Discussion Detail - XAZXOI

问答
何意味？？？
qsq LV 5 @ 2025-10-29 20:36:38

#include <bits/stdc++.h>
#define ll long long
using namespace std;
unsigned int n,m,a[100001],b,ans,cnt;
int main(){
	ios::sync_with_stdio();
	cin.tie();
	cin>>n>>m;
	for(unsigned int i=1;i<=n;i++){
		cin>>a[i];
	}
	for(unsigned int i=1;i<=m;i++){
		cnt=0;
		cin>>b;
		for(unsigned int j=1;j<=n;j++){
			unsigned int c=b&a[j];
			if(c!=a[j]){
				break;
			}
			if(c==a[j]){
				cnt++;
			}
		}
		if(cnt==n) ans++;
	}
	cout<<ans;
	return 0;
}

qsq LV 5 @ 2025-12-20 11:36:20

行

Csvoner SU @ 2025-12-20 11:32:56

数据范围是 $1 \le n,m \le 10^5$ ， $O(n^2)$ 超时了，得换个思路

#include <bits/stdc++.h>
using namespace std;
int n, m;
int a[100000 + 5];
int b[100000 + 5];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= m; i++)
        cin >> b[i];
    int now = 0;
    for (int i = 1; i <= n; i++)
        for (int d = 0; d <= 29; d++)
            if (a[i] & (1 << d))
                now = now | (1 << d);
    int ans = 0;
    for (int i = 1; i <= m; i++)
        if ((b[i] & now) == now)
            ans++;
    cout << ans;
    return 0;
}