【省内训练2019-07-01】Match

【思路要点】

• 考虑方差的期望的公式，记 $F_0=1,F_i=(2i-1)F_{i-1}$ ， $Met$ 表示可能的权值序列的集合，则有
  $$\begin{gathered} Ans=\frac{1}{|Met|}\sum _{a_{\in }Met}\frac{{\sum }_{i=1}^N(a_i-\frac{\sum a_i}{N}{)}^2}{N} \\ Ans=\frac{1}{N\times F_N}\sum _{a\in Met}\sum _{i=1}^N(a_i^2-2\frac{a_i(\sum a_i{)}^2}{N}+\frac{(\sum a_i{)}^2}{N^2}) \\ Ans=\frac{1}{N\times F_N}\sum _{a\in Met}\sum _{i=1}^N{a}_i^2-\frac{1}{N^2\times F_N}\sum _{a\in Met}(\sum a_i{)}^2 \end{gathered}$$
• 因此答案由两部分组成，对于第一部分，可以考虑交换求和顺序，计算
  $$\begin{gathered} \frac{1}{N\times F_N}\sum _{i=1}^N{a}_i^2\sum _{a\in Met,a_i\in a}1 \\ =\frac{1}{N\times F_N}\sum _{i=1}^{2N}\sum _{j=i+1}^{2N}(j-i{)}^2\times F_{N-1} \\ =\frac{1}{N\times (2N-1)}\sum _{i=1}^{2N}\sum _{j=i+1}^{2N}(j-i{)}^2 \end{gathered}$$
• 可以发现求和符号内的部分是一个多项式，可以用插值或 $Berlekamp-Massey$ 算法 解决。
• 对于第二部分，考虑 $(\sum a_i{)}^2$ 的组合意义，即在 $\sum a_i$ 个元素中依次选择两个的方案数，考虑枚举这两个元素，有
  $$\begin{gathered} \frac{1}{N^2\times F_N}\sum _{a\in Met}(\sum a_i{)}^2 \\ =\frac{1}{N^2\times F_N}\sum _{i=1}^{2N}\sum _{j=i+1}^{2N}((j-i{)}^2\times F_{N-1}+\sum _{k=1,k̸=i,j}^{2N}\sum _{l=k+1,l̸=i,j}^{2N}(j-i)(l-k)\times F_{N-2}) \\ =\frac{1}{N^2\times (2N-1)\times (2N-3)}\sum _{i=1}^{2N}\sum _{j=i+1}^{2N}((j-i{)}^2\times (2N-1)+\sum _{k=1,k̸=i,j}^{2N}\sum _{l=k+1,l̸=i,j}^{2N}(j-i)(l-k)) \end{gathered}$$
• 可以发现求和符号内的部分同样是一个多项式，可以用插值或 $Berlekamp-Massey$ 算法 解决。
• 时间复杂度 $O(M^5+M^{2}LogN)$ ，其中 $M$ 为递推阶数，有 $M=7$ 。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 5005;
const int MAXLOG = 62;
const int P = 1e9 + 7;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
struct LinearSequence {
	vector <int> a[MAXN];
	int cnt, delta[MAXN], fail[MAXN];
	int k, h[MAXN], now[MAXLOG][MAXN];
	int power(int x, int y) {
		if (y == 0) return 1;
		int tmp = power(x, y / 2);
		if (y % 2 == 0) return 1ll * tmp * tmp % P;
		else return 1ll * tmp * tmp % P * x % P;
	}
	void times(int *res, int *x, int *y) {
		static int tmp[MAXN];
		memset(tmp, 0, sizeof(tmp));
		for (int i = 0; i <= k - 1; i++)
		for (int j = 0; j <= k - 1; j++)
			tmp[i + j] = (tmp[i + j] + 1ll * x[i] * y[j]) % P;
		for (int i = 2 * k - 2; i >= k; i--) {
			int val = tmp[i]; tmp[i] = 0;
			for (unsigned j = 0; j < a[cnt].size(); j++)
				tmp[i - j - 1] = (tmp[i - j - 1] + 1ll * val * a[cnt][j]) % P;
		}
		memcpy(res, tmp, sizeof(tmp));
	}
	void init(int n, int *val) {
		for (int i = 0; i <= cnt; i++)
			a[i].clear();
		cnt = 0;
		for (int i = 1; i <= n; i++) {
			delta[i] = val[i];
			for (unsigned j = 0; j < a[cnt].size(); j++)
				delta[i] = (delta[i] - 1ll * a[cnt][j] * val[i - j - 1] % P + P) % P;
