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
題目連結：
http://acm.hdu.edu.cn/showproblem.php?pid=1724
題目大意：
求橢圓
被x=l,x=r兩條線所圍區域面積。
題目思路：
【自适應Simpson積分】
首先易得上半部分面積積分公式為sqrt(b2(1-x2/a2))
接下來就是套用自适應Simpson積分即可。eps一開始設為1e-4就行。
一道模闆題。
/****************************************************
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)//原函數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)//自适應simpson積分過程
{
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為精度需求
return simpson(l,mid,eps/2,L)+simpson(mid,r,eps/2,R);
}
double simpson(double l,double r,double eps)//自适應simpson積分
{
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;
}
/*
//
//
*/