War

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Special Judge

Problem Description

Long long ago there are two countrys in the universe. Each country haves its own manor in 3-dimension space. Country A‘s manor occupys x^2+y^2+z^2<=R^2. Country B‘s manor occupys x^2+y^2<=HR^2 && |z|<=HZ. There may be a war between them. The occurrence of a
war have a certain probability.

We calculate the probability as follow steps.

1. VC=volume of insection manor of A and B.

2. VU=volume of union manor of A and B.

3. probability=VC/VU

Input

Multi test cases(about 1000000). Each case contain one line. The first line contains three integers R,HR,HZ. Process to end of file.

[Technical Specification]

0< R,HR,HZ<=100

Output

For each case，output the probability of the war which happens between A and B. The answer should accurate to six decimal places.

Sample Input

1 1 1
2 1 1

Sample Output

0.666667
0.187500

Source

BestCoder Round #12

题解及代码：

这道题的意思很简单：给定中心重合的一个球和一个圆柱，求出其重合体积占所有体积的比例。

这题写起来很麻烦，因为要分成5类分别写（可耻de把官方的图扣下来 = =！）

分类大致就是分成这5类，积分的方式这里使用的是simpson积分法，只要知道被积函数和上下限就可以了，不用自己做不定积分。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
const double pi=3.14159265358979,eps=1e-7;
double r,hr,hz;

double f(double n)
{
    return pi*(r*r-n*n);
}

double simpson(double a,double b)
{
    return (b-a)/6.0*(f(a)+4*f((a+b)/2.0)+f(b));
}

double cal(double a,double b)
{
    double sum=simpson(a,b),mid=(a+b)/2.0;
    double t=simpson(a,mid)+simpson(mid,b);

    if(fabs(t-sum)<eps) return sum;

    return cal(a,mid)+cal(mid,b);
}

int main()
{

    while(scanf("%lf%lf%lf",&r,&hr,&hz)!=EOF)
    {
        double v=0,hv=0;
        if(hr>=r&&hz>=r)
        {
            v=4.0/3.0*pi*r*r*r;
            hv=2*pi*hr*hr*hz;
            printf("%.6lf\n",v/hv);
            continue;
        }
        if(hr>=r&&hz<r)
        {
            v=4.0/3.0*pi*r*r*r;
            double t=2*cal(hz,r);
            hv=2*pi*hr*hr*hz;
            printf("%.6lf\n",(v-t)/(hv+t));
            continue;
        }
        if(r*r>=hr*hr+hz*hz)
        {
            v=4.0/3.0*pi*r*r*r;
            hv=2*pi*hr*hr*hz;
            printf("%.6lf\n",hv/v);
            continue;
        }
        if(hr<r&&hz>=r)
        {
            v=4.0/3.0*pi*r*r*r;
            double t=2*cal(sqrt(r*r-hr*hr),r)+2*sqrt(r*r-hr*hr)*pi*hr*hr;
            hv=2*pi*hr*hr*hz;
            printf("%.6lf\n",t/(hv+v-t));
            continue;
        }
        v=4.0/3.0*pi*r*r*r;
        hv=2*pi*hr*hr*hz;
        double t=2*cal(sqrt(r*r-hr*hr),hz)+2*sqrt(r*r-hr*hr)*pi*hr*hr;
        printf("%.6lf\n",t/(hv+v-t));
    }
    return 0;
}