# ALDS 1 1-D - Maximum Profit

## 抄题

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 $R_t$ at a certain time $t$ ($t = 0, 1, 2, ... n-1$), and reports the maximum value of $R_j - R_i$ where $j > i$ .

### Input

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

### Output

Print the maximum value in a line.

### Constraints

$2 \leq n \leq 200,000$

$1 \leq R_t \leq 10^9$

## 翻译

### 限制

$2 \leq n \leq 200,000$

$1 \leq R_t \leq 10^9$

## 题解

• longest-segment（即为 $R_j-R_i$ ）
• lowest-point（即为 $i$，因为有 $i<j$ ）

• longest-segment = $-2\cdot 10^9$ （足够小，以至于任何一个向下/向上的线段都能够覆盖该值）
• lowest-point = $R[0]$ （遍历时用更小的点来覆盖之）

• longest-segment = $\max($ longest-segment, $R[i]$ - lowest-point $)$
• lowest-point = $\min($ lowest-point, $R[i])$

#include <iostream>

using namespace std;

#define MAXN 200005

int main()
{
int n=0;
cin >> n;

int R[MAXN];

for(int i=0; i<n; i++)
cin >> R[i];

auto point_min = R[0];
auto segment_max = -2000000000;

for(auto i=1; i<n; i++)
{
segment_max = max(segment_max, R[i] - point_min);
point_min = min(point_min, R[i]);
}

cout << segment_max << endl;

return 0;
}