UVa 10294(polya 翻转与旋转)

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);
     }
 }