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 ;

 }