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:

 1 /*
 2  * Author:  Yzcstc
 3  * Created Time:  2014/10/2 13:59:16
 4  * File Name: poj3929.cpp
 5  */
 6 #include<cstdio>
 7 #include<iostream>
 8 #include<cstring>
 9 #include<cstdlib>
10 #include<cmath>
11 #include<algorithm>
12 #include<string>
13 #include<map>
14 #include<set>
15 #include<vector>
16 #include<queue>
17 #include<stack>
18 #include<ctime>
19 #define repf(i, a, b) for (int i = (a); i <= (b); ++i)
20 #define repd(i, a, b) for (int i = (a); i >= (b); --i)
21 #define M0(x)  memset(x, 0, sizeof(x))
22 #define Inf  0x7fffffff
23 #define MP make_pair
24 #define PB push_back
25 #define eps 1e-8
26 #define pi acos(-1.0)
27 typedef long long LL;
28 using namespace std;
29 double H, D, V;
30 double R;
31 double f1(double x){//(R*R - x*x)^(1/2)积分
32        return 0.5 * (x * sqrt(R*R - x*x) + R*R * asin(x / R));
33 }
34
35 double f2(double x){ //x^2*ln(x)积分
36        return x * x * x * (1.0/3 * log(x) - 1.0/9);
37 }
38
39 double f3(double x){ //x^2*ln(R + (R*R-x*x)^(1/2))积分
40        double s = sqrt(R*R-x*x);
41        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));
42 }
43
44 double volume(double l){
45       double r = R;
46       double s1 = f1(r) - f1(l);
47       double s2 = 0.5 * (f2(r) - f2(l));
48       double s3 = 0.5 * R * (f1(r) - f1(l));
49       double s4 = 0.5 * (f3(r) - f3(l));
50       return H * s1 + (s2 - s3 - s4) * H / R;
51 }
52
53 void solve(){
54       scanf("%lf%lf%lf", &H, &D, &V);
55       R = D / 2;
56       double l = 0, r = R, mid;
57       for (int i = 0; i < 200; ++i){
58             mid = (l + r) / 2;
59             if (2 * volume(mid) < V) r = mid;
60             else l = mid;
61       }
62       printf("%.5f\n", l + R);
63 }
64
65 int main(){
66 //    freopen("a.in", "r", stdin);
67 //    freopen("a.out", "w", stdout);
68     int cas = 0;
69     scanf("%d", &cas);
70     while (cas--){
71          solve();
72     }
73     return 0;
74 }