2020牛客NOIP赛前集训营-提高组（第二场）- B.包含 （FWT）

题面

题解

这题就是个快速沃尔什变换的模板题，输入ai时，令s[ai]=1，对s[]做一遍DWT_AND(s)（快速沃尔什正变换，按位与），然后直接访问s[x]完事。

#include<map>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define MAXN 10000005
#define LL long long
#define DB double
#define ENDL putchar('\n')
#define lowbit(x) ((-x)&(x))
LL read() {
	LL f = 1,x = 0;char s = getchar();
	while(s < '0' || s > '9') {if(s=='-')f=-f;s = getchar();}
	while(s >= '0' && s <= '9') {x=x*10+(s-'0');s=getchar();}
	return f*x;
}
const int MOD = 998244353;
const int N = 1048576;
int n,m,i,j,s,o,k;
bool a[N+5];
void DWT_AND(bool *s) {
	for(int k = 1;k < N;k <<= 1) {
		for(int i = 0;i < N;i += (k<<1)) { 
			for(int j = i;j < i+k;j ++) {
				s[j] |= s[j+k];
			}
		}
	}
	return ;
}
int main() {
	n = read();m = read();
	for(int i = 1;i <= n;i ++) {
		s = read();
		a[s] = 1;
	}
	DWT_AND(a);
	for(int i = 1;i <= m;i ++) {
		s = read();
		printf(a[s] ? "yes\n":"no\n");
	}
	return 0;
}