pku3289Moonshine（模拟问题）

真是一道让人头疼的几何模拟题。。。wa了n多遍。。。注意圆台部分上底半径的求解，否则圆台体积会求错。。。横放时圆台部分体积要微元求解。。。最终s的确定要靠二分逼近。。。

#include<stdio.h>
#include<stdlib.h> 
#include<string.h>
#include<math.h>
#define pi 3.1415926535898
#define N 1000
double k,hb,db,hn,dn,h,s,v,V,r1,r2,r,delta;
double area(double r,double s)  //横放时弓形底面积 
{
  double t,s1,s2,ang;
  //if(s>r)  return (pi*(r*r)-area(r,2*r-s));
  t=db/2-s;
  ang=t<r?acos(t/r):0;
  s1=ang*r*r;
  s2=r*cos(ang)*r*sin(ang);
  //s2=t*sqrt(r*r-t*t);
  return s1-s2;
}
double volume(double s)  //横放液面高度为s时的体积 
{
   if(s*2>db)   return V-volume(db-s);     
   double vb,vn,vc;
   vb=area(db/2,s)*hb;
   vn=area(dn/2,s)*hn;
   vc=area(db/2,s)+area(dn/2,s); 
   int i;
   for(i=1;i<N;i+=2)
     vc+=4*area((db+(dn-db)*i/N)/2,s); 
   for(i=2;i<N;i+=2)
     vc+=2*area((db+(dn-db)*i/N)/2,s);  
   vc*=(h-hb-hn)/3/N;  
   return vb+vc+vn;    
} 
int main()
{
  double i,d; 
  while(scanf("%lf%lf%lf%lf%lf%lf",&k,&hb,&db,&hn,&dn,&h),!(k==0&&hb==0&&db==0&&hn==0&&dn==0&&h==0)){
    r1=db/2;   r2=dn/2;
    //V=pi*r1*r1*hb+pi*(r1*r1+r1*r2+r2*r2)*(h-hb-hn)/3+pi*r2*r2*hn;   //瓶子总体积 
    V = 2 * volume(r1);
    if(k<=hb){
      v=pi*r1*r1*k;                           //液体体积       
    }
    else if(k<(h-hn)){
      d=r2*(h-hn-hb)/(r1-r2);                         //wa了两天，就因为此处d和r的推导有错误 
      r=(d+h-k-hn)/(d+h-hb-hn)*r1;
      v=pi*r1*r1*hb+pi*(r1*r1+r1*r+r*r)*(k-hb)/3;       
    }
    else{
      v=pi*r1*r1*hb;    
      v+=pi*(r1*r1+r1*r2+r2*r2)*(h-hb-hn)/3;
      v+=pi*r2*r2*(hn+k-h);
    } 
    s=db/2;
    for(i=s/2;i>.00001;i/=2){            //二分逼近 
      if(volume(s)>v)    s-=i;
      else               s+=i;                          
    }
    printf("%.2lf\n",s);
  } 
  //system("pause");
  return 0;     
}