UVA 11255 - Necklace （等价置换）

Problem C - Necklace

Once upon a time, three girls - Winnie, Grace and Bonnie - owned a large number of pearls. However, each of them only had a single color of pearls. Winnie had white
pearls, Grace had grey pearls and Bonnie had black pearls. One day, after a long discussion, they decided to make necklaces using the pearls. They are interested in knowing how many patterns can be formed using a certain number of pearls of each color, and
have asked you to solve this problem for them.

Note that rotating or flipping over a necklace cannot produce a different kind of necklace. i.e. The following figure shows three equivalent necklaces.



The following figure shows all possible necklaces formed by 3 white pearls, 2 grey pearls and 1 black pearl.



Input

The input begins with an integer N ( ≤ 2500)
which indicates the number of test cases followed. Each of the following test cases consists of three non-negative integers a, b, c,
where 3 ≤a +b + c≤ 40.

Output

For each test case, print out the number of different necklaces that formed by a white
pearls, b grey pearls and c black
pearls in a single line.

Sample input

2

3 2 1

2 2 2

Sample output

6

11

Problem setter: Cho

Special thanks: Michael (for making up the story)

Source: Tsinghua-HKUST Programming Contest 2007

题意：有一个项链要染上三种颜色，每种颜色的数量分别是a，b，c， 其中通过旋转和翻转得到的情况算为一种，问一共有多少种染色情况。

思路：可以直接用Burnside定理求。 训练指南上面似乎也有类似的讲解。。。所以就不多讲了。注意置换里面有(1,2,3,...n)这个置换，别漏了。

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <queue>
#include <set>
using namespace std;
#define eps 1e-8
#define rep(i,a,b) for(int i = (a); i < (b); ++i)
#define rrep(i,b,a) for(int i = (b); i >= (a); --i)
#define clr(a,x) memset(a,(x),sizeof(a))
#define mp make_pair
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ll long long
#define ld long double
const int maxn = 42;
ll C[maxn][maxn];
int dp[maxn][maxn][maxn];
int a,b,c,n;

void pre_init()
{
rep(i,0,maxn) {
C[i][0] = 1;
rep(j,1,i+1) C[i][j] = C[i-1][j] + C[i-1][j-1];
}
}

void solve()
{
int fenmu = n * 2;
ll fenzi = 0;
//1 2 3 .... n
fenzi += C
[a] * C[n-a][b];
//旋转
rep(d,1,n) {
int len = n / __gcd(d,n);
int num = n / len;
if (a % len || b % len || c % len) continue;
fenzi += C[num][a / len] * C[num - a / len][b / len];
}
//翻转
if (n % 2) {
int num = (n - 1) >> 1;
int cnt = a % 2 + b % 2 + c % 2;
if (cnt == 1) {
fenzi += C[num][a/2] * C[num-a/2][b/2] * n;
}
} else {
int num = n >> 1;
int cnt = a % 2 + b % 2 + c % 2;
if (cnt == 0) fenzi += C[num][a/2] * C[num-a/2][b/2] * (n / 2);
--num;
if (cnt == 2) fenzi += 2 * C[num][a / 2] * C[num - a/2][b/2] * (n / 2);
else if (cnt == 0) {
if (a >= 2) fenzi += C[num][(a - 2)/2] * C[num - (a - 2) / 2][b / 2] * (n / 2);
if (b >= 2) fenzi += C[num][(b - 2) / 2] * C[num - (b-2) / 2][a / 2] * (n / 2);
if (c >= 2) fenzi += C[num][(c - 2) / 2] * C[num - (c - 2) / 2][a / 2] * (n / 2);
}
}
printf("%lld\n",fenzi / fenmu);
}

int main()
{
#ifdef ACM
freopen("in.txt","r",stdin);
//        freopen("out.txt","w",stdout);
#endif // ACM
pre_init();
int T; cin >> T;
while (T--) {
scanf("%d%d%d",&a,&b,&c);
n = a + b + c;
solve();
}
}