【BZOJ 1815】【SHOI 2006】color 有色图

http://www.lydsy.com/JudgeOnline/problem.php?id=1815
这道题好难啊，组合数学什么根本不会啊qwq
题解详见08年的Pólya计数论文。
主要思想是只枚举具有代表性的点的置换，算出这些点的置换造成的边的置换的保持不变的着色数（边的置换的保持不变的着色数我想了一天啊_(:з」∠)_），最后再乘上与这种具有代表性的点的置换同类的点的置换总数就可以了。
WA了好几次，中间一个地方忘取模了qwq

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 60;

int n, m, p;

int GCD(int a, int b) {return b ? GCD(b, a % b) : a;}

int ipow(int a, int b) {
	int ret = 1, w = a;
	while (b) {
		if (b & 1) ret = 1ll * ret * w % p;
		w = 1ll * w * w % p;
		b >>= 1;
	}
	return ret;
}

int gcd[N][N], powm[N * N], jc[N], njc[N], ni[N];

int L[N], ans = 0;

void solve(int tot) {
	int res = jc[n], cnt = 1, ret = 0;
	for (int i = 1; i <= tot; ++i)
		res = 1ll * res * ni[L[i]] % p;
	for (int i = tot - 1; i >= 0; --i) {
		if (L[i] != L[i + 1]) {
			res = 1ll * res * njc[cnt] % p;
			cnt = 1;
			continue;
		}
		++cnt;
	}
	
	for (int i = 1; i <= tot; ++i)
		ret += (L[i] >> 1);
	for (int i = 1; i <= tot; ++i)
		for (int j = 1; j < i; ++j)
			ret += gcd[L[i]][L[j]];
	(ans += 1ll * res * powm[ret] % p) %= p;
}

void dfs(int tmp, int last, int rest) {
	if (rest == 0) {
		solve(tmp - 1);
		return;
	}
	for (int i = last; i <= rest; ++i) {
		L[tmp] = i;
		dfs(tmp + 1, i, rest - i);
	}
}

int main() {
	scanf("%d%d%d", &n, &m, &p);
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= i; ++j)
			gcd[i][j] = GCD(i, j);
	
	powm[0] = 1;
	for (int i = 1, top = n * n; i <= top; ++i)
		powm[i] = 1ll * powm[i - 1] * m % p;
	
	jc[0] = njc[0] = ni[0] = 1;
	for (int i = 1; i <= n; ++i) {
		jc[i] = 1ll * jc[i - 1] * i % p;
		ni[i] = ipow(i, p - 2);
		njc[i] = 1ll * njc[i - 1] * ni[i] % p;
	}
	
	dfs(1, 1, n);
	printf("%d\n", 1ll * ans * njc[n] % p);
	return 0;
}