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
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
[Technical Specification]
0< R,HR,HZ<=100
題解及代碼:
這道題的意思很簡單:給定中心重合的一個球和一個圓柱,求出其重合體積佔所有體積的比例。
這題寫起來很麻煩,因爲要分成5類分別寫(可恥de把官方的圖扣下來 = =!)
分類大致就是分成這5類,積分的方式這裏使用的是simpson積分法,只要知道被積函數和上下限就可以了,不用自己做不定積分。
#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;
}