Maximum Profit

Maximum Profit

You can obtain profits from foreign exchange margin transactions. For example, if you buy 1000 dollar at a rate of 100 yen per dollar, and sell them at a rate of 108 yen per dollar, you can obtain (108 - 100) × 1000 = 8000 yen.

Write a program which reads values of a currency RtRt at a certain time tt (t=0,1,2,...n−1t=0,1,2,...n−1), and reports the maximum value of Rj−RiRj−Ri where j>ij>i .

Input

The first line contains an integer nn. In the following nn lines, RtRt (t=0,1,2,...n−1t=0,1,2,...n−1) are given in order.

Output

Print the maximum value in a line.

Constraints

• 2≤n≤200,000
• 1≤Rt≤109

Sample Input 1

6
5
3
1
3
4
3

Sample Output 1

3

Sample Input 2

3
4
3
2

Sample Output 2

-1

一开始想到两重循环
for(int j = 1; j < n; ++ j)
　　for(int i = 0; i < j; ++ i)
　　　　maxn = max(maxn , a[j] - a[i]);

但是 n≤200,000 若采用两重循环(O(n^2))会超时, 所以在i自增的过程中, 将现阶段a[i]的最小值(记为minn)保存下来, 此时只需要O(1)便可求出i时刻的最大利益
for(int i = 1; i < n; ++ i)
{
　　maxn = maxn(a[i] - minn);　　// minn初始化为a[0]
　　minn = min(minn, a[i]);
}
注意 maxn的初始值不能是-1, 因为如果序列单调递减, 则最大值有可能小于-1(-1反而比该序列的maxn还大)
方便起见, maxn初值为a[1] - a[0]

边扫描边记录

#include <iostream>
#include <algorithm>
using namespace std;
const int MAX = 200010;
int a[MAX];
int main()
{
	int n;
	cin >> n;
	for(int i = 0; i < n; ++ i)
	{
		cin >> a[i];
	}

	int maxn = a[1] - a[0], minn = a[0];

	for(int i = 1; i < n; ++ i)
	{
		maxn = max(maxn, a[i] - minn);
		minn = min(minn, a[i]);
	}
	cout << maxn << endl;
	return 0;
}