codeforces 865 C. Buy Low Sell High_codeforces865c-优快云博客

本文链接：https://blog.youkuaiyun.com/Egqawkq/article/details/78150452

本文介绍了一个股票交易策略问题，目标是在给定一系列每日股价的情况下，通过最优买卖操作实现最大利润。文章提供了一段C++代码示例，展示了如何通过特定算法计算出最大可能的收益。

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You can perfectly predict the price of a certain stock for the next N days. You would like to profit on this knowledge, but only want to transact one share of stock per day. That is, each day you will either buy one share, sell one share, or do nothing. Initially you own zero shares, and you cannot sell shares when you don't own any. At the end of the N days you would like to again own zero shares, but want to have as much money as possible.

Input

Input begins with an integer N (2 ≤ N ≤ 3·105), the number of days.

Following this is a line with exactly N integers p1, p2, ..., pN (1 ≤ pi ≤ 106). The price of one share of stock on the i-th day is given by pi.

Output

Print the maximum amount of money you can end up with at the end of N days.

 
  Examples 
 
    input 
   
9
10 5 4 7 9 12 6 2 10
 
    output 
   
20
 
    input 
   
20
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4
 
    output 
   
41

Note

In the first example, buy a share at 5, buy another at 4, sell one at 9 and another at 12. Then buy at 2 and sell at 10. The total profit is - 5 - 4 + 9 + 12 - 2 + 10 = 20.

题意：

已知每一天的某种股票的价钱，每天只能处理一股，希望最后手里没有股票，可以出卖股票和买进股票。

问你最后最多可以获得多少钱。

代码：

#include <bits/stdc++.h>

using namespace std;

int main()
{
	int n;
	cin>>n;
	long long ans=0;
	int u;
	priority_queue<int>q;
	for(int i=1;i<=n;i++)
	{
		cin>>u;
		q.push(-u);
		q.push(-u);
		ans+=u+q.top();
		cout<<ans<<endl;
		q.pop();
	}
	cout<<ans<<endl;
	return 0;
}