Cylinder
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 224 Accepted Submission(s): 88
Problem Description
Using a sheet of paper and scissors, you can cut out two faces to form a cylinder in the following way:
1. Cut the paper horizontally (parallel to the shorter side) to get two rectangular parts.
2. From the first part, cut out a circle of maximum radius. The circle will form the bottom of the cylinder.
3. Roll the second part up in such a way that it has a perimeter of equal length with the circle's circumference, and attach one end of the roll to the circle. Note that the roll may have some overlapping parts in order to get the required length of the perimeter.
Given the dimensions of the sheet of paper, can you calculate the biggest possible volume of a cylinder which can be constructed using the procedure described above?
Input
The input consists of several test cases. Each test case consists of two numbers w and h (1 ≤ w ≤ h ≤ 100), which indicate the width and height of the sheet of paper.
The last test case is followed by a line containing two zeros.
Output
For each test case, print one line with the biggest possible volume of the cylinder. Round this number to 3 places after the decimal point.
Sample Input
10 10
10 50
10 30
0 0
Sample Output
54.247
785.398
412.095
Hint
In the first case, the optimal cylinder has a radius of about 1.591549,
in the second case, the optimal cylinder has a radius of 5,
and in the third case, the optimal cylinder has a radius of about 3.621795.
Source
2007~2008 University of Ulm Local Contest
结题报告：这道题的题意是求圆柱的最大体积，当时内部比赛时看错题意了，以为是封闭的圆柱呢！哎！但是真正题意是圆柱没有上盖；给我们一张纸，w, h(h > w);平行于w把这张纸分成两半，一半画一个圆，另一部分当圆柱的侧面，分两种情况，第一种情况是让w当高，其半径为应满足2*PI*r <= h- 2r,并且w >= 2 * r即：r=w/2或h/(2*PI + 2)中选出最小的；第二种情况是让h-2*r当高，极限情况下V= (h- 2* r) * PI* r*r，求导后r=h/3并且还得满足w>= 2*PI*r;r应去最小的情况；
代码如下：
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
using namespace std;
double PI = acos(-1);
double Min(double a, double b)
{
return a < b ? a : b;
}
double Max(double a, double b)
{
return a > b ? a : b;
}
int main()
{
double w, h, r, ans1, ans2, ans;
while (scanf("%lf%lf", &w, &h) != EOF && w &&h)
{
r = Min(w / 2.0, h /(2.0 * PI + 2.0));//以w为圆柱的高
ans1 = PI * r * r * w;
r = Min(w / (2.0 * PI), h / 3.0);
ans2 = PI * r * r * (h - 2.0 * r);//以h-2r为圆柱的高
ans = Max(ans1, ans2);
printf("%.3lf\n", ans);
}
return 0;
}
