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 * (1.0/3 * log(x) - 1.0/9);
}

double f3(double x){ //x^2*ln(R + (R*R-x*x)^(1/2))积分 
       double s = sqrt(R*R-x*x); 
       return 1.0/18 * (-2*x*x*x-3*R*x*s + 3*R*R*R*atan(x/s) + 6*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 / 2;
      double l = 0, r = R, mid;
      for (int i = 0; i < 200; ++i){
            mid = (l + r) / 2;
            if (2 * volume(mid) < V) r = mid; 
            else l = mid;
      }
      printf("%.5f
", l + R);
}

int main(){
//    freopen("a.in", "r", stdin);
//    freopen("a.out", "w", stdout);
    int cas = 0;
    scanf("%d", &cas);
    while (cas--){
         solve();
    }
    return 0;
}