P1843奶牛晒衣服11分求调

#include<bits/stdc++.h>
#define int long long 
using namespace std;
int n,a,b,s[512345];
int l=1,r=0,ans=-1;
int p[512345],pos=0;
void clean(int st)
{
	if(st>pos)  return ;
	for(int i=st;i>=1;i--)
	{
		p[i]=0;
	}
	pos=0;
	return ;
}
bool check(int x)  //x：允许给予x分钟晒衣服 
{
	for(int i=n;i>=1;i--)
	{
		if(s[i]-x*a>0)
		{
			p[++pos]=s[i]; 
		}
		else  break;
		//统计x分钟晒不干的 
	}
	int nowt=0;
	for(int i=pos;i>=1;i--)
	{
		int cut=p[i]-nowt*a;
		int nl=0,nr=cut,num=0;
		while(nl<=nr)
		{
			int mid=(nl+nr)/2;  //烘干时间 
			int t=x-nowt-mid;  //继续晒 
			if(mid*(a+b)+t*a>=cut)
			{
				num=mid;
				nl=mid+1;
			}
			else   nr=mid-1;
		}
		nowt+=cut;
		if(nowt>x)  
		{
			clean(i);
			return false;
		}
	}
	return true;
}

signed main()
{
	cin>>n>>a>>b;
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",&s[i]);
		r+=s[i]/a+(s[i]%a!=0);
	}
	sort(s+1,s+n+1);
	//cout<<check(1)<<endl;
	while(l<=r)
	{
		//cout<<l<<" "<<r<<" ";
		int mid=(l+r)/2;
		if(check(mid))
		{
			r=mid-1;
			ans=mid;
		}
		else  l=mid+1;
		//cout<<mid<<endl;
	}
	//单调性：只有衣服晒的干和晒不干的2种选择 
	cout<<ans;
	return 0;
}