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。

/**************************************************** Author : Coolxxx Copyright 2017 by Coolxxx. All rights reserved. BLOG : http://blog.csdn.net/u010568270 ****************************************************/ #include<bits/stdc++.h> #pragma comment(linker,"/STACK:1024000000,1024000000") #define abs(a) ((a)>0?(a):(-(a))) #define lowbit(a) (a&(-a)) #define sqr(a) ((a)*(a)) #define mem(a,b) memset(a,b,sizeof(a)) const double eps=1e-8; const int J=10000; const int mod=1000000007; const int MAX=0x7f7f7f7f; const double PI=3.14159265358979323; const int N=104; using namespace std; typedef long long LL; double anss; LL aans; int cas,cass; int n,m,lll,ans; double a,b; double F(double x)//原函数f(x) { return sqrt(b*b*(1-x*x/(a*a))); } double simpson(double a,double b)//求simpson公式S(a,b) { double mid=(a+b)/2; return (b-a)/6*(F(a)+4*F(mid)+F(b)); } double simpson(double l,double r,double eps,double A)//自适应simpson积分过程 { double mid=(l+r)/2; double L=simpson(l,mid); double R=simpson(mid,r); if(abs(L+R-A)<=15*eps)return L+R+(L+R-A)/15.0;//eps为精度需求 return simpson(l,mid,eps/2,L)+simpson(mid,r,eps/2,R); } double simpson(double l,double r,double eps)//自适应simpson积分 { return simpson(l,r,eps,simpson(l,r)); } int main() { #ifndef ONLINE_JUDGE // freopen("1.txt","r",stdin); // freopen("2.txt","w",stdout); #endif int i,j,k; double x,y,z; // for(scanf("%d",&cass);cass;cass--) for(scanf("%d",&cas),cass=1;cass<=cas;cass++) // while(~scanf("%s",s)) // while(~scanf("%d",&n)) { scanf("%lf%lf%lf%lf",&a,&b,&x,&y); printf("%.3lf\n",simpson(x,y,1e-4)*2); } return 0; } /* // // */