[Ybt] [Basic calculation greedy class example 1] Cows drying clothes

Cows drying clothes

Topic link: Cow drying clothes

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Title description

Problem solving ideas

Obviously this is a greed.
We use the dryer for the wettest clothes every time.
But this is $O\left(n^2\right)$The algorithm will time out.
We use the heap to optimize, the time complexity is$O(n\log n)$。

code

#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;

priority_queue<int>q;

int n,a,b,t;

int main()
{
    
    
	cin>>n>>a>>b;
	for(int i=1;i<=n;i++)
	{
    
    
		int t;
		scanf("%d",&t);
		q.push(t);
	}
	while(q.top()>t*a)
	{
    
    
		int s=q.top();
		q.pop();
		q.push(s-b);
		t++;
	}
	cout<<t<<endl;
}