C2. Potions (Hard Version) Codeforces Round 723 (Div. 2)

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于 2024-03-14 15:55:04 首次发布

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文章讨论了如何在有限的药水中最大化保持非负健康状态，通过使用优先队列策略来确定在给定条件下的最大可喝药水数量。

This is the hard version of the problem. The only difference is that in this version n≤200000�≤200000. You can make hacks only if both versions of the problem are solved.

There are n� potions in a line, with potion 11 on the far left and potion n� on the far right. Each potion will increase your health by ai�� when drunk. ai�� can be negative, meaning that potion will decrease will health.

You start with 00 health and you will walk from left to right, from first potion to the last one. At each potion, you may choose to drink it or ignore it. You must ensure that your health is always non-negative.

What is the largest number of potions you can drink?

Input

The first line contains a single integer n� (1≤n≤2000001≤�≤200000) — the number of potions.

The next line contains n� integers a1�1, a2�2, ... ,an�� (−109≤ai≤109−109≤��≤109) which represent the change in health after drinking that potion.

Output

Output a single integer, the maximum number of potions you can drink without your health becoming negative.

Example

input

Copy

6
4 -4 1 -3 1 -3

output

Copy

Note

For the sample, you can drink 55 potions by taking potions 11, 33, 44, 55 and 66. It is not possible to drink all 66 potions because your health will go negative at some point

题目大致意思：给n瓶药水，照顺序喝，不能喝到负数，问最多喝几瓶？

这题考点是优先队列的使用。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl "\n"

ll n,m;

void solve(){
	cin >> n;
	ll sum=0,c=0,x;
	priority_queue<ll,vector<ll>,greater<ll> >q;
	while(n--){
		cin >> x;
		if(sum+x >= 0){
			c++;
			sum+=x;
			q.push(x);
		}else{
			if(q.empty() || x <= q.top())continue;
			else{
				sum-=q.top();
				q.pop();
				sum+=x;
				q.push(x);
			}
		}
	}
	cout << c << endl;
	return;
}

int main(){
	ll t=1;//cin >> t;
	while(t--)solve();
	return 0;
}