Equity interest [non DP]

Title Description ：

 Title Description 

 Xiao Ke's father is interested in studying the stock market recently , Xiao Ke thinks naively , If you can predict the daily price of stocks , It's easy to figure out when to buy and when to sell to maximize profits . Suppose that the time sequence of a stock is given n Sky price , Excuse me at n What is the maximum profit that can be made by buying and selling the stock only once in a day ？
 Input format 

 first line , An integer n, Days ,1≤n≤10^5

 Next line ,n It's an integer , The first i It's an integer , It means the first one i Day's share price . All integers are 1 To 10000 Positive integers within 
 Output format 
 Input and output sample Columns 
 sample input 1：

6
7 1 5 3 6 4

 sample output 1：

5

 sample input 2：

5
7 6 4 3 1

 sample output 2：

0

 explain 

【 Sample explanation 1】 In the  2  God （ Stock price  = 1） Buy when , In the  5  God （ Stock price  = 6） Sell when , Maximum profit  = 6-1 = 5 .

    Note that profit cannot be  7-1 = 6,  Because you can only buy first and then sell 

【 Sample explanation 2】 There is no need to trade in this case , In this way, the profit can be maximized 0

It's a little bit easier , about a[i], stay 1 ~ i - 1 In the process of finding a[1] To a[i - 1] Minimum of , With the a[i] Subtracting the , take max, But pay attention to the special judgment < 0 The situation of ：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
#include <sstream>
#include <vector>
#include <map>

using namespace std;

const int N = 100010 ;
int a[N] , n ;
int minn , maxn , c ;

int main()
{
    


    cin >> n ;
    if (n < 2)
    {
    
        cout << 0 ;
        return 0 ;
    }
    for (int i = 1; i <= n; i++) cin >> a[i] ;
    minn = a[1] , maxn = a[2] - a[1] ;
    for (int i = 3; i <= n; i++)
    {
    
        if (a[i - 1] < minn)minn = a[i - 1] ;
        c = a[i] - minn ;
        maxn = max (maxn , c) ;
    }
    cout << max (0 , maxn) ;


    return 0;
}