利用 bitset 进行位运算优化即可，时间复杂度为$O(\dfrac{n}{\omega})$。 代码中判断子集时运用了$A\subseteq B\Leftrightarrow (A\cap B )=A$的技巧。

#include <iostream>
#include <bitset>

using namespace std;

bitset<64> a, b, c;
int m, n;

void print(bitset<64> bs) {
	for (int i = 0; i < 64; ++i)
		if (bs[i])
			cout << i << ' ';
	cout << endl;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr), cout.tie(nullptr);

	cin >> m;
	for (int i = 1; i <= m; ++i){
		int x;
		cin >> x;
		a[x] = true;
	}
	cin >> n;
	for (int i = 1; i <= n; ++i) {
		int x;
		cin >> x;
		b[x] = true;
	}

	cout << m << endl;
	c = a & b;
	print(c);
	print(a | b);
	print(~a);
	cout << (a == b) << endl;
	cout << (a == c) << endl;
	cout << a[0] << endl;

	return 0;
}