UVa10294 - Arif in Dhaka (First Love Part 2)

UVa10294 - Arif in Dhaka (First Love Part 2)

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 t·N (t multiplied by N), of which exactly N can be used at a time.
Fig. 1: 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 a bracelet 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.

題目大意：n個珠子的項鍊，染t種顏色
輸出只能旋轉的方案數和還能翻轉的方案數。

解：這道題是polya的裸體

定理前提：
（1）定義置換f是一個對C的置換
（2）定義置換羣G
（3）定義羣作用“。”，並誘導出羣等價關係“~”
（4）由（1）（2）（3），我們得出找G中置換的方法：使旋轉和翻轉意義下本質相同的着色在置換羣G的作用下保持等價關係
（5）Burnside引理：P = 1/|G|∑|S(g)|
其中S（g）爲C中對置換g保持不變的着色集合。

polya計數定理
設G是集合X上的一個置換羣，X中每個元素可以被染成k種顏色，則不等價的着色數爲：P = 1/|G|∑k^nc(g)
其中nc(g)爲置換g中循環節的個數。

對這道題，旋轉的循環節個數爲：對位置1—>s的珠子，這個置換的循環節個數爲gcd（n，s-1）
翻轉的循環節個數爲：
（1）n & 1 = 1： n / 2 + 1
（2）n & 1 = 0： [(n-2) / 2 + 2] 和 n / 2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;

LL p[100];

LL gcd(LL a, LL b) {
    return b == 0 ? a : gcd(b, a % b);
}

int main()
{
    int n, t;
    while (~scanf("%d%d", &n, &t)) {
      p[0] = 1;
      for (int i = 1; i <= n; i++)
        p[i] = p[i-1] * t;
      LL sum = 0, G = n;
      for (int i = 1; i <= n; i++) 
        sum += p[gcd(n, i-1)];
      printf("%lld ", sum / G);
      if (n & 1) 
        sum += n * p[(n >> 1) + 1], G += n;
      else
        sum += n * (p[(n-2)/2+2] + p[n/2])/2, G += n;
      printf("%lld\n", sum / G);
    }
}