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POJ3737UmBasketella【圆锥体积与面积+求导】

2015-09-05 23:43 260 查看

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[align=center]UmBasketella[/align]

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6448		Accepted: 2516

Description

In recent days, people always design new things with multifunction. For instance, you can not only use cell phone to call your friends, but you can also use your cell phone take photographs or listen to MP3. Another example is the combination between watch
and television. These kinds of multifunction items can always improve people's daily life and are extremely favored by users.

The company Mr. Umbrella invented a new kind umbrella "UmBasketella" for people in Rainbow city recently and its idea also comes from such multifunction--the combination of umbrella and daily necessities. This kind of umbrella can be used as a basket and
you can put something you want to carry in it. Since Rainbow city rains very often, such innovative usage is successful and "UmBasketella" sells very well. Unfortunately, the original "UmBasketella" do not have an automatic volume control technology so that
it is easily damaged when users try to put too many things in it. To solve this problem, you are needed to design an "UmBasketella" with maximum volume. Suppose that "UmBasketella" is a cone-shape container and its surface area (include the bottom) is known,
could you find the maximum value of the cone?

Input

Input contains several test cases. Eash case contains only one real number S, representing the surface area of the cone. It is guaranteed that 1≤S≤10000.

Output

For each test case, output should contain three lines.

The first line should have a real number representing the maximum volume of the cone.

Output the height of the cone on the second line and the radius of the bottom area of the cone on the third line.

All real numbers should rounded to 0.01.

Sample Input

Sample Output

10.93
4.37
1.55

题意：给出一个圆锥的表面积包括底面积；求当该圆锥体体积最大时该圆锥的高，体积，底面半径；

圆锥体面积：S=PI*L*R+PI*R*R;L=sqrt(h*h+R*R);

既有R*R=S*S/(PI*PI*h*h+2*PI*S);

V=PI*R*R*h/3;将R*R带入求导函数零点即为最大值点；即可求出h，r，v

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#define PI 3.1415926
using namespace std;
int main()
{
double s,h,r,v;
while(scanf("%lf",&s)!=EOF){
r=sqrt(s/PI)/2.0;
h=sqrt((s*s)/(PI*PI*r*r)-2.0*s/PI);
v=PI*r*r*h/3.0;
printf("%.2lf\n%.2lf\n%.2lf\n",v,h,r);
}
return 0;
}

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标签： POJ3737UmBasketella圆

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