题解 | #买卖股票的最好时机(一)# Python3

import sys

# dp[i]为第i天卖出可获得的最高收益
# dp[i] = prices[i] - pre_min (之前天里面的最低值)
# 最后输出max[dp]

n = int(input())

if n<=1:
    print(0)

else:
    dp = [0] * (n+1)
    dp[0] = 10**5 + 5
    pre_min = dp[0]
    prices = list(map(int,input().strip().split(' ')))
    for i in range(1,n+1):
        dp[i] = prices[i-1] - pre_min
        pre_min = min(pre_min, prices[i-1])
    max_profit = max(dp[1:])
    print(max_profit if max_profit >0 else 0)