题目

In this problem, you are initially given an empty multiset. You have to process two types of queries:

1. ADD \(x\) — add an element equal to \(2^{x}\) to the multiset;
2. GET \(w\) — say whether it is possible to take the sum of some subset of the current multiset and get a value equal to \(w\).

Input

The first line contains one integer \(m\) (\(1 \le m \le 10^5\)) — the number of queries.

Then \(m\) lines follow, each of which contains two integers \(t_i\), \(v_i\), denoting the \(i\)-th query. If \(t_i = 1\), then the \(i\)-th query is ADD \(v_i\) (\(0 \le v_i \le 29\)). If \(t_i = 2\), then the \(i\)-th query is GET \(v_i\) (\(0 \le v_i \le 10^9\)).

7
1 0
1 1
1 2
1 10
2 4
2 6
2 7

Output

For each GET query, print YES if it is possible to choose a subset with sum equal to \(w\), or NO if it is impossible.

YES
YES
YES

题解

解题思路

由于每次给的数是\(2^n\)，因此对于一个合法的数必然是由不同的\(2^n\)组合出来的，同时由于\(n\)的规模比较小，因此可以暴力枚举。
枚举过程中使用二分加速搜索，找最大小于查找值\(x\)的指数\(n\)，最后判断是否能将\(x\)归零即可

Code

#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false); cin.tie(nullptr);
using namespace std;
typedef long long LL;
const int N=1e5+5;
int v[N];
bool ck(int x){
	for(int i=29;i>=0;i--){
		int l=0,r=v[i],ans=0;
		while(l<=r){
			int mid=l+r>>1;
			if((mid<<i)<=x){
				ans=mid;l=mid+1;
			}
			else r=mid-1;
		}
		x-=ans<<i;
	}
	return !x;
}

signed main(){
	IOS;
	int _;cin>>_;
	while(_--){
		int op,x;cin>>op>>x;
		if(op==1) v[x]++;
		else{
			if(ck(x)){
				cout<<"YES"<<endl;
			}
			else cout<<"NO"<<endl;
		}
	}
	return 0;
}