#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=2e5+10;
int r[N];
signed main()
{
	int n,q;scanf("%lld%lld",&n,&q);
	for(int i=1;i<=n;i++)scanf("%lld",&r[i]);
	sort(r+1,r+n+1);
	for(int i=1;i<=n;i++)r[i]+=r[i-1];
	while(q--)
	{
		int x;scanf("%lld",&x);
		int id=lower_bound(r+1,r+n+1,x)-r-1;
		if(r[id+1]==x)id++;
		printf("%lld\n",id);
	}
	return 0;
}

注意暴力O(nq)会超时

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll n,q,a[200010],x,s[200010];
int main()
{
	scanf("%lld%lld",&n,&q);
	memset(s,0,sizeof s);
	for(ll i=1;i<=n;i++)scanf("%lld",&a[i]);
	sort(a+1,a+n+1);
	for(ll i=1;i<=n;i++)s[i]=s[i-1]+a[i];//前缀和 
	while(q--)
	{
		scanf("%lld",&x);
		if(x>=s[n]){printf("%lld\n",n);continue;}
		if(x<s[1]){puts("0");continue;}//临界状态 
		ll id=upper_bound(s+1,s+n+1,x)-s-1;
		//直接在前缀和中查找能拉多少
		printf("%lld\n",id);
	}
	return 0;
}