【NOI Online 3】Sequence

題目解法

我們希望計算出 $Ans_i$ 表示併爲 $i$ 的合法子序列的個數。
計算出 $Ans_i$ 之後，只需要通過線性篩處理歐拉函數後即可輕鬆算出答案。

首先刪去序列中所有的 $0$ ，特殊考慮，令剩餘序列中存在 $A_i$ 個 $i$ 。
將 $A_i$ 看做集合冪級數，乘法看做子集卷積，則應當有
$Ans(x)=1+A(x)+\frac{A^2(x)}{2}+\frac{A^3(x)}{6}+\dots$

也即
$Ans(x)=e^{A(x)}$

則通過 FWT 將子集卷積轉化爲點積後，計算每一位對應多項式的指數函數即可。由於這裏，我們可以接受 $O(N^2)$ 計算多項式的指數函數 $g(x)=e^{f(x)}$ ，可以考慮利用如下恆等式進行計算：
$g'(x)=f'(x)g(x)$

時間複雜度 $O(N+VLog^2V)$ 。

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 262144 + 5;
const int MAXM = 20;
const int P    = 1e9 + 7;
typedef long long ll;
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;
}
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 tot, prime[MAXN], f[MAXN], phi[MAXN];
void sieve(int n) {
	phi[1] = 1;
	for (int i = 2; i <= n; i++) {
		if (f[i] == 0) {
			prime[++tot] = f[i] = i;
			phi[i] = i - 1;
		}
		for (int j = 1; j <= tot && prime[j] <= f[i]; j++) {
			int tmp = prime[j] * i;
			if (tmp > n) break;
			f[tmp] = prime[j];
			if (f[i] == prime[j]) phi[tmp] = phi[i] * prime[j];
			else phi[tmp] = phi[i] * (prime[j] - 1);
		}
	}
}
void FWTOR(int *a, int N) {
	for (int len = 2; len <= N; len <<= 1)
	for (int s = 0; s < N; s += len)
	for (int i = s, j = s + len / 2; j < s + len; i++, j++)
		a[j] = (a[i] + a[j]) % P;
}
void UFWTOR(int *a, int N) {
	for (int len = 2; len <= N; len <<= 1)
	for (int s = 0; s < N; s += len)
	for (int i = s, j = s + len / 2; j < s + len; i++, j++)
		a[j] = (a[j] - a[i] + P) % P;
}
int n, inv[MAXN], cnt[MAXN], bits[MAXN], dp[MAXM][MAXN];
int main() {
	freopen("sequence.in", "r", stdin);
	freopen("sequence.out", "w", stdout);
	sieve(1 << 18), read(n);
	int u = (1 << 18) - 1;
	for (int i = 1; i <= u; i++)
		bits[i] = bits[i / 2] + i % 2;
	for (int i = 1; i <= n; i++) {
		int x; read(x);
		dp[bits[x]][x]++;
	}
	inv[0] = 1;
	for (int i = 1; i <= 18; i++) {
		FWTOR(dp[i], u + 1);
		inv[i] = power(i, P - 2);
	}
	for (int i = 1; i <= u; i++) {
		static int f[MAXM], d[MAXM], g[MAXM];
		for (int j = 1; j <= 18; j++) {
			f[j] = dp[j][i];
			d[j - 1] = 1ll * f[j] * j % P;
		}
		g[0] = 1;
		for (int j = 1; j <= 18; j++) {
			int tmp = 0;
			for (int k = 0; k <= j - 1; k++)
				update(tmp, 1ll * d[k] * g[j - 1 - k] % P);
			g[j] = 1ll * tmp * inv[j] % P;
		}
		for (int j = 1; j <= 18; j++)
			dp[j][i] = g[j];
	}
	for (int i = 1; i <= 18; i++)
		UFWTOR(dp[i], u + 1);
	int ans = 1;
	for (int i = 1; i <= u; i++)
		update(ans, 1ll * dp[bits[i]][i] * phi[i + 1] % P);
	cout << 1ll * ans * power(2, dp[0][0]) % P << endl;
	return 0;
}