code forces 630 Q. Pyramids

最新推荐文章于 2017-10-08 12:12:45 发布

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#code forces 630 Q #Pyramids #几何数学

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本文介绍了一个计算三种不同形状金字塔（三角形、四边形和五边形底面）体积的程序。通过输入各金字塔的边长，利用数学公式计算总体积，并确保误差不超过10^-9。代码使用C语言实现。

IT City administration has no rest because of the fame of the Pyramids in Egypt. There is a project of construction of pyramid complex near the city in the place called Emerald Walley. The distinction of the complex is that its pyramids will be not only quadrangular as in Egypt but also triangular and pentagonal. Of course the amount of the city budget funds for the construction depends on the pyramids' volume. Your task is to calculate the volume of the pilot project consisting of three pyramids — one triangular, one quadrangular and one pentagonal.

The first pyramid has equilateral triangle as its base, and all 6 edges of the pyramid have equal length. The second pyramid has a square as its base and all 8 edges of the pyramid have equal length. The third pyramid has a regular pentagon as its base and all 10 edges of the pyramid have equal length.

$\text{[math]}$

Input

The only line of the input contains three integers l3, l4, l5 (1 ≤ l3, l4, l5 ≤ 1000) — the edge lengths of triangular, quadrangular and pentagonal pyramids correspondingly.

Output

Output one number — the total volume of the pyramids. Absolute or relative error should not be greater than 10 - 9.

Examples

input
2 5 3

output

38.546168065709




数学推公式！

AC code 

#include<stdio.h>
#include<math.h>
#define PI acos(-1.0)
int main()
{
    int x,y,z;
    while(scanf("%d%d%d",&x,&y,&z)!=EOF)
    {
    	double s3,s4,s5,t,h,t1,t2;
    	s3=sqrt(2)/12*pow(x,3);
    	s4=sqrt(2)/6*pow(y,3);
    	t=sin(54*1.0/180*PI)*(z*1.0/2)/sin(36*1.0/180*PI);
    	h=sqrt((sqrt(3)/2)*z*1.0*(sqrt(3)/2)*z*1.0-t*t);
    	s5=(1.0/3)*(5.0/2)*z*t*h;
    	printf("%.9lf\n",s3+s4+s5);
	}
	return  0;
 }