题解 CF914G 【Sum the Fibonacci】

经典板子题？

令 \(i = s_a\&s_b, j = s_c, k = s_d \oplus s_e\)

\(\sum_{p}\sum_{i\&j\&k = 2^p} f_i \times f_j \times f_k \times (\sum_{s_a|s_b = i, s_a \&s_b = 0}1) \times (\sum_{s_c = j}1)\times (\sum_{s_d \oplus s_e = k}1)\)

令 \(A_i = (\sum_{s_a|s_b = i, s_a \&s_b = 0}1)\)。

\(B_j = (\sum_{s_c = j}1)\)。

\(C_k = (\sum_{s_d \oplus s_e = k}1)\)。

\(A_i\) 可以通过子集卷积来做，复杂度 \(N \log^2 N\)。

\(B_j\) 显然。

\(C_k\) 可以直接 FWTXOR 来做，复杂度 \(N \log N\)。

最后 FWTAND 卷一下三个就行了。

#include <bits/stdc++.h>
const int BL = 2 << 20;
struct streambufcin {
	std::streambuf *in;
	streambufcin() {
		std::ios::sync_with_stdio(false);
		std::cin.tie(NULL);
		in = std::cin.rdbuf();
		bg = ed = NULL;
#ifdef LOCAL
		freopen("in.in", "r", stdin);
#endif
	}
	char buf[BL], *bg, *ed, ch;
	char gc() { return bg == ed ? ed = buf + in -> sgetn(bg = buf, BL), (bg == ed ? -1 : *bg++) : *bg++; }
	template<class T>
	streambufcin &operator >> (T &x) {
		bool neg = false;
		for (ch = gc(); !isdigit(ch); ch = gc());
		if (ch == '-')
			neg = true;
		for (x = 0; isdigit(ch); ch = gc())
			x = x * 10 + ch - 48;
		if (neg)
			x = -x;
		return *this;
	}
	streambufcin &operator >> (char &c) {
		for (c = gc(); isspace(c); c = gc());
		return *this;
	}
} cin;

struct streambufcout {
	std::streambuf *out = std::cout.rdbuf();
	char buf[BL], *optr = buf;
	streambufcout() { out = std::cout.rdbuf(); }
	~streambufcout() { flush(); }
	void flush() { out -> sputn(buf, optr - buf); optr = buf; }
	void pc(char c) {
		if (optr == buf + BL)
			flush();
		*optr++ = c;
	}
	template<class T>
	streambufcout &operator << (T x) {
		static int st[233], top;
		if (x < 0) {
			x = -x;
			pc('-');
		}
		do {
			st[++top] = x % 10;
			x /= 10;
		} while (x);
		while (top)
			pc(st[top--] ^ 48);
		return *this;
	}
	streambufcout &operator << (char c) { pc(c); return *this; }
} cout;

int n;
const int N = 262144;
const int mod = 1e9 + 7;
const int inv2 = mod + 1 >> 1;
int s[N], fib[N], cnt[N];
int A[N], B[N], C[N], F[19][N], h[N], answer[N];

int add(int x, int y) {
	if (x + y >= mod)
		return x + y - mod;
	else
		return x + y;
}
int dec(int x, int y) {
	if (x - y < 0)
		return x - y + mod;
	else
		return x - y;
}

void mul(int &x, int y) { x = 1ll * x * y % mod; }
void _ad(int &x, int y) { x += y; if (x >= mod) x -= mod; }
void _de(int &x, int y) { x -= y; if (x < 0) x += mod; }

void FWTOR(int *f, int inv) {
	for (int len = 1; len < N; len <<= 1) {
		for (int i = 0; i < N; i += len << 1) {
			for (int j = 0; j < len; j++) {
				if (inv > 0)
					_ad(f[i + j + len], f[i + j]);
				else
					_de(f[i + j + len], f[i + j]);
			}
		}
	}
}

void FWTAND(int *f, int inv) {
	for (int len = 1; len < N; len <<= 1) {
		for (int i = 0; i < N; i += len << 1) {
			for (int j = 0; j < len; j++) {
				if (inv > 0)
					_ad(f[i + j], f[i + j + len]);
				else
					_de(f[i + j], f[i + j + len]);
			}
		}
	}
}

void FWTXOR(int *f, int inv) {
	for (int len = 1; len < N; len <<= 1) {
		for (int i = 0; i < N; i += len << 1) {
			for (int j = 0; j < len; j++) {
				int x = f[i + j], y = f[i + j + len];
				f[i + j] = add(x, y), f[i + j + len] = dec(x, y);
				if (inv < 0) 
					mul(f[i + j], inv2), mul(f[i + j + len], inv2);
			}
		}
	}
}

int main() {
	for (int i = 1; i < 262144; i++)
		cnt[i] = cnt[i >> 1] + (i & 1);
	fib[0] = 0; fib[1] = 1;
	for (int i = 2; i < 262144; i++)
		fib[i] = add(fib[i - 1], fib[i - 2]);
	cin >> n;
	for (int i = 1; i <= n; i++) { int x; cin >> x; A[x]++; B[x]++; C[x]++; }
	for (int i = 0; i < 262144; i++) 
		F[cnt[i]][i] = A[i], A[i] = 0;
	for (int i = 0; i < 19; i++)
		FWTOR(F[i], 1);
	for (int i = 0; i < 19; i++) {
		for (int j = 0; j < 262144; j++)
			h[j] = 0;
		for (int j = 0; j <= i; j++)
			for (int k = 0; k < 262144; k++)
				_ad(h[k], 1ll * F[j][k] * F[i - j][k] % mod);
		FWTOR(h, -1);
		for (int j = 0; j < 262144; j++)
			if (cnt[j] == i)
				_ad(A[j], h[j]);
	}
	FWTXOR(C, 1);
	for (int i = 0; i < 262144; i++)
		mul(C[i], C[i]);
	FWTXOR(C, -1);
	for (int i = 0; i < 262144; i++)
		mul(A[i], fib[i]), mul(B[i], fib[i]), mul(C[i], fib[i]);
	FWTAND(A, 1), FWTAND(B, 1), FWTAND(C, 1);
	for (int i = 0; i < 262144; i++)
		answer[i] = 1ll * A[i] * B[i] % mod * C[i] % mod;
	FWTAND(answer, -1);
	int Answer = 0;
	for (int i = 1; i < 262144; i <<= 1)
		_ad(Answer, answer[i]);
	cout << Answer << '\n';
	return 0;
}