HDU 1724 Ellipse 自适应Simpson积分

Posted 2020-09-08 Coolxxx

Ellipse

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1977    Accepted Submission(s): 832

Problem Description

Math is important!! Many students failed in 2+2’s mathematical test, so let‘s AC this problem to mourn for our lost youth..
Look this sample picture:



A ellipses in the plane and center in point O. the L,R lines will be vertical through the X-axis. The problem is calculating the blue intersection area. But calculating the intersection area is dull, so I have turn to you, a talent of programmer. Your task is tell me the result of calculations.(defined PI=3.14159265 , The area of an ellipse A=PI*a*b )

 

 

Input

Input may contain multiple test cases. The first line is a positive integer N, denoting the number of test cases below. One case One line. The line will consist of a pair of integers a and b, denoting the ellipse equation , A pair of integers l and r, mean the L is (l, 0) and R is (r, 0). (-a <= l <= r <= a).

 

 

Output

For each case, output one line containing a float, the area of the intersection, accurate to three decimals after the decimal point.

 

 

Sample Input

2 2 1 -2 2 2 1 0 2

 

 

Sample Output

6.283 3.142

 

 

Author

威士忌

 

 

Source

HZIEE 2007 Programming Contest

 

 

题目链接：

　　http://acm.hdu.edu.cn/showproblem.php?pid=1724

题目大意：

　　求椭圆被x=l,x=r两条线所围区域面积。

题目思路：

　　【自适应Simpson积分】

　　首先易得上半部分面积积分公式为sqrt(b²(1-x²/a²))

　　接下来就是套用自适应Simpson积分即可。eps一开始设为1e-4就行。

　　一道模板题。

 

 1 /****************************************************
 2     
 3     Author : Coolxxx
 4     Copyright 2017 by Coolxxx. All rights reserved.
 5     BLOG : http://blog.csdn.net/u010568270
 6     
 7 ****************************************************/
 8 #include<bits/stdc++.h>
 9 #pragma comment(linker,"/STACK:1024000000,1024000000")
10 #define abs(a) ((a)>0?(a):(-(a)))
11 #define lowbit(a) (a&(-a))
12 #define sqr(a) ((a)*(a))
13 #define mem(a,b) memset(a,b,sizeof(a))
14 const double eps=1e-8;
15 const int J=10000;
16 const int mod=1000000007;
17 const int MAX=0x7f7f7f7f;
18 const double PI=3.14159265358979323;
19 const int N=104;
20 using namespace std;
21 typedef long long LL;
22 double anss;
23 LL aans;
24 int cas,cass;
25 int n,m,lll,ans;
26 double a,b;
27 double F(double x)//原函数f(x)
28 {
29     return sqrt(b*b*(1-x*x/(a*a)));
30 }
31 double simpson(double a,double b)//求simpson公式S(a,b)
32 {
33     double mid=(a+b)/2;
34     return (b-a)/6*(F(a)+4*F(mid)+F(b));
35 }
36 double simpson(double l,double r,double eps,double A)//自适应simpson积分过程
37 {
38     double mid=(l+r)/2;
39     double L=simpson(l,mid);
40     double R=simpson(mid,r);
41     if(abs(L+R-A)<=15*eps)return L+R+(L+R-A)/15.0;//eps为精度需求
42     return simpson(l,mid,eps/2,L)+simpson(mid,r,eps/2,R);
43 }
44 double simpson(double l,double r,double eps)//自适应simpson积分
45 {
46     return simpson(l,r,eps,simpson(l,r));
47 }
48 int main()
49 {
50     #ifndef ONLINE_JUDGE
51 //    freopen("1.txt","r",stdin);
52 //    freopen("2.txt","w",stdout);
53     #endif
54     int i,j,k;
55     double x,y,z;
56 //    for(scanf("%d",&cass);cass;cass--)
57     for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
58 //    while(~scanf("%s",s))
59 //    while(~scanf("%d",&n))
60     {
61         scanf("%lf%lf%lf%lf",&a,&b,&x,&y);
62         printf("%.3lf\n",simpson(x,y,1e-4)*2);
63     }
64     return 0;
65 }
66 /*
67 //
68 
69 //
70 */

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