hdu 5060 War

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War

Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 98 Accepted Submission(s): 28
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 12 1 1

Sample Output

0.6666670.187500

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;}

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