[daily question in summer] first time, second time, deal!

Topic link ：P2637 for the first time , The second time , Clinch a deal ！ - Luogu | New ecology of computer science education (luogu.com.cn)

Title Description

Because the diet exercise of cows gives FJ There's a lot of hay left that can't be handled , So he's going to open an auction to sell his hay .

He has n  A batch of hay （ Each batch is about 100  bundle ）. His clients are m  individual , All the farmers next to him . The first i  A farmer will tell FJ He will do it for FJ Pay for each batch of hay pi  The money . Every farmer wants to buy （ I just want to buy ）FJ A batch of forage .

To make sure farmers don't envy each other , therefore FJ Decided to sell his forage at a fixed price . Each bid is better than FJ Farmers who ask higher prices will buy forage , The rest will be rejected .

Please help FJ Find out the lowest price of a single batch of forage that can make him the most money .

Input format

first line ： Two integers separated by spaces ,n  and m.

Line two to line two m+1  That's ok ： The first i+1  The line contains only one integer ：pi.

Output format

All in one line , Contains two integers separated by spaces ：FJ The lowest price of each batch of forage that can be offered , And the most money he can make .

Examples #1

The sample input #1

5 4
2
8
10
7

Sample output #1

7 21

Tips

FJ Yes 5  Batch forage ,4  A farmer wants to buy . Their bids are ： Each batch of forage is 2,8,10  and 7.

FJ The price should be set to 7, There will be 3  A farmer will pay for forage ,FJ I'll earn it 21  The money .

---

about 100%  The data of ,1<= n <= 1000,1<= m <= 1000,1<= pi <= 1,000,000.

AC code:

#include<iostream>
#include<algorithm>

using namespace std;

int main()
{
	int n,m;
	cin>>n>>m;
	int a[m];
	for(int i=0;i<m;i++)
		cin>>a[i];
	sort(a,a+m);
	int res=0;
	int r;
	for(int i=0;i<m;i++)
	{
	    if(a[i]*(m-i)>res)
	        r=a[i];  
	    if((m-i)<=n)
			res=max(a[i]*(m-i),res);
		else
			res=max(a[i]*n,res);
	}
	cout<<r<<" "<<res;
	
	return 0;
}