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Codeforces Round #199 (Div. 二) C. Cupboard and Balloons

2013-09-13 
Codeforces Round #199 (Div. 2) C. Cupboard and BalloonsC. Cupboard and Balloonstime limit per test2

Codeforces Round #199 (Div. 2) C. Cupboard and Balloons
C. Cupboard and Balloonstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard 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?+?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).

Codeforces Round #199 (Div. 二) C. Cupboard and Balloons

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 Codeforces Round #199 (Div. 二) C. Cupboard and Balloons. 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.

Sample test(s)input
1 1
output
3
input
1 2
output
5
input
2 1
output
2
几何题,有点坑,有个地方没想清楚,一直wrong,我们可以先把长方形的那部分尽可能的多放,然后,我们可以,找到剩下的空间,刚好放两个球,和刚好放三个球的情况,这样,我们就可以得出公式了!
#include <iostream>#include <stdio.h>#include <string.h>#include <math.h>using namespace std;int main(){    double h,r;double temp=sqrt(3.0)/2.0;    while(scanf("%lf%lf",&r,&h)!=EOF){        int ans=2*(int)(h/r);        double t=h-ans/2.0*1.0*r;        //printf("%d %.6f %.6f\n",ans,t,temp*r);        if(t<r/2.0)ans++;        else if(t<temp*r)ans+=2;        else ans+=3;        printf("%d\n",ans);    }    return 0;}


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