Codeforces Round 723 (Div. 2)
C1
Potions (Easy Version)
Codeforces
1500
原题
← BC2 →
翻译中…

This is the easy version of the problem. The only difference is that in this version $n \leq 2000$. You can make hacks only if both versions of the problem are solved.

翻译中…

There are $n$ potions in a line, with potion $1$ on the far left and potion $n$ on the far right. Each potion increases your health by $a_i$ when drunk. $a_i$ can be negative, meaning that the potion decreases your health.

翻译中…

You start with $0$ health, and you will walk from left to right, from the 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.

翻译中…

What is the largest number of potions you can drink?

翻译中…

Input

翻译中…

The first line contains a single integer $n$ ($1 \leq n \leq 2000$) — the number of potions.

翻译中…

The next line contains $n$ integers $a_1$, $a_2$, ... ,$a_n$ ($-10^9 \leq a_i \leq 10^9$) which represent the change in health after drinking that potion.

翻译中…

Output

翻译中…

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

翻译中…

Note

翻译中…

For the sample, you can drink $5$ potions by taking potions $1$, $3$, $4$, $5$ and $6$. It is not possible to drink all $6$ potions because your health will go negative at some point

翻译中…

样例

Input #1
6
4 -4 1 -3 1 -3
Output #1
5
InputCopy OutputCopy
题目信息
题号
1526C1
时间限制
1 s
内存限制
256 MB
标签
暴力
数据结构
动态规划
贪心

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。