in a set. Nbeads of each type are there in the box. So the total number of beads istN (t multiplied byN), ofwhich exactlyN can be used at atime.

 

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

 

The figure above shows necklaces for some different values of N (Here,t is always2). Nowlet’s turn out attentions tobracelets.A bracelet is a necklacethat can be turned over (A junior programmer in Bangladesh says that wristwatch is anecklace (Boys!!! Don’tmind :-))). So for abracelet thefollowing two arrangements are equivalent. Similarly, all other oppositeorientation or mirror images are equivalent.

 

 

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

 

Input

The input file contains severallines of input. Each line contains two positive integersN(0<N<51)andt(0<t<11) as described inthe problem statement. Also note that within this input range inputs will besuch that no final result will exceed11digits. Input is terminated by end of file.  

 

Output

For each line of input produceone line of output which contains two round numbersNN andNB separated by asingle space, where NN is the numberof total possible necklaces andNBis the number of total possible bracelets for the corresponding input set. 

 

Sample Input

5 2

5 3

5 4

5 5

 

SampleOutput

8 8

51 39

208 136

629 377

有两种置换：旋转和翻转

讨论旋转：n为奇数或偶数是一样的。

而旋转（把本来的看作旋转次数为0）可以是1个间隔，2个间隔......n-1个间隔。即有n种置换（暂讨论旋转）。

假设某个置换，每个珠子旋转i个间隔，可以证明，这个置换中每个循环包含n/gcd(i,n)个元素，因此，这个置换有gcd(i,n)个循环。

不动点的总数为Σ(i=0,n-1) (k^(gcd(i,n)))。（暂讨论旋转）

讨论翻转：

当n为奇数，只有穿过一个点和另一条边的中点的对称轴。一共有n条，即有n种置换。

对于每种置换，都有(n-1)/2个两个元素的循环和一个1个元素的循环，共(n+1)/2个循环。不动点数为n*k^((n+1)/2)

当n为偶数，有穿过两个点的对称轴和不穿过点的对称轴。

前者：n/2种置换。每种置换n/2+1个循环。不动点数为n/2*t^(n/2+1)

后者：n/2种置换。每种置换n/2个循环。不动点数为n/2*t^(n/2)

根据Burnside引理，求一下平均值即为答案。

#include <cstdio>#include <iostream>using std::cin;using std::cout;typedef long long ll;ll power[27];int gcd(int a,int b){//cout << "gcd(" << a << "," << b << ")=";while (b){int t = b;b = a%b;a = t;}//cout << a << '\n';return a;}int main(){freopen("10294.in","r",stdin);freopen("10294.out","w",stdout);int n,t;while (scanf("%d%d",&n,&t)!=EOF){power[0] = 1;for (int i=1;i<=n;i++)power[i] = power[i-1] * t;ll a = 0;ll b = 0;for (int i=0;i<n;i++){a += power[gcd(i,n)];}if (n & 1){b = n * power[(n+1)>>1];}else{b = n/2*power[n/2+1] + n/2*power[n/2];}cout << a/n << " " << (a+b)/2/n << "\n";}return 0;}