Ellipse
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1977 Accepted Submission(s): 832Problem 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 2Sample Output
6.283
3.142Author
威士忌
Source
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 */