			if (delta[i] == 0) continue;
			fail[cnt] = i;
			if (cnt == 0) {
				a[++cnt].resize(i);
				continue;
			}
			int mul = 1ll * delta[i] * power(delta[fail[cnt - 1]], P - 2) % P;
			a[cnt + 1].resize(i - fail[cnt - 1] - 1);
			a[cnt + 1].push_back(mul);
			for (unsigned j = 0; j < a[cnt - 1].size(); j++)
				a[cnt + 1].push_back(1ll * a[cnt - 1][j] * (P - mul) % P);
			if (a[cnt + 1].size() < a[cnt].size()) a[cnt + 1].resize(a[cnt].size());
			for (unsigned j = 0; j < a[cnt].size(); j++)
				a[cnt + 1][j] = (a[cnt + 1][j] + a[cnt][j]) % P;
			cnt++;
		}
		//cerr << a[cnt].size() << endl;
		memset(now, 0, sizeof(now));
		k = a[cnt].size();
		if (k == 1) now[0][0] = a[cnt][0];
		else now[0][1] = 1;
		for (int i = 1; i <= 2 * k; i++)
			h[i] = val[i];
		for (int p = 1; p < MAXLOG; p++)
			times(now[p], now[p - 1], now[p - 1]);
	}
	int query(long long n) {
		if (n <= k) return h[n]; n -= k;
		static int res[MAXN];
		memset(res, 0, sizeof(res));
		res[0] = 1;
		for (int p = 0; p < MAXLOG; p++) {
			long long tmp = 1ll << p;
			if (n & tmp) times(res, res, now[p]);
		}
		int ans = 0;
		for (int i = 0; i <= k - 1; i++)
			ans = (ans + 1ll * res[i] * h[i + k]) % P;
		return ans;
	}
} Sqr, Ind;
int f[MAXN], sqr[MAXN], ind[MAXN];
void update(int &x, int y) {
	x += y;
	if (x >= P) x -= P;
}
int power(int x, int y) {
	if (y == 0) return 1;
	int tmp = power(x, y / 2);
	if (y % 2 == 0) return 1ll * tmp * tmp % P;
	else return 1ll * tmp * tmp % P * x % P;
}
int getsqr(int n) {
	int ans = 0;
	for (int i = 1; i <= 2 * n; i++)
	for (int j = i + 1; j <= 2 * n; j++) {
		update(ans, 1ll * (j - i) * (j - i) * (2 * (n - 1) - 1) % P);
		for (int k = 1; k <= 2 * n; k++)
			if (k != i && k != j) {
				for (int l = k + 1; l <= 2 * n; l++)
					if (l != i && l != j)
						update(ans, 1ll * (j - i) * (l - k) % P);
			}
	}
	return ans;
}
int getind(int n) {
	int ans = 0;
	for (int i = 1; i <= 2 * n; i++)
	for (int j = i + 1; j <= 2 * n; j++)
		update(ans, 1ll * (j - i) * (j - i) % P);
	return ans;
}
int main() {
	freopen("match.in", "r", stdin);
	freopen("match.out", "w", stdout);
	int n, m = 20; read(n);
	for (int i = 1; i <= m; i++) {
		sqr[i] = getsqr(i);
		ind[i] = getind(i);
	}
	Sqr.init(m, sqr);
	Ind.init(m, ind);
	int ts = Sqr.query(n), ti = Ind.query(n);
	int ans = 1ll * ti * power(2 * n - 1, P - 2) % P * power(n, P - 2) % P;
	update(ans, P - 1ll * ts * power(2 * n - 1, P - 2) % P * power(2 * n - 3, P - 2) % P * power(n, P - 2) % P * power(n, P - 2) % P);
	writeln(ans);
	return 0;
}