POJ 2405 Beavergnaw （计算几何-简单的问题）

时间：2025-02-02 14:06:08
Beavergnaw
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 6203   Accepted: 4089

Description

POJ 2405 Beavergnaw （计算几何-简单的问题）POJ 2405 Beavergnaw （计算几何-简单的问题）When chomping a tree the beaver cuts a very specific shape out of the tree trunk.
What is left in the tree trunk looks like two frustums of a cone joined by a cylinder with the diameter the same as its height. A very curious beaver tries not to demolish a tree but rather sort out what should be the diameter of the cylinder joining the frustums
such that he chomped out certain amount of wood. You are to help him to do the calculations. 

We will consider an idealized beaver chomping an idealized tree. Let us assume that the tree trunk is a cylinder of diameter D and that the beaver chomps on a segment of the trunk also of height D. What should be the diameter d of the inner cylinder such that
the beaver chmped out V cubic units of wood?

Input

Input contains multiple cases each presented on a separate line. Each line contains two integer numbers D and V separated by whitespace. D is the linear units and V is in cubic units. V will not exceed the maximum volume of wood that the beaver can chomp. A
line with D=0 and V=0 follows the last case.

Output

For each case, one line of output should be produced containing one number rounded to three fractional digits giving the value of d measured in linear units.

Sample Input

10 250

20 2500

25 7000

50 50000

0 0

Sample Output

8.054

14.775

13.115

30.901

Source

题目大意：告诉你圆柱直径D，以及啃掉的面积V， 求d

解题思路：

简单的几何问题，够造体积相等，求未知数

V=直径为D的圆柱的体积-两个园台的体积-直径为d的圆柱的体积。

圆台体积公式 = 1/3* pi * (r1*r1 + r2*r2 + r1*r2)*h     r1,r2,h分别为圆台上低半径、下底半径和高

V=pi*(D/2)*(D/2)*D -     1/3 *( D*s1-d*s2   )     - pi*(d/2)*(d/2)*d

V=pi*(D/2)*(D/2)*D -     1/3 *pi(   D*D/4 + d*d/4 + D*d/4   )*( (D-d)/2)     - pi*(d/2)*(d/2)*d

V=pi*D*D*D/4 -      1/3 *pi(   D*D/4 + d*d/4 + D*d/4   )*(D/2 - d/2 )     - pi*d*d*d/4

V=pi*D*D*D/4 -      1/24 *pi(   D*D + d*d + D*d   )*(D - d )     - pi*d*d*d/4

V=pi*D*D*D/4 -      1/24 *pi(   D*D *D+ d*d*D + D*d*D - D*D*d - d*d*d - D *d *d)      - pi*d*d*d/4

V=pi*D*D*D/4 -      1/24 *pi(   D*D *D - d*d*d)      - pi*d*d*d/4

V=pi*D*D*D/6 - pi*d*d*d/6

d*d*d = D*D*D - 6*V/pi

d=( D*D*D - 6*V/pi )^(1/3)

解题代码：

#include <iostream>

#include <cmath>

#include <cstdio>

using namespace std;

const double pi=acos(double(-1));

int main(){

    int d,v;

    while(cin>>d>>v && (d||v) ){

        double D=(double)d,V=(double)v;

        double tmp=D*D*D-6*V/pi;

        printf("%.3f\n",pow(tmp,1.0/3.0));

    }

    return 0;

}

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