C. Cupboard and Balloons
time limit per test
memory limit per test
input
output
A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius 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 + r
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
The single line contains two integers r, h (1 ≤ r, h ≤ 107).
Output
Print a single integer — the maximum number of balloons Xenia can put in the cupboard.
Examples
Input
1 1
Output
3
Input
1 2
Output
5
Input
2 1
Output
#include<cstdio>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<list>
typedef long long ll;
typedef unsigned long long llu;
const int maxn = 100000 + 10;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int main(){
int r,h;
while(scanf("%d %d",&r,&h) == 2){
int ans = h/r*2;
h%=r;
if (2*h < r)ans += 1;
else if (2*h >= sqrt(3)*r)ans += 3;
else ans += 2;
printf("%d\n",ans);
}
return 0;
}