[省选联考2022]卡牌

呆滞，卡了一天的常发现 \(umap\) 部分要跑 \(3s\) 多，被演了。

考虑寿司晚宴的熟悉套路：
大于根号的质数因子最多只有一个。

我们考虑按其分类：
并对每一个大质数类，都把内的数字按：
\(a_0 = 1,a_{now} = 2^{cnt} - 1\)
其中\(now\)对应小质数的位置集合。

这样实际上最终的答案即为这些物品的\(or\)背包的答案。

考虑先对每个大质数类的物品都\(FMT\)乘起来。

先求出全局无限制答案。当我们强制大质数\(x\)要被选时：实际上是其对应的背包结果的\(a_0 - 1\),对应到\(FMT\)数组上即全局\(-1\).

那么只要对大质数类的结果求出\(\frac{F_i - 1}{F_i}\)即可。

考虑特判\(43*43\),将\(43\)也归为大质数

值得注意是卡常时可以将\(FMT\)的过程手动正向展开少一个\(log\)（因为初数组只有两个元素）。

这样可以做到复杂度 \(O((2000 + \sum c_i)\times 2^{13} + m2^{13}\times 13 + (C) * 2^{} \times log mod)\)

其中\(C\)是大质数数量。

点击查看代码

//晦暗的宇宙，我们找不到光，看不见尽头，但我们永远都不会被黑色打倒。——Quinn葵因
#include<bits/stdc++.h>
#define ll long long
#define N 4005
#define mod 998244353
#define LIM 43

int vis[N];

using std::vector;

vector<int>P;

int w[N],t[N];
int tcnt;

int F[N][(1ll << 14)];
int G[N][(1ll << 14)];
int S[(1ll << 14)];
int R[(1ll << 14)];
int cnt[N];

inline void FWT_or(int *f,int type){
	int n = (1ll << 13);
	for(int mid = 1;mid < n;mid <<= 1)
	for(int block = mid << 1,j = 0;j < n;j += block)
	for(int i = j;i < j + mid;++i)
	f[i + mid] = (f[i + mid] + f[i] * type + mod) % mod;
}

inline void FWT_and(int *f,int type){
	int n = (1ll << 13);
	for(int mid = 1;mid < n;mid <<= 1)
	for(int block = mid << 1,j = 0;j < n;j += block)
	for(int i = j;i < j + mid;++i)	
	f[i] = (f[i] + f[i + mid] * type + mod) % mod;
}


using std::unordered_map;

int INV[N][(1ll << 13)];

inline ll qpow(ll a,ll b){
	ll res = 1;
	while(b){
		if(b & 1)res = res * a % mod;
		b >>= 1;
		a = a * a % mod;
	}
	return res;
}

inline void print(int x){for(int i = 1;i <= 13;++i)std::cout<<((x >> (i - 1) & 1))<<" ";}

#define M 2000 

inline void init(){
	for(int i = 2;i <= 2000;++i){
		if(!vis[i])P.push_back(i);
		for(int j = 2;j * i <= 2000;++j)
		vis[i * j] = 1;
	} 
	for(int i = 1;i <= M;++i)for(int j = 0;j < (1ll << 13);++j)F[i][j] = 1;
	for(auto v : P)if(v < LIM)w[v] = tcnt++;
	for(int i = 1;i <= M;++i){	
		if(cnt[i]){
		int now = 0;
		int res = i;
		for(auto v : P){
			if(v < LIM && res % v == 0){now = now | (1ll << w[v]);res /= v;}			
			while(v < LIM && res % v == 0)res /= v;
		}
		for(int j = 0;j < (1ll << 13);++j)F[i][j] = 0;
		if(res > 1)t[i] = res;	
		if(res == 43 * 43)t[i] = 43;			
		int r = (qpow(2,cnt[i]) - 1) % mod;
		for(int j = 0;j < (1ll << 13);++j){
			F[i][j] = 1;
			if((j & now) == now)
			F[i][j] = (F[i][j] + r) % mod;
		}
		}
	}
	for(int i = 0;i < (1ll << 13);++i)S[i] = 1;
	for(int i = 1;i <= M;++i)
	for(int j = 0;j < (1ll << 13);++j)
	S[j] = (1ll * S[j] * F[i][j]) % mod;
	for(int i = 1;i <= M;++i)for(int j = 0;j < (1ll << 13);++j)G[i][j] = 1;
	for(int i = 1;i <= M;++i){
		if(t[i]){
			for(int j = 0;j < (1ll << 13);++j)
			G[t[i]][j] = (1ll * G[t[i]][j] * F[i][j]) % mod;
		}	
	}
} 


int n,m;

using std::vector;

vector<int>p,d;

inline int read(){int x;scanf("%d",&x);return x;}

signed main(){
 	freopen("card.in","r",stdin);
 	freopen("card.out","w",stdout);
	scanf("%d",&n);
	for(int i = 1;i <= n;++i){int x;scanf("%d",&x);cnt[x] ++ ;}
	init();scanf("%d",&m);
	while(m -- ){
		int q;scanf("%d",&q);
		p.clear();d.clear();
		for(int i = 1;i <= q;++i)p.push_back(read());
		std::sort(p.begin(),p.end());p.erase(std::unique(p.begin(),p.end()),p.end());
		ll now = 0;
		for(int i = 0;i < (1ll << 13);++i)R[i] = S[i];
		for(auto v : p){
			if(v < LIM)now = (now) | (1ll << w[v]);
			else
			d.push_back(v);
		}
		for(int i = 0;i < (1ll << 13);++i){
			R[i] = S[i];
			for(auto v : d){
			if(!INV[v][i])INV[v][i] = 1ll * (G[v][i] - 1) * qpow(G[v][i],mod - 2) % mod;
			R[i] = 1ll * R[i] * INV[v][i] % mod;
			}
		}
		ll res = 0;
		FWT_or(R,-1);		
		for(int i = 0;i < (1ll << 13);++i)
		if((now & i) == now){
			res = (res + R[i]);
			if(res > mod) res -= mod;
		} 		
		std::cout<<res<<"\n";
	}
}