【POJ2888】Magic Bracelet

\(\text{Solution}\)

考虑眼切，你就\(AC\)了
对于求本质不同的方案数，显然可以用\(Burnside\)引理。
接着\(DP\)求方案数，对于头尾的限制，可以在最后在加一个和头部一样的珠子。用矩阵优化即可。

\(\text{Code}\)

#include<cstdio>
#include<cstring>
using namespace std;
const int P = 9973,N = 1e6 + 5;
int n,ans,m,T,k,vis[N],p[N],tot,cnt;

struct nd{int s[10][10];}a,O,I,b;
struct nid{int z,s;}c[N];
void init()
{
	for (int i = 0; i < 10; i++) I.s[i][i] = 1;
	for (int i = 2; i <= 1000000; i++) {
		if (!vis[i]) p[++tot] = i;
		for (int j = 1; j <= tot && p[j] * i <= 1000000; j++)
		{
			vis[p[j] * i] = 1;
			if (i % p[j] == 0) break;
		}
	}
}
nd Mul(nd x,nd y)
{
	nd z = O;
	for (int k = 0; k < m; k++)
		for (int i = 0; i < m; i++) if (x.s[i][k])
			for (int j = 0; j < m; j++)
				z.s[i][j] = (z.s[i][j] + x.s[i][k] * y.s[k][j] % P) % P;
	return z;
}
int fpow(int x,long long y)
{
	long long res = 1;
	for (; x; x >>= 1,y = y * y % P)
		if (x & 1) res = res * y % P;	
	return res;
}
void dfs(int x,int sum,int phi)
{
	if (x > cnt)
	{
		b = I; int id = sum; nd y = a;
		for (; id; id >>= 1,y = Mul(y,y))
			if (id & 1) b = Mul(b,y);
		long long res = 0;
		for (int i = 0; i < m; i++) res = (res + b.s[i][i]) % P;
		res = (long long)res * phi % P;
		ans = (ans + res) % P;
		return;
	}
	int g = 1;
	for (int i = 0; i < c[x].s; i++) dfs(x + 1,sum * g,phi / g),g = g * c[x].z;
	dfs(x + 1,sum * g,phi / (g / c[x].z) / (c[x].z - 1));
}
int main()
{
	init(),scanf("%d",&T);
	for (; T; T--)
	{
		scanf("%d%d%d",&n,&m,&k),ans = 0,cnt = 0; 
		int g = n; memset(c,0,sizeof c);
		for (int i = 0; i < m; i++)
			for (int j = 0; j < m; j++) a.s[i][j] = 1;
		for (int i = 1; i <= tot && p[i] <= g; i++)
		{
			if (g % p[i] == 0) c[++cnt].z = p[i];
			while (g % p[i] == 0) c[cnt].s++,g /= p[i];
		}
		if (g != 1) c[++cnt] = nid{g,1};
		for (int i = 1,q,p; i <= k; i++)
			scanf("%d%d",&q,&p),q--,p--,a.s[q][p] = a.s[p][q] = 0;
		int Phi = n;
		for (int i = 1; i <= cnt; i++) Phi = Phi / c[i].z * (c[i].z - 1); 
		dfs(1,1,Phi),ans = ans * fpow(P - 2,n) % P;
		printf("%d\n",ans);
	}
 }