poj3929

题意：

如上图放置的一个圆锥，告诉你从圆锥顶的洞中流出多少体积的水，求现在水面高度。。

思路:

无聊时做的一道题，实际上就是一道高数题，重积分，可惜我高数本来也不好而且还忘光了，积了很久，而且错了很多遍。。mark一下。。

本来还想偷懒最难积分的最后一重想用自适应的simpson积分公式。。无奈精度要求太高一直都是TLE。。

code:

 /*
  * Author:  Yzcstc
  * Created Time:  2014/10/2 13:59:16
  * File Name: poj3929.cpp
  */
 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<map>
 #include<set>
 #include<vector>
 #include<queue>
 #include<stack>
 #include<ctime>
 #define repf(i, a, b) for (int i = (a); i <= (b); ++i)
 #define repd(i, a, b) for (int i = (a); i >= (b); --i)
 #define M0(x)  memset(x, 0, sizeof(x))
 #define Inf  0x7fffffff
 #define MP make_pair
 #define PB push_back
 #define eps 1e-8
 #define pi acos(-1.0)
 typedef long long LL;
 using namespace std;
 double H, D, V;
 double R;
 double f1(double x){//(R*R - x*x)^(1/2)积分
        return 0.5 * (x * sqrt(R*R - x*x) + R*R * asin(x / R));
 }

 double f2(double x){ //x^2*ln(x)积分
        return x * x * x * (1.0/ * log(x) - 1.0/);
 }

 double f3(double x){ //x^2*ln(R + (R*R-x*x)^(1/2))积分
        double s = sqrt(R*R-x*x);
        return 1.0/ * (-*x*x*x-*R*x*s + *R*R*R*atan(x/s) + *x*x*x*log(R + s));
 }

 double volume(double l){
       double r = R;
       double s1 = f1(r) - f1(l);
       double s2 = 0.5 * (f2(r) - f2(l));
       double s3 = 0.5 * R * (f1(r) - f1(l));
       double s4 = 0.5 * (f3(r) - f3(l));
       return H * s1 + (s2 - s3 - s4) * H / R;
 }

 void solve(){
       scanf("%lf%lf%lf", &H, &D, &V);
       R = D / ;
       double l = , r = R, mid;
       for (int i = ; i < ; ++i){
             mid = (l + r) / ;
             if ( * volume(mid) < V) r = mid;
             else l = mid;
       }
       printf("%.5f\n", l + R);
 }

 int main(){
 //    freopen("a.in", "r", stdin);
 //    freopen("a.out", "w", stdout);
     int cas = ;
     scanf("%d", &cas);
     while (cas--){
          solve();
     }
     return ;
 }