CF 342C(Cupboard and Balloons-难得对的贪心)

C. Cupboard and Balloons

time limit per test

memory limit per test

input

output

r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + rfrom the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).

Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius 

. Help Xenia calculate the maximum number of balloons she can put in her cupboard.

You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.

Input

r, h (1 ≤ r, h ≤ 10⁷).

Output

Print a single integer — the maximum number of balloons Xenia can put in the cupboard.

Sample test(s)

input

1 1

output

3

input

1 2

output

5

input

2 1

output

2

贪心题，难得对一次。。。

首先由于球直径等于厚度，本题等价于【平面扔圆】

我的贪心思路：

只有2种情况

1.最上面放一个，然后并排放（被Hack，要考虑3个球间的缝隙）

2.并排放

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int h,r;
int main()
{
//  freopen("Balloons.in","r",stdin);
  while (scanf("%d%d",&r,&h)==2)
  {
    int d=r;
    int ans=1+((h+(int)((2.0-sqrt(3))/2.0*d))/d)*2;
    if (h*2>=d) ans=max(ans,2+(2*h-d)/(2*d)*2);
    cout<<ans<<endl;
    
    
  }  
  
  
  return 0;
